Bloomberg¶
Table of Contents¶
- 1. Two Sum (Easy)
- 56. Merge Intervals (Medium)
- 3. Longest Substring Without Repeating Characters (Medium)
- 430. Flatten a Multilevel Doubly Linked List (Medium)
- 20. Valid Parentheses (Easy)
- 380. Insert Delete GetRandom O(1) (Medium)
- 2. Add Two Numbers (Medium)
- 79. Word Search (Medium)
- 121. Best Time to Buy and Sell Stock (Easy)
- 88. Merge Sorted Array (Easy)
- 42. Trapping Rain Water (Hard)
- 242. Valid Anagram (Easy)
- 4. Median of Two Sorted Arrays (Hard)
- 1169. Invalid Transactions (Medium)
- 31. Next Permutation (Medium)
- 5. Longest Palindromic Substring (Medium)
- 394. Decode String (Medium)
- 78. Subsets (Medium)
- 146. LRU Cache (Medium)
- 14. Longest Common Prefix (Easy)
- 128. Longest Consecutive Sequence (Medium)
- 34. Find First and Last Position of Element in Sorted Array (Medium)
- 217. Contains Duplicate (Easy)
- 22. Generate Parentheses (Medium)
- 560. Subarray Sum Equals K (Medium)
- 21. Merge Two Sorted Lists (Easy)
- 2667. Create Hello World Function (Easy)
- 13. Roman to Integer (Easy)
- 53. Maximum Subarray (Medium)
- 1244. Design A Leaderboard (Medium) 👑
1. Two Sum¶
from typing import List
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
"""Hash Map: O(n) time, O(n) space.
Store seen values so each number can find its complement in one pass.
"""
hashmap = {} # val: idx
for idx, val in enumerate(nums):
comp = target - val
if comp in hashmap:
return [hashmap[comp], idx]
hashmap[val] = idx
return []
def test_two_sum():
s = Solution()
assert s.twoSum([2, 7, 11, 15], 9) == [0, 1]
assert s.twoSum([3, 2, 4], 6) == [1, 2]
assert s.twoSum([3, 3], 6) == [0, 1]
assert s.twoSum([1, 2, 3, 4, 5], 10) == []
assert s.twoSum([-1, -2, -3, -4, -5], -8) == [2, 4]
#include <cassert>
#include <unordered_map>
#include <vector>
using namespace std;
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int, int> map;
for (size_t i = 0; i < nums.size(); i++) {
int diff = target - nums[i];
if (map.find(diff) != map.end()) {
return {map[diff], (int)i};
}
map[nums[i]] = (int)i;
}
return {-1, -1};
}
};
int main() {
Solution solution;
vector<int> nums = {2, 7, 11, 15};
int target = 9;
vector<int> result = solution.twoSum(nums, target);
assert((result == vector<int>{0, 1}));
return 0;
}
56. Merge Intervals¶
def merge(intervals: list[list[int]]) -> list[list[int]]:
"""
Merge all overlapping intervals.
| G----H
| E===========F
| C--------D
| A========B
|
|-----------------------------------
Draw the above to explain the process to the interviewer.
Time complexity: O(nlogn) due to sorting.
Space complexity: O(n) for the result list.
"""
n = len(intervals)
if n <= 1:
return intervals
intervals.sort(key=lambda x: x[0])
res = [intervals[0]]
for i in range(1, n):
if intervals[i][0] <= res[-1][1]:
res[-1][1] = max(res[-1][1], intervals[i][1])
else:
res.append(intervals[i])
return res
print(merge([[1, 3], [2, 6], [8, 10], [15, 18]]))
# [[1, 6], [8, 10], [15, 18]]
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
// Interval
vector<vector<int>> merge(vector<vector<int>>& intervals) {
sort(intervals.begin(), intervals.end());
vector<vector<int>> res;
for (auto& range : intervals) {
if (!res.empty() && range[0] <= res.back()[1]) {
res.back()[1] = max(res.back()[1], range[1]);
} else {
res.emplace_back(range);
}
}
return res;
}
int main() {
vector<vector<int>> intervals = {{1, 3}, {2, 6}, {8, 10}, {15, 18}};
vector<vector<int>> res = merge(intervals);
for (auto& range : res) {
cout << range[0] << ", " << range[1] << endl;
}
return 0;
}
3. Longest Substring Without Repeating Characters¶
class Solution:
def lengthOfLongestSubstring(self, s: str) -> int:
"""Sliding Window: O(n) time, O(min(n, charset)) space.
Jump the left edge past each repeated character's previous index.
"""
left = 0
res = 0
seen = {}
for right, ch in enumerate(s):
if ch in seen and seen[ch] >= left:
left = seen[ch] + 1
seen[ch] = right
res = max(res, right - left + 1)
return res
def lengthOfLongestSubstringSet(self, s: str) -> int:
"""Sliding Window Set: O(n) time, O(min(n, charset)) space.
Shrink until the current character is unique in the window.
