LeetCode #1901 — MEDIUM

Find a Peak Element II

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

A peak element in a 2D grid is an element that is strictly greater than all of its adjacent neighbors to the left, right, top, and bottom.

Given a 0-indexed m x n matrix mat where no two adjacent cells are equal, find any peak element mat[i][j] and return the length 2 array [i,j].

You may assume that the entire matrix is surrounded by an outer perimeter with the value -1 in each cell.

You must write an algorithm that runs in O(m log(n)) or O(n log(m)) time.

Example 1:

Input: mat = [[1,4],[3,2]]
Output: [0,1]
Explanation: Both 3 and 4 are peak elements so [1,0] and [0,1] are both acceptable answers.

Example 2:

Input: mat = [[10,20,15],[21,30,14],[7,16,32]]
Output: [1,1]
Explanation: Both 30 and 32 are peak elements so [1,1] and [2,2] are both acceptable answers.

Constraints:

m == mat.length
n == mat[i].length
1 <= m, n <= 500
1 <= mat[i][j] <= 10⁵
No two adjacent cells are equal.

Step 01

Brute Force Baseline

Problem summary: A peak element in a 2D grid is an element that is strictly greater than all of its adjacent neighbors to the left, right, top, and bottom. Given a 0-indexed m x n matrix mat where no two adjacent cells are equal, find any peak element mat[i][j] and return the length 2 array [i,j]. You may assume that the entire matrix is surrounded by an outer perimeter with the value -1 in each cell. You must write an algorithm that runs in O(m log(n)) or O(n log(m)) time.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search

Example 1

[[1,4],[3,2]]

Example 2

[[10,20,15],[21,30,14],[7,16,32]]

Core Insight

What unlocks the optimal approach

Let's assume that the width of the array is bigger than the height, otherwise, we will split in another direction.
Split the array into three parts: central column left side and right side.
Go through the central column and two neighbor columns and look for maximum.
If it's in the central column - this is our peak.
If it's on the left side, run this algorithm on subarray left_side + central_column.
If it's on the right side, run this algorithm on subarray right_side + central_column

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1901: Find a Peak Element II
class Solution {
    public int[] findPeakGrid(int[][] mat) {
        int l = 0, r = mat.length - 1;
        int n = mat[0].length;
        while (l < r) {
            int mid = (l + r) >> 1;
            int j = maxPos(mat[mid]);
            if (mat[mid][j] > mat[mid + 1][j]) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return new int[] {l, maxPos(mat[l])};
    }

    private int maxPos(int[] arr) {
        int j = 0;
        for (int i = 1; i < arr.length; ++i) {
            if (arr[j] < arr[i]) {
                j = i;
            }
        }
        return j;
    }
}

// Accepted solution for LeetCode #1901: Find a Peak Element II
func findPeakGrid(mat [][]int) []int {
	maxPos := func(arr []int) int {
		j := 0
		for i := 1; i < len(arr); i++ {
			if arr[i] > arr[j] {
				j = i
			}
		}
		return j
	}
	l, r := 0, len(mat)-1
	for l < r {
		mid := (l + r) >> 1
		j := maxPos(mat[mid])
		if mat[mid][j] > mat[mid+1][j] {
			r = mid
		} else {
			l = mid + 1
		}
	}
	return []int{l, maxPos(mat[l])}
}

# Accepted solution for LeetCode #1901: Find a Peak Element II
class Solution:
    def findPeakGrid(self, mat: List[List[int]]) -> List[int]:
        l, r = 0, len(mat) - 1
        while l < r:
            mid = (l + r) >> 1
            j = mat[mid].index(max(mat[mid]))
            if mat[mid][j] > mat[mid + 1][j]:
                r = mid
            else:
                l = mid + 1
        return [l, mat[l].index(max(mat[l]))]

// Accepted solution for LeetCode #1901: Find a Peak Element II
impl Solution {
    pub fn find_peak_grid(mat: Vec<Vec<i32>>) -> Vec<i32> {
        let mut l: usize = 0;
        let mut r: usize = mat.len() - 1;
        while l < r {
            let mid: usize = (l + r) >> 1;
            let j: usize = mat[mid]
                .iter()
                .position(|&x| x == *mat[mid].iter().max().unwrap())
                .unwrap();
            if mat[mid][j] > mat[mid + 1][j] {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        let j: usize = mat[l]
            .iter()
            .position(|&x| x == *mat[l].iter().max().unwrap())
            .unwrap();
        vec![l as i32, j as i32]
    }
}

// Accepted solution for LeetCode #1901: Find a Peak Element II
function findPeakGrid(mat: number[][]): number[] {
    let [l, r] = [0, mat.length - 1];
    while (l < r) {
        const mid = (l + r) >> 1;
        const j = mat[mid].indexOf(Math.max(...mat[mid]));
        if (mat[mid][j] > mat[mid + 1][j]) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return [l, mat[l].indexOf(Math.max(...mat[l]))];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.