My website

```python class Solution: def connectSticks(self, A: List[int]) -> int: “”” main idea: to minimize cost ==> to connect smallest stick pair as possible => PQ? tc O(NlgN) sc O(N) “”” n = len(A) if n <2: return 0 heapify(A) res = 0 while len(A)>1: x = heappop(A) y = heappop(A)... [Read More]

Three round ```python class Solution: def findLengthOfShortestSubarray(self, arr: List[int]) -> int: “”” tc O(N) sc O(1) main idea: since we can remove only one subarray to make whole array sorted, we eigher remove (1) left sorted subarray’s right side (2) right sorted subarray’s left side (3) combined sorted left, right... [Read More]

Two round way ```python class Solution: def maxSatisfied(self, customers: List[int], grumpy: List[int], X: int) -> int: “”” 1. get min happy customers by default, mark them to -1 2. sliding window left - right get the max additional customer tc O(N) sc O(1) “”” base = 0 for idx,(c,g) in... [Read More]

Brute Force solution """ tc O(N) sc O(N) """ class Solution: def findErrorNums(self, nums: List[int]) -> List[int]: n = len(nums) seen = set() res = [] for x in nums: if x in seen: res.append(x) else: seen.add(x) for i in range(1,n+1): if i not in seen: res.append(i) return res [Read More]

```python “”” tc O(N^2) sc O(N) step1 :get unique x, sorted step2:record idx of each x position, initialize cnt for tracking overlaps in at [x[i],x[i+1]] step3: sort by y , use +1,-1 as mark to check if exist overlaps area S = (cur_y - prev_y) * cur_x_sum [Read More]