"""
left = 0
res = 0
window = set()
for right, ch in enumerate(s):
while ch in window:
window.remove(s[left])
left += 1
window.add(ch)
res = max(res, right - left + 1)
return res
def test_length_of_longest_substring():
s = Solution()
cases = [
("abcabcbb", 3),
("bbbbb", 1),
("pwwkew", 3),
("", 0),
(" ", 1),
("abba", 2),
("dvdf", 3),
]
for text, expected in cases:
assert s.lengthOfLongestSubstring(text) == expected
assert s.lengthOfLongestSubstringSet(text) == expected
#include <iostream>
#include <string>
#include <unordered_set>
using namespace std;
int lengthOfLongestSubstring(string s) {
int n = s.length();
int res = 0;
int left = 0;
unordered_set<char> window;
for (int right = 0; right < n; right++) {
char ch = s[right];
while (window.find(ch) != window.end()) {
window.erase(s[left]);
left++;
}
window.insert(ch);
res = max(res, right - left + 1);
}
return (int)res;
}
int main() {
string s = "abcabcbb";
cout << lengthOfLongestSubstring(s) << endl; // 3
s = "bbbbb";
cout << lengthOfLongestSubstring(s) << endl; // 1
s = "pwwkew";
cout << lengthOfLongestSubstring(s) << endl; // 3
return 0;
}
430. Flatten a Multilevel Doubly Linked List¶
20. Valid Parentheses¶
# Stack
def is_valid(s: str) -> bool:
if len(s) % 2:
return False
pairs = {
"(": ")",
"{": "}",
"[": "]",
}
stack = []
for ch in s:
if ch in pairs:
stack.append(ch)
elif not stack or ch != pairs[stack.pop()]:
return False
return True if not stack else False
def test_is_valid():
assert is_valid("()[]{}")
assert not is_valid("(]")
assert not is_valid("([)]")
assert is_valid("{[]}")
#include <cassert>
#include <stack>
#include <string>
#include <unordered_map>
using namespace std;
class Solution {
public:
bool isValid(string s) {
unordered_map<char, char> map{{')', '('}, {'}', '{'}, {']', '['}};
stack<char> stack;
if (s.length() % 2 == 1) return false;
for (char& ch : s) {
if (stack.empty() || map.find(ch) == map.end()) {
stack.push(ch);
} else {
if (map[ch] != stack.top()) {
return false;
}
stack.pop();
}
}
return stack.empty();
}
};
int main() {
Solution s;
assert(s.isValid("()") == true);
assert(s.isValid("()[]{}") == true);
assert(s.isValid("(]") == false);
assert(s.isValid("([)]") == false);
assert(s.isValid("{[]}") == true);
return 0;
}
380. Insert Delete GetRandom O(1)¶
import random
class RandomizedSet:
def __init__(self):
self.nums = []
self.pos = {} # num: idx
def insert(self, val: int) -> bool:
if val in self.pos:
return False
self.pos[val] = len(self.nums)
self.nums.append(val)
return True
def remove(self, val: int) -> bool:
if val not in self.pos:
return False
idx = self.pos[val]
last_val = self.nums[-1]
self.nums[idx] = last_val
self.pos[last_val] = idx
self.nums.pop()
del self.pos[val]
return True
def getRandom(self) -> int:
return random.choice(self.nums)
def test_RandomizedSet():
obj = RandomizedSet()
assert obj.insert(1)
assert not obj.remove(2)
assert obj.insert(2)
assert obj.getRandom() in [1, 2]
assert obj.remove(1)
assert not obj.insert(2)
assert obj.getRandom() == 2
2. Add Two Numbers¶
from typing import Optional
from leetpattern.utils import LinkedList, ListNode
class Solution:
def addTwoNumbers(
self, l1: Optional[ListNode], l2: Optional[ListNode]
) -> Optional[ListNode]:
"""Digit Carry Simulation: O(max(m, n)) time, O(max(m, n)) space.
Walk both reversed lists while carrying overflow into the next digit.
"""
dummy = ListNode()
cur = dummy
carry = 0
while l1 or l2 or carry:
v1 = l1.val if l1 else 0
v2 = l2.val if l2 else 0
carry, val = divmod(v1 + v2 + carry, 10)
cur.next = ListNode(val)
cur = cur.next
if l1:
l1 = l1.next
if l2:
l2 = l2.next
return dummy.next
def test_add_two_numbers():
s = Solution()
l1 = LinkedList([2, 4, 3]).head
l2 = LinkedList([5, 6, 4]).head
assert LinkedList(s.addTwoNumbers(l1, l2)).to_array() == [7, 0, 8]
l1 = LinkedList([0]).head
l2 = LinkedList([0]).head
assert LinkedList(s.addTwoNumbers(l1, l2)).to_array() == [0]
l1 = LinkedList([9, 9, 9, 9, 9, 9, 9]).head
l2 = LinkedList([9, 9, 9, 9]).head
assert LinkedList(s.addTwoNumbers(l1, l2)).to_array() == [
8,
9,
9,
9,
0,
0,
0,
1,
]
#include <cassert>
#include <vector>
using namespace std;
struct ListNode {
int val;
ListNode* next;
ListNode() : val(0), next(nullptr) {}
ListNode(int x) : val(x), next(nullptr) {}
ListNode(int x, ListNode* next) : val(x), next(next) {}
};
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode dummy;
ListNode* cur = &dummy;
int carry = 0;
while (l1 || l2 || carry) {
if (l1) {
carry += l1->val;
l1 = l1->next;
}
if (l2) {
carry += l2->val;
l2 = l2->next;
}
cur->next = new ListNode(carry % 10);
cur = cur->next;
carry /= 10;
}
return dummy.next;
}
};
int main() {
Solution sol;
ListNode* l1 = new ListNode(2, new ListNode(4, new ListNode(3)));
ListNode* l2 = new ListNode(5, new ListNode(6, new ListNode(4)));
ListNode* result = sol.addTwoNumbers(l1, l2);
vector<int> expected = {7, 0, 8};
for (int val : expected) {
assert(result != nullptr && result->val == val);
result = result->next;
}
assert(result == nullptr); // End of list
return 0;
}
79. Word Search¶
from typing import List
def exist(board: List[List[str]], word: str) -> bool:
m, n = len(board), len(board[0])
path = set()
dirs = ((0, 1), (1, 0), (0, -1), (-1, 0))
def dfs(r, c, i):
if i == len(word):
return True
if r < 0 or r >= m or c < 0 or c >= n or board[r][c] != word[i] or (r, c) in path:
return False
path.add((r, c))
for dr, dc in dirs:
if dfs(r + dr, c + dc, i + 1):
return True
path.remove((r, c))
return False
for i in range(m):
for j in range(n):
if dfs(i, j, 0):
return True
return False
board = [
["A", "B", "C", "E"],
["S", "F", "C", "S"],
["A", "D", "E", "E"],
]
word = "ABCCED"
print(exist(board, word)) # True
121. Best Time to Buy and Sell Stock¶
"""
- Return the maximum profit that can be achieved from buying on one day and selling on another day.
