162. Find Peak Element - cocoder39/coco39_LC GitHub Wiki

162. Find Peak Element

this is a special scenario to apply binary search. Though the input is not sorted, it still works because it could reduce the search space by half each iteration

class Solution:
    def findPeakElement(self, nums: List[int]) -> int:
        n = len(nums)
        #if n == 1:
        #    return 0
        
        low, high = 0, n-1
        
        while low + 1 < high:
            mid = low + (high - low) // 2 # low+1 <= mid <= high-1
            if nums[mid-1] < nums[mid] > nums[mid+1]:
                return mid
            elif nums[mid] < nums[mid-1]: # from -1 to mid-1 to mid: up then down, so mid-1 is peak or there is peak between (-1, mid-1)
                high = mid
            else: # nums[mid] < nums[mid+1]: # from mid to mid+1 to n: up then down, so mid+1 is peak or there is peak between (mid+1, n)
                low = mid
        
        if nums[low] < nums[high]:
        # if (low < 1 or nums[low-1] < nums[low]) and (low+1 >= n or nums[low] > nums[low+1]):
            return high
        return low

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since nums[-1] and nums[n] are -INF. the trend of the sequence must be going up then going down.

  • if (nums[mid] > nums[mid - 1]), since nums[mid - 1] < nums[mid] > nums[n], the trend within [mid - 1, n] must be going up then going down, there exists a peek
  • if (nums[mid] < nums[mid - 1]), since nums[-1] < nums[mid - 1] > nums[mid], the trend within [-1, mid] must be going up then going down, there exists a peek

tip: it is safe to access nums[mid - 1] since mid - 1 >= start as we have discussed in Binary search (bug)

int findPeakElement(vector<int>& nums) {
        int start = 0, end = nums.size() - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] > nums[mid - 1] && nums[mid] > nums[mid + 1]) {
                return mid;
            }
            //nums[mid - 1] < nums[mid] > nums[n], there exists a peek within [mid, n - 1]
            else if (nums[mid] > nums[mid - 1]) { //mid - 1 >= start is valid
                start = mid;
            }
            //nums[-1] < nums[mid - 1] > nums[mid], there exists a peek within [0, mid - 1]
            else { //nums[mid] < nums[mid - 1]
                //a peek is within [-1, mid]
                end = mid;
            }
        }
        
        if(nums[start] >= nums[end]){
            return start;
        }
        return end;
    }
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