LeetCode #2951 — EASY

Find the Peaks

Build confidence with an intuition-first walkthrough focused on array fundamentals.

Solve on LeetCode

The Problem

Problem Statement

You are given a 0-indexed array mountain. Your task is to find all the peaks in the mountain array.

Return an array that consists of indices of peaks in the given array in any order.

Notes:

A peak is defined as an element that is strictly greater than its neighboring elements.
The first and last elements of the array are not a peak.

Example 1:

Input: mountain = [2,4,4]
Output: []
Explanation: mountain[0] and mountain[2] can not be a peak because they are first and last elements of the array.
mountain[1] also can not be a peak because it is not strictly greater than mountain[2].
So the answer is [].

Example 2:

Input: mountain = [1,4,3,8,5]
Output: [1,3]
Explanation: mountain[0] and mountain[4] can not be a peak because they are first and last elements of the array.
mountain[2] also can not be a peak because it is not strictly greater than mountain[3] and mountain[1].
But mountain [1] and mountain[3] are strictly greater than their neighboring elements.
So the answer is [1,3].

Constraints:

3 <= mountain.length <= 100
1 <= mountain[i] <= 100

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed array mountain. Your task is to find all the peaks in the mountain array. Return an array that consists of indices of peaks in the given array in any order. Notes: A peak is defined as an element that is strictly greater than its neighboring elements. The first and last elements of the array are not a peak.

Baseline thinking

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

Pattern signal: Array

Example 1

[2,4,4]

Example 2

[1,4,3,8,5]

Core Insight

What unlocks the optimal approach

If <code>nums[i] > num[i - 1]</code> and <code>nums[i] > nums[i + 1]</code> <code>nums[i]</code> is a peak.

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 #2951: Find the Peaks
class Solution {
    public List<Integer> findPeaks(int[] mountain) {
        List<Integer> ans = new ArrayList<>();
        for (int i = 1; i < mountain.length - 1; ++i) {
            if (mountain[i - 1] < mountain[i] && mountain[i + 1] < mountain[i]) {
                ans.add(i);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2951: Find the Peaks
func findPeaks(mountain []int) (ans []int) {
	for i := 1; i < len(mountain)-1; i++ {
		if mountain[i-1] < mountain[i] && mountain[i+1] < mountain[i] {
			ans = append(ans, i)
		}
	}
	return
}

# Accepted solution for LeetCode #2951: Find the Peaks
class Solution:
    def findPeaks(self, mountain: List[int]) -> List[int]:
        return [
            i
            for i in range(1, len(mountain) - 1)
            if mountain[i - 1] < mountain[i] > mountain[i + 1]
        ]

// Accepted solution for LeetCode #2951: Find the Peaks
fn find_peaks(mountain: Vec<i32>) -> Vec<i32> {
    let mut ret = vec![];
    for i in 1..(mountain.len() - 1) {
        if mountain[i - 1] < mountain[i] && mountain[i] > mountain[i + 1] {
            ret.push(i as i32);
        }
    }

    ret
}

fn main() {
    let mountain = vec![1, 4, 3, 8, 5];
    let ret = find_peaks(mountain);
    println!("ret={ret:?}");
}

#[test]
fn test() {
    {
        let mountain = vec![2, 4, 4];
        let expected = vec![];
        let ret = find_peaks(mountain);
        assert_eq!(ret, expected);
    }
    {
        let mountain = vec![1, 4, 3, 8, 5];
        let expected = vec![1, 3];
        let ret = find_peaks(mountain);
        assert_eq!(ret, expected);
    }
}

// Accepted solution for LeetCode #2951: Find the Peaks
function findPeaks(mountain: number[]): number[] {
    const ans: number[] = [];
    for (let i = 1; i < mountain.length - 1; ++i) {
        if (mountain[i - 1] < mountain[i] && mountain[i + 1] < mountain[i]) {
            ans.push(i);
        }
    }
    return ans;
}

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(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.