"""
from typing import List
# Brute Force
def maxProfitBF(prices: List[int]) -> int:
max_profit = 0
n = len(prices)
for i in range(n):
for j in range(i + 1, n):
max_profit = max(max_profit, prices[j] - prices[i])
return max_profit
# DP
def maxProfitDP(prices: List[int]) -> int:
dp = [[0] * 2 for _ in range(len(prices))]
dp[0][0] = -prices[0] # buy
dp[0][1] = 0 # sell
for i in range(1, len(prices)):
dp[i][0] = max(dp[i - 1][0], -prices[i]) # the lowest price to buy
dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] + prices[i])
return dp[-1][1]
# Greedy
def maxProfitGreedy(prices: List[int]) -> int:
max_profit = 0
seen_min = prices[0]
for i in range(1, len(prices)):
max_profit = max(max_profit, prices[i] - seen_min)
seen_min = min(seen_min, prices[i])
return max_profit
# Fast Slow Pointers
def maxProfitFS(prices: List[int]) -> int:
max_profit = 0
slow, fast = 0, 1
while fast < len(prices):
if prices[fast] > prices[slow]:
max_profit = max(max_profit, prices[fast] - prices[slow])
else:
slow = fast
fast += 1
return max_profit
# |------------|------- |---------|
# | Approach | Time | Space |
# |------------|--------|---------|
# | Brute Force| O(n^2)| O(1) |
# | DP | O(n) | O(n) |
# | Greedy | O(n) | O(1) |
# | Fast Slow | O(n) | O(1) |
# |------------|--------|---------|
prices = [7, 1, 5, 3, 6, 4]
print(maxProfitBF(prices)) # 5
print(maxProfitDP(prices)) # 5
print(maxProfitGreedy(prices)) # 5
print(maxProfitFS(prices)) # 5
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
int maxProfitMemo(vector<int> &prices) {
if (prices.size() <= 1) return 0;
int seen_min = prices[0];
int res = 0;
for (int &price : prices) {
res = max(res, price - seen_min);
seen_min = min(seen_min, price);
}
return res;
}
};
int main() {
vector<int> prices = {7, 1, 5, 3, 6, 4};
Solution obj;
cout << obj.maxProfitMemo(prices) << endl;
return 0;
}
88. Merge Sorted Array¶
from typing import List
class merge:
@staticmethod
def lr(nums1: List[int], m: int, nums2: List[int], n: int) -> None:
"""
Do not return anything, modify nums1 in-place instead.
"""
p1, p2, t = m - 1, n - 1, m + n - 1
while p1 >= 0 or p2 >= 0:
if p1 == -1:
nums1[t] = nums2[p2]
p2 -= 1
elif p2 == -1:
nums1[t] = nums1[p1]
p1 -= 1
elif nums1[p1] > nums2[p2]:
nums1[t] = nums1[p1]
p1 -= 1
else:
nums1[t] = nums2[p2]
p2 -= 1
t -= 1
if __name__ == "__main__":
nums1 = [1, 2, 3, 0, 0, 0]
m = 3
nums2 = [2, 5, 6]
n = 3
merge.lr(nums1, m, nums2, n)
assert nums1 == [1, 2, 2, 3, 5, 6]
#include <cassert>
#include <vector>
using namespace std;
class Solution {
public:
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
int p1 = m - 1, p2 = n - 1, t = m + n - 1;
while (p1 >= 0 || p2 >= 0) {
if (p1 == -1) {
nums1[t] = nums2[p2];
p2--;
} else if (p2 == -1) {
nums1[t] = nums1[p1];
p1--;
} else if (nums1[p1] > nums2[p2]) {
nums1[t] = nums1[p1];
p1--;
} else {
nums1[t] = nums2[p2];
p2--;
}
t--;
}
}
};
int main() {
Solution solution;
vector<int> nums1 = {1, 3, 6, 0, 0, 0};
vector<int> nums2 = {2, 5, 6};
vector<int> output = {1, 2, 3, 5, 6, 6};
solution.merge(nums1, 3, nums2, 3);
assert(nums1 == output);
return 0;
}
42. Trapping Rain Water¶
from typing import List
class Solution:
def trap(self, height: List[int]) -> int:
"""Two Pointers: O(n) time, O(1) space.
Track max from each side; shorter side determines water level.
"""
left, right = 0, len(height) - 1
max_l = max_r = res = 0
while left < right:
if height[left] < height[right]:
max_l = max(max_l, height[left])
res += max_l - height[left]
left += 1
else:
max_r = max(max_r, height[right])
res += max_r - height[right]
right -= 1
return res
def trapDP(self, height: List[int]) -> int:
"""DP Prefix Max: O(n) time, O(n) space.
