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Problem 4 — Find Peak Element

Problem 4 — Find Peak Element

Section titled “Problem 4 — Find Peak Element”

LeetCode 162 | Difficulty: 🟡 Medium

🎯 Problem Statement

Section titled “🎯 Problem Statement”

A peak element is an element that is strictly greater than its neighbors. Find any peak index. Assume nums[-1] = nums[n] = -∞.

Input:  nums = [1, 2, 3, 1]
Output: 2   (nums[2] = 3 is a peak)


Input:  nums = [1, 2, 1, 3, 5, 6, 4]
Output: 5   (nums[5] = 6 is a peak — index 1 also works)

🧠 Approach: Pattern 5 (Monotonic Function)

Section titled “🧠 Approach: Pattern 5 (Monotonic Function)”

Key insight: If nums[mid] < nums[mid + 1], a peak must exist to the right (we’re climbing uphill). Otherwise, a peak exists at mid or to the left.

nums = [1, 2, 3, 1]


mid=1 → nums[1]=2 < nums[2]=3 → climbing up → peak RIGHT
mid=2 → nums[2]=3 > nums[3]=1 → peak at mid or LEFT


Peak found at index 2 ✓

Why this works: The array boundaries are -∞. If we always move toward the larger neighbor, we must eventually reach a peak (you can’t go uphill forever).

💻 Solution

Section titled “💻 Solution”
function findPeakElement(nums) {
  let lo = 0, hi = nums.length - 1;


  while (lo < hi) {
    const mid = lo + Math.floor((hi - lo) / 2);


    if (nums[mid] < nums[mid + 1]) {
      lo = mid + 1; // Peak is to the right (climbing up)
    } else {
      hi = mid;     // Peak is at mid or to the left
    }
  }


  return lo; // lo === hi is a peak
}


console.log(findPeakElement([1, 2, 3, 1]));          // 2
console.log(findPeakElement([1, 2, 1, 3, 5, 6, 4])); // 5
console.log(findPeakElement([1, 2, 3, 4, 5]));        // 4 (strictly increasing)
console.log(findPeakElement([5, 4, 3, 2, 1]));        // 0 (strictly decreasing)

🧪 Walkthrough

Section titled “🧪 Walkthrough”
nums = [1, 2, 3, 1]


Step 1: lo=0, hi=3, mid=1
  nums[1]=2 < nums[2]=3 → climbing up → peak is RIGHT
  lo = mid + 1 = 2


Step 2: lo=2, hi=3, mid=2
  nums[2]=3 > nums[3]=1 → peak at mid or LEFT
  hi = mid = 2


Step 3: lo=2, hi=2 → loop exits, return 2 ✓

📊 Complexity

Section titled “📊 Complexity”
MetricValue
TimeO(log n) — binary search
SpaceO(1) — no extra memory

🎯 Variations

Section titled “🎯 Variations”

Variation: Find Peak in 2D Matrix (LeetCode 1901)

Section titled “Variation: Find Peak in 2D Matrix (LeetCode 1901)”
function findPeakGrid(mat) {
  let lo = 0, hi = mat[0].length - 1;


  while (lo <= hi) {
    const mid = lo + Math.floor((hi - lo) / 2);


    // Find max element in this column
    let maxRow = 0;
    for (let r = 0; r < mat.length; r++) {
      if (mat[r][mid] > mat[maxRow][mid]) maxRow = r;
    }


    const leftIsBigger = mid > 0 && mat[maxRow][mid - 1] > mat[maxRow][mid];
    const rightIsBigger = mid < mat[0].length - 1 && mat[maxRow][mid + 1] > mat[maxRow][mid];


    if (!leftIsBigger && !rightIsBigger) {
      return [maxRow, mid]; // Found a peak
    }


    if (leftIsBigger) {
      hi = mid - 1; // Peak is to the left
    } else {
      lo = mid + 1; // Peak is to the right
    }
  }


  return [-1, -1];
}

🔑 Key Takeaways

Section titled “🔑 Key Takeaways”
  • Compare with the next element (nums[mid] < nums[mid + 1])
  • lo < hi — converge to the answer
  • Any peak works — we don’t need the highest peak
  • The algorithm is guaranteed to find a peak because boundaries are -∞
  • This is a Pattern 5 (Monotonic Function) problem disguised as array search

🔗 Related

Section titled “🔗 Related”