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162. Find Peak Element

8 June 2025·91 words·1 min· loading

Table of Contents

Table of Contents

IDEA
#

If a problem requries $O(\log_n)$ time complexity, It’s binary serach.

SOLUTION/CODE
#

Compare nums[mid] and nums[mid+1]
- If nums[mid] is less than nums[mid + 1], it means there must be a peak on the right side.
- Else, it means that might be a peak on left side. (mid can be a peak itself)

class Solution {
    public int findPeakElement(int[] nums) {
        int l =0, r=nums.length-1, mid=0;

        while (l < r){
            mid = (l+r)/2;

            if(nums[mid] < nums[mid+1]){
                l = mid+1;
            }else{
                r = mid;
            }
        }

        return l;
    }
}