Precompute max heights from left and right.
"""
n = len(height)
max_left = [0] * n
max_right = [0] * n
for i in range(1, n):
max_left[i] = max(max_left[i - 1], height[i - 1])
for i in range(n - 2, -1, -1):
max_right[i] = max(max_right[i + 1], height[i + 1])
return sum(max(0, min(max_left[i], max_right[i]) - height[i]) for i in range(n))
def trapStack(self, height: List[int]) -> int:
"""Monotonic Stack: O(n) time, O(n) space.
Pop shorter bars and accumulate trapped water by layer.
"""
stack = []
res = 0
for i in range(len(height)):
while stack and height[i] > height[stack[-1]]:
top = stack.pop()
if not stack:
break
w = i - stack[-1] - 1
h = min(height[i], height[stack[-1]]) - height[top]
res += w * h
stack.append(i)
return res
def test_trap():
s = Solution()
for fn in (s.trap, s.trapDP, s.trapStack):
assert fn([0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1]) == 6
assert fn([4, 2, 0, 3, 2, 5]) == 9
assert fn([1, 0, 1]) == 1
assert fn([]) == 0
#include <vector>
#include <algorithm>
#include <iostream>
using namespace std;
class Solution
{
public:
int trap(vector<int> &height)
{
if (height.empty())
return 0;
int res = 0;
int left = 0, right = height.size() - 1;
int maxL = height[left], maxR = height[right];
while (left < right)
{
if (maxL < maxR)
{
left++;
maxL = max(maxL, height[left]);
res += maxL - height[left];
}
else
{
right--;
maxR = max(maxR, height[right]);
res += maxR - height[right];
}
}
return res;
}
};
int main()
{
vector<int> height = {0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1};
Solution solution;
cout << solution.trap(height) << endl;
return 0;
}
242. Valid Anagram¶
"""
- Return true if an input string is an anagram of another string.
- An anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once, e.g., `listen` is an anagram of `silent`.
"""
from collections import Counter
# Hashmap
def isAnagramHash(s: str, t: str) -> bool:
"""Return True if t is an anagram of s, False otherwise."""
if len(s) != len(t):
return False
count = dict()
for i in s:
if i in count:
count[i] += 1
else:
count[i] = 1
for j in t:
if j in count:
count[j] -= 1
else:
return False
for count in count.values():
if count != 0:
return False
return True
# Array
def isAnagramArray(s: str, t: str) -> bool:
if len(s) != len(t):
return False
count = [0 for _ in range(26)]
for i in s:
count[ord(i) - ord("a")] += 1
for j in t:
count[ord(j) - ord("a")] -= 1
for i in count:
if i != 0:
return False
return True
# Counter
def isAnagramCounter(s: str, t: str) -> bool:
return Counter(s) == Counter(t)
# |-------------|-----------------|--------------|
# | Approach | Time | Space |
# |-------------|-----------------|--------------|
# | Hashmap | O(n) | O(1) |
# | Array | O(n) | O(1) |
# | Counter | O(n) | O(1) |
# |-------------|-----------------|--------------|
s = "anagram"
t = "nagaram"
print(isAnagramHash(s, t)) # True
print(isAnagramArray(s, t)) # True
print(isAnagramCounter(s, t)) # True
#include <cassert>
#include <string>
#include <vector>
using namespace std;
class Solution {
public:
bool isAnagram(string s, string t) {
if (s.size() != t.size()) return false;
vector<int> count(26, 0);
for (char ch : s) count[ch - 'a']++;
for (char ch : t) count[ch - 'a']--;
for (int c : count) {
if (c != 0) return false;
}
return true;
}
};
int main() {
Solution solution;
assert(solution.isAnagram("anagram", "nagaram") == true);
assert(solution.isAnagram("rat", "car") == false);
assert(solution.isAnagram("a", "ab") == false);
assert(solution.isAnagram("a", "a") == true);
return 0;
}
4. Median of Two Sorted Arrays¶
from typing import List
# Brute Force
def findMedianSortedArraysBF(nums1: List[int], nums2: List[int]) -> float:
nums = sorted(nums1 + nums2)
n = len(nums)
if n % 2 == 0:
return (nums[n // 2 - 1] + nums[n // 2]) / 2
else:
return nums[n // 2]
# Binary Search
def findMedianSortedArraysBS(nums1: List[int], nums2: List[int]) -> float:
if len(nums1) > len(nums2):
nums1, nums2 = nums2, nums1
m, n = len(nums1), len(nums2)
imin, imax, half_len = 0, m, (m + n + 1) // 2
while imin <= imax:
i = (imin + imax) // 2
j = half_len - i
if i < m and nums2[j - 1] > nums1[i]:
imin = i + 1
elif i > 0 and nums1[i - 1] > nums2[j]:
imax = i - 1
else:
if i == 0:
max_of_left = nums2[j - 1]
elif j == 0:
max_of_left = nums1[i - 1]
else:
max_of_left = max(nums1[i - 1], nums2[j - 1])
if (m + n) % 2 == 1:
return max_of_left
if i == m:
min_of_right = nums2[j]
elif j == n:
min_of_right = nums1[i]
else:
min_of_right = min(nums1[i], nums2[j])
return (max_of_left + min_of_right) / 2
# |--------------|-----------------|--------------|
# | Approach | Time | Space |
# |--------------|-----------------|--------------|
# | Brute Force | O((n+m)log(n+m))| O(n+m) |
# | Binary Search| O(log(min(n,m)))| O(1) |
# |--------------|-----------------|--------------|
nums1 = [1, 3]
nums2 = [2]
print(findMedianSortedArraysBF(nums1, nums2)) # 2.0
print(findMedianSortedArraysBS(nums1, nums2)) # 2.0
1169. Invalid Transactions¶
31. Next Permutation¶
from typing import List
def nextPermutation(nums: List[int]) -> None:
"""
Do not return anything, modify nums in-place instead.
