LC 162: Find peak element

[dbkuppagiri]

No description provided.

[super30admin]

Strengths:

• The student has a good understanding of the binary search approach and its time complexity.
• The code is clean, readable, and well-commented, which is excellent for maintainability.
• The student correctly identifies the space complexity and the need for O(log n) time complexity.

Areas for Improvement:

• The logic for adjusting the pointers in the binary search is incorrect. When nums[mid] < nums[mid + 1], moving left to mid + 1 is correct, but when nums[mid] >= nums[mid + 1], right should be set to mid instead of mid - 1 to ensure the peak is not skipped.
• The student should test the solution with edge cases, such as arrays with a single element or arrays where the peak is at the beginning or end.
• The comment about removing the equal condition in the while loop is misleading; the condition left < right is sufficient for this problem, but the explanation could be clearer.

Suggested Correction:

var findPeakElement = function (nums) {
    let left = 0, right = nums.length - 1;
    while (left < right) {
        const mid = Math.floor(left + (right - left) / 2);
        if (nums[mid] < nums[mid + 1]) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    return left;
};

[dbkuppagiri]

Updated as per the suggestions.