FIND PEAK ELEMENT | LEETCODE # 162 | PYTHON BINARY SEARCH SOLUTION

[@lepakshiagarwal6118]

This solution gives a time limit exceeded on leetcode but was really easy to understand. Thank you for explaining

[@aminebesbes6202]

The solution could be more simplified as this (Easy Binary Search with no need to check boundaries): class Solution: def findPeakElement(self, nums: List[int]) -> int: left, mid, right = 0, 0, len(nums)-1 while leftnums[mid+1]: right = mid else: left = mid +1 return left

[@rdnetala]

easy recursrive sol: class Solution: def findPeakElement(self, nums: List[int]) -> int: def bs(left, right): if left == right: return left # Base case: single element, so it’s a peak mid = (left + right) // 2 # Compare middle with the next element to decide which side to search if nums[mid] > nums[mid + 1]: # Peak is in the left half return bs(left, mid) else: # Peak is in the right half return bs(mid + 1, right) return bs(0, len(nums) - 1)

[@rsKayiira]

Good solution, Could you do LC 987?

[@subee128]

Thanks