"""
n = len(nums)
i = n - 1
while i > 0 and nums[i - 1] >= nums[i]:
i -= 1
if i != 0:
j = n - 1
while nums[j] <= nums[i - 1]:
j -= 1
nums[i - 1], nums[j] = nums[j], nums[i - 1]
left, right = i, n - 1
while left < right:
nums[left], nums[right] = nums[right], nums[left]
left += 1
right -= 1
nums = [1, 2, 3]
nextPermutation(nums)
print(nums) # [1, 3, 2]
nums = [1, 2, 3, 4, 6, 5]
nextPermutation(nums)
print(nums) # [1, 2, 3, 5, 4, 6]
5. Longest Palindromic Substring¶
class longestPalindrome:
"""Return the longest palindromic substring in s."""
@staticmethod
def dp(s: str) -> str:
n = len(s)
if n <= 1:
return s
start, maxLen = 0, 1
# Init
dp = [[0] * n for _ in range(n)]
for i in range(n):
dp[i][i] = 1
# update
for j in range(1, n):
for i in range(j):
if s[i] == s[j]:
if j - i <= 2:
dp[i][j] = 1
else:
dp[i][j] = dp[i + 1][j - 1]
if dp[i][j] and j - i + 1 > maxLen:
maxLen = j - i + 1
start = i
return s[start : start + maxLen]
@staticmethod
def expand_around_center(s: str) -> str:
def expand(left, right):
while left >= 0 and right < len(s) and s[left] == s[right]:
left -= 1
right += 1
return right - left - 1
if len(s) <= 1:
return s
start, end = 0, 0
for i in range(len(s)):
odd = expand(i, i)
even = expand(i, i + 1)
maxLen = max(odd, even)
if maxLen > end - start:
start = i - (maxLen - 1) // 2
end = i + maxLen // 2
return s[start : end + 1]
if __name__ == "__main__":
s = "babad"
assert longestPalindrome.dp(s) in ["bab", "aba"]
assert longestPalindrome.expand_around_center(s) in ["bab", "aba"]
394. Decode String¶
# Stack
def decodeString(s: str) -> str:
stack = [] # (str, int)
num = 0
res = ""
for c in s:
if c.isdigit():
num = num * 10 + int(c)
elif c == "[":
stack.append((res, num))
res, num = "", 0
elif c == "]":
top = stack.pop()
res = top[0] + res * top[1]
else:
res += c
return res
s = "3[a2[c]]"
print(decodeString(s)) # accaccacc
78. Subsets¶
from typing import List
# Iterative Inclusion Backtracking
def subsets_iterative_inclusion(nums: List[int]) -> List[List[int]]:
n = len(nums)
res, path = [], []
def dfs(i):
res.append(path.copy())
for j in range(i, n):
path.append(nums[j])
dfs(j + 1)
path.pop()
dfs(0)
return res
# Binary Decision Backtracking
def subsets_binary_decision(nums: List[int]) -> List[List[int]]:
n = len(nums)
res, path = [], []
def dfs(i):
if i == n:
res.append(path.copy())
return
# Exclude
dfs(i + 1)
# Include
path.append(nums[i])
dfs(i + 1)
path.pop()
dfs(0)
return res
print(subsets_iterative_inclusion([1, 2, 3]))
# [[], [1], [1, 2], [1, 2, 3], [1, 3], [2], [2, 3], [3]]
print(subsets_binary_decision([1, 2, 3]))
# [[], [3], [2], [2, 3], [1], [1, 3], [1, 2], [1, 2, 3]]
146. LRU Cache¶
from collections import OrderedDict
class DoublyListNode:
def __init__(self, key=0, val=0):
self.key = key
self.val = val
self.prev = None
self.next = None
class LRUCache:
"""
Design and implement a data structure for Least Recently Used (LRU) cache.
It should support the following operations: get and put.
"""
def __init__(self, capacity: int):
self.cap = capacity
self.cache = {}
self.head = DoublyListNode()
self.tail = DoublyListNode()
self.head.next = self.tail
self.tail.prev = self.head
def remove(self, node):
node.prev.next = node.next
node.next.prev = node.prev
def add_to_last(self, node):
self.tail.prev.next = node
node.prev = self.tail.prev
node.next = self.tail
self.tail.prev = node
def get(self, key: int) -> int:
if key not in self.cache:
return -1
node = self.cache[key]
self.remove(node)
self.add_to_last(node)
return node.val
def put(self, key: int, value: int) -> None:
if key in self.cache:
node = self.cache[key]
node.val = value
self.remove(node)
self.add_to_last(node)
else:
node = DoublyListNode(key=key, val=value)
self.cache[key] = node
self.add_to_last(node)
if len(self.cache) > self.cap:
lru = self.head.next
del self.cache[lru.key]
self.remove(lru)
class LRUCacheOrderedDict:
def __init__(self, capacity: int):
self.cache = OrderedDict()
self.cap = capacity
def get(self, key: int):
if key not in self.cache:
return -1
self.cache.move_to_end(key, last=True)
return self.cache[key]
def put(self, key: int, value: int):
if key in self.cache:
self.cache.move_to_end(key, last=True)
elif len(self.cache) >= self.cap:
self.cache.popitem(last=False)
self.cache[key] = value
def test_lru_cache():
cache = LRUCache(2)
cache.put(1, 1)
cache.put(2, 2)
assert cache.get(1) == 1
cache.put(3, 3)
assert cache.get(2) == -1
cache.put(4, 4)
assert cache.get(1) == -1
assert cache.get(3) == 3
assert cache.get(4) == 4
updated = LRUCache(2)
updated.put(1, 1)
updated.put(2, 2)
updated.put(1, 10)
updated.put(3, 3)
assert updated.get(1) == 10
assert updated.get(2) == -1
single = LRUCache(1)
single.put(1, 1)
single.put(2, 2)
assert single.get(1) == -1
assert single.get(2) == 2
def test_lru_cache_ordered_dict():
cache = LRUCacheOrderedDict(2)
cache.put(1, 1)
cache.put(2, 2)
assert cache.get(1) == 1
cache.put(3, 3)
assert cache.get(2) == -1
cache.put(4, 4)
assert cache.get(1) == -1
assert cache.get(3) == 3
assert cache.get(4) == 4
#include <cassert>
#include <iostream>
#include <unordered_map>
using namespace std;
class Node {
public:
int key;
int val;
Node *prev;
Node *next;
Node(int k = 0, int v = 0) : key(k), val(v), prev(nullptr), next(nullptr) {}
};
class LRUCache {
private:
int cap;
unordered_map<int, Node *> cache;
Node *head;
Node *tail;
void remove(Node *node) {
node->prev->next = node->next;
node->next->prev = node->prev;
}
void insert_to_last(Node *node) {
tail->prev->next = node;
node->prev = tail->prev;
tail->prev = node;
node->next = tail;
}
void move_to_last(Node *node) {
remove(node);
insert_to_last(node);
}
public:
LRUCache(int capacity) {
this->cap = capacity;
head = new Node();
tail = new Node();
head->next = tail;
tail->prev = head;
}
int get(int key) {
if (cache.find(key) != cache.end()) {
Node *node = cache[key];
move_to_last(node);
return node->val;
}
return -1;
}
void put(int key, int value) {
if (cache.find(key) != cache.end()) {
Node *node = cache[key];
node->val = value;
move_to_last(node);
} else {
Node *newNode = new Node(key, value);
cache[key] = newNode;
insert_to_last(newNode);
if ((int)cache.size() > cap) {
Node *removed = head->next;
remove(removed);
cache.erase(removed->key);
delete removed;
}
}
}
};
int main() {
LRUCache lru(2);
lru.put(1, 1);
lru.put(2, 2);
assert(lru.get(1) == 1); // returns 1
lru.put(3, 3); // evicts key 2
assert(lru.get(2) == -1); // returns -1 (not found)
lru.put(4, 4); // evicts key 1
assert(lru.get(1) == -1); // returns -1 (not found)
assert(lru.get(3) == 3); // returns 3
assert(lru.get(4) == 4); // returns 4
return 0;
}
14. Longest Common Prefix¶
from typing import List
class longestCommonPrefix:
def horizontal_scan(self, strs: List[str]) -> str:
if not strs:
return ""
prefix = strs[0]
for i in range(1, len(strs)):
while not strs[i].startswith(prefix):
prefix = prefix[:-1]
if not prefix:
return ""
return prefix
def vertical_scan(self, strs: List[str]) -> str:
if not strs:
return ""
for i in range(len(strs[0])):
char = strs[0][i]
for j in range(1, len(strs)):
if i >= len(strs[j]) or strs[j][i] != char:
return strs[0][:i]
return strs[0]
def divide_conquer(self, strs: List[str]) -> str:
if not strs:
return ""
def merge(left, right):
n = min(len(left), len(right))
for i in range(n):
if left[i] != right[i]:
return left[:i]
return left[:n]
def find(strs, start, end):
if start == end:
return strs[start]
mid = start + (end - start) // 2
left = find(strs, start, mid)
right = find(strs, mid + 1, end)
return merge(left, right)
return find(strs, 0, len(strs) - 1)
def binary_search(self, strs: List[str]) -> str:
if not strs:
return ""
def isCommonPrefix(strs, length):
prefix = strs[0][:length]
return all(s.startswith(prefix) for s in strs)
minLen = min(len(s) for s in strs)
low, high = 0, minLen
while low < high:
mid = low + (high - low) // 2
if isCommonPrefix(strs, mid + 1):
low = mid + 1
else:
high = mid
return strs[0][:low]
if __name__ == "__main__":
solution = longestCommonPrefix()
strs = ["flower", "flow", "flight"]
assert solution.horizontal_scan(strs) == "fl"
assert solution.vertical_scan(strs) == "fl"
assert solution.divide_conquer(strs) == "fl"
assert solution.binary_search(strs) == "fl"
128. Longest Consecutive Sequence¶
from typing import List
# Set
def longestConsecutiveSet(nums: List[int]) -> int:
num_set = set(nums) # O(n)
longest = 0
for n in nums:
if (n - 1) not in num_set: # left boundary
length = 1
while (n + length) in num_set:
length += 1
longest = max(longest, length)
return longest
# Union Find
def longestConsecutiveUF(nums: List[int]) -> int:
if not nums:
return 0
par = {num: num for num in nums}
rank = {num: 1 for num in nums}
def find(num):
p = par[num]
while p != par[p]:
p = par[p]
return p
def union(num1, num2):
p1, p2 = find(num1), find(num2)
if p1 == p2:
return
if rank[p1] < rank[p2]:
par[p1] = p2
rank[p2] += rank[p1]
else:
par[p2] = p1
rank[p1] += rank[p2]
for num in nums:
if num - 1 in par:
union(num, num - 1)
return max(rank.values())
# |------------|------- |---------|
# | Approach | Time | Space |
# |------------|--------|---------|
# | Set | O(N) | O(N) |
# | Union Find | O(N) | O(N) |
# |------------|--------|---------|
nums = [100, 4, 200, 1, 3, 2]
print(longestConsecutiveSet(nums)) # 4
print(longestConsecutiveUF(nums)) # 4
34. Find First and Last Position of Element in Sorted Array¶
from bisect import bisect_left
from typing import List
class searchRange:
"""
找 lower bound 和 upper bound
看灵神对这道题的题解，分类讨论区间的写法
target 的 upper bound 是 target + 1 的 lower bound - 1
这样就能统一用 lower bound 的写法
"""
# [left, right]
def bisect_left_closed(self, nums, target):
"""
闭区间写法
"""
left, right = 0, len(nums) - 1
while left <= right:
mid = left + (right - left) // 2
if nums[mid] < target:
left = mid + 1
else:
right = mid - 1
return left
# [left, right)
def bisect_left_right_open(self, nums, target):
"""
左闭右开区间写法
"""
left, right = 0, len(nums)
while left < right:
mid = left + (right - left) // 2
if nums[mid] >= target:
right = mid
else:
left = mid + 1
return left
# (left, right)
def bisect_left_open(self, nums, target):
"""
推荐开区间写法
"""
left, right = -1, len(nums)
while left + 1 < right:
mid = left + (right - left) // 2
if nums[mid] < target:
left = mid
else:
right = mid
return right
def search_range(self, nums: List[int], target: int) -> List[int]:
# edge case
if not nums:
return [-1, -1]
lower = self.bisect_left_closed(nums, target)
upper = self.bisect_left_closed(nums, target + 1) - 1
return [lower, upper] if lower <= upper else [-1, -1]
def search_range_bisect(self, nums: List[int], target: int) -> List[int]:
"""用 python bisect 库函数"""
# edge case
if not nums:
return [-1, -1]
lower = bisect_left(nums, target)
upper = bisect_left(nums, target + 1) - 1
return [lower, upper] if lower <= upper else [-1, -1]
if __name__ == "__main__":
nums = [5, 7, 7, 8, 8, 10]
target = 8
sol = searchRange()
assert sol.search_range(nums, target) == [3, 4]
assert sol.search_range_bisect(nums, target) == [3, 4]
#include <vector>
#include <iostream>
using namespace std;
class Solution
{
int bisect_left(vector<int> &nums, int target)
{
int left = 0, right = (int)nums.size() - 1;
while (left <= right)
{
int mid = left + (right - left) / 2;
if (nums[mid] < target)
{
left = mid + 1;
}
else
{
right = mid - 1;
}
}
return left;
}
public:
vector<int> searchRange(vector<int> &nums, int target)
{
int left = bisect_left(nums, target);
if (left == (int)nums.size() || nums[left] != target)
{
return {-1, -1};
}
int right = bisect_left(nums, target + 1) - 1;
return {left, right};
}
};
int main()
{
vector<int> nums = {5, 7, 7, 8, 8, 10};
int target = 8;
Solution s;
vector<int> res = s.searchRange(nums, target);
cout << res[0] << ", " << res[1] << endl;
return 0;
}
217. Contains Duplicate¶
from typing import List
class Solution:
def containsDuplicate(self, nums: List[int]) -> bool:
"""Set: O(n) time, O(n) space.
Track seen elements; return True on first repeat.
"""
seen = set()
for n in nums:
if n in seen:
return True
seen.add(n)
return False
def containsDuplicateSort(self, nums: List[int]) -> bool:
"""Sort: O(n log n) time, O(1) space.
Duplicates become adjacent after sorting.
"""
nums.sort()
for i in range(1, len(nums)):
if nums[i] == nums[i - 1]:
return True
return False
def containsDuplicateBF(self, nums: List[int]) -> bool:
"""Brute Force: O(n^2) time, O(1) space.
Compare every pair.
"""
for i in range(len(nums)):
for j in range(i + 1, len(nums)):
if nums[i] == nums[j]:
return True
return False
def test_contains_duplicate():
s = Solution()
for fn in (s.containsDuplicate, s.containsDuplicateSort, s.containsDuplicateBF):
assert fn([1, 2, 3, 1]) is True
assert fn([1, 2, 3, 4]) is False
assert fn([1, 1, 1, 3, 3, 4, 3, 2, 4, 2]) is True
assert fn([]) is False
22. Generate Parentheses¶
from typing import List
# Backtracking
def generateParenthesis1(n: int) -> List[str]:
path, res = [], []
def dfs(openN, closeN):
if openN == closeN == n:
res.append("".join(path))
return
if openN < n:
path.append("(")
dfs(openN + 1, closeN)
path.pop()
if closeN < openN:
path.append(")")
dfs(openN, closeN + 1)
path.pop()
dfs(0, 0)
return res
# Backtracking
def generateParenthesis2(n: int) -> List[str]:
m = n * 2
res, path = [], [""] * m
def dfs(i, left):
if i == m:
res.append("".join(path))
return
if left < n:
path[i] = "("
dfs(i + 1, left + 1)
if i - left < left:
path[i] = ")"
dfs(i + 1, left)
dfs(0, 0)
return res
if __name__ == "__main__":
print(generateParenthesis1(3))
# ['((()))', '(()())', '(())()', '()(())', '()()()']
print(generateParenthesis2(3))
# ['((()))', '(()())', '(())()', '()(())', '()()()']
560. Subarray Sum Equals K¶
from collections import defaultdict
from typing import List
# Prefix Sum
def subarraySum(nums: List[int], k: int) -> int:
preSums = defaultdict(int)
preSums[0] = 1
curSum = 0
res = 0
for num in nums:
curSum += num
res += preSums[curSum - k]
preSums[curSum] += 1
return res
nums = [1, 1, 1]
k = 2
print(subarraySum(nums, k)) # 2
#include <iostream>
#include <unordered_map>
#include <vector>
using namespace std;
int subarraySum(vector<int>& nums, int k) {
int n = nums.size();
vector<int> prefixSum(n + 1);
for (int i = 0; i < n; i++) {
prefixSum[i + 1] = prefixSum[i] + nums[i];
}
int res = 0;
unordered_map<int, int> cnt;
for (int ps : prefixSum) {
if (cnt.find(ps - k) != cnt.end()) res += cnt[ps - k];
cnt[ps]++;
}
return res;
}
int main() {
vector<int> nums = {1, 1, 1};
int k = 2;
cout << subarraySum(nums, k) << endl; // 2
return 0;
}
21. Merge Two Sorted Lists¶
"""
- Task: Merge the two linked lists into one sorted list.
"""
from typing import Optional
from leetpattern.utils import ListNode
# Linked List
def merge_two_lists(
list1: Optional[ListNode], list2: Optional[ListNode]
) -> Optional[ListNode]:
dummy = ListNode()
cur = dummy
while list1 and list2:
if list1.val < list2.val:
cur.next = list1
list1 = list1.next
else:
cur.next = list2
list2 = list2.next
cur = cur.next
if list1:
cur.next = list1
elif list2:
cur.next = list2
return dummy.next
#include <cassert>
#include <iostream>
#include <vector>
using namespace std;
struct ListNode {
int val;
ListNode* next;
ListNode() : val(0), next(nullptr) {}
ListNode(int x) : val(x), next(nullptr) {}
ListNode(int x, ListNode* next) : val(x), next(next) {}
};
class Solution {
public:
ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) {
ListNode dummy;
ListNode* cur = &dummy;
while (list1 && list2) {
if (list1->val < list2->val) {
cur->next = list1;
list1 = list1->next;
} else {
cur->next = list2;
list2 = list2->next;
}
cur = cur->next;
}
cur->next = list1 ? list1 : list2;
return dummy.next;
}
};
int main() {
Solution sol;
ListNode* list1 = new ListNode(1, new ListNode(2, new ListNode(4)));
ListNode* list2 = new ListNode(1, new ListNode(3, new ListNode(4)));
ListNode* merged = sol.mergeTwoLists(list1, list2);
vector<int> res;
while (merged) {
res.push_back(merged->val);
merged = merged->next;
}
assert(res == std::vector<int>({1, 1, 2, 3, 4, 4}));
return 0;
}
2667. Create Hello World Function¶
13. Roman to Integer¶
from itertools import pairwise
# Arrays
def romanToInt(s: str) -> int:
ROMAN = {
"I": 1,
"V": 5,
"X": 10,
"L": 50,
"C": 100,
"D": 500,
"M": 1000,
}
res = 0
for x, y in pairwise(s):
x, y = ROMAN[x], ROMAN[y]
res += x if x >= y else -x
return res + ROMAN[s[-1]]
if __name__ == "__main__":
assert romanToInt("III") == 3
assert romanToInt("IV") == 4
assert romanToInt("IX") == 9
assert romanToInt("LVIII") == 58
assert romanToInt("MCMXCIV") == 1994
assert romanToInt("MMXXIII") == 2023
53. Maximum Subarray¶
from typing import List
# DP Kadane
def maxSubArrayDP(nums: List[int]) -> int:
n = len(nums)
dp = [0 for _ in range(n)]
dp[0] = nums[0]
res = nums[0]
for i in range(1, n):
dp[i] = max(
dp[i - 1] + nums[i], # continue the previous subarray
nums[i], # start a new subarray
)
res = max(res, dp[i])
return res
# Greedy
def maxSubArrayGreedy(nums: List[int]) -> int:
max_sum = nums[0]
cur_sum = 0
for num in nums:
cur_sum = max(cur_sum + num, num)
max_sum = max(max_sum, cur_sum)
return max_sum
# Prefix Sum
def maxSubArrayPrefixSum(nums: List[int]) -> int:
prefix_sum = 0
prefix_sum_min = 0
res = float("-inf")
for num in nums:
prefix_sum += num
res = max(res, prefix_sum - prefix_sum_min)
prefix_sum_min = min(prefix_sum_min, prefix_sum)
return res
nums = [-2, 1, -3, 4, -1, 2, 1, -5, 4]
print(maxSubArrayDP(nums)) # 6
print(maxSubArrayGreedy(nums)) # 6
print(maxSubArrayPrefixSum(nums)) # 6
1244. Design A Leaderboard 👑¶
class Leaderboard:
def __init__(self):
self.scores = {}
def addScore(self, playerId: int, score: int) -> None:
if playerId in self.scores:
self.scores[playerId] += score
else:
self.scores[playerId] = score
def top(self, K: int) -> int:
topK = sorted(self.scores.values(), reverse=True)[:K]
return sum(topK)
def reset(self, playerId: int) -> None:
if playerId in self.scores:
self.scores[playerId] = 0
board = Leaderboard()
board.addScore(1, 73)
board.addScore(2, 56)
board.addScore(3, 39)
board.addScore(4, 51)
print(board.top(1)) # 73
board.reset(1)
board.reset(2)
print(board.top(2)) # 90