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Find the Peaks — Easy Problem & Solution

You are given an array mountain of heights. An index i is a peak when mountain[i] is strictly greater than both of its neighbours.

  • Difficulty: Easy
  • Topics: Arrays, Enumeration
  • Asked at: TCS, Accenture
  • Time limit: 2 s
  • Memory limit: 256 MB
  • Languages: JavaScript, TypeScript, Python, Java, C++, C, C#, Go, Kotlin, Swift, Rust, PHP and Ruby

Problem statement

You are given an array mountain of heights. An index i is a peak when mountain[i] is strictly greater than both of its neighbours.

The first and last indices are never peaks — they have only one neighbour. Return every peak index in increasing order.

Example 1

Input: mountain = [2,4,4]
Output: []
Explanation: Index 1 is not a peak because 4 is not strictly greater than the 4 on its right.

Example 2

Input: mountain = [1,4,3,8,5]
Output: [1,3]
Explanation: 4 beats 1 and 3; 8 beats 3 and 5. Indices 0 and 4 are excluded by the rule.

Example 3

Input: mountain = [5,5,5,5]
Output: []

Constraints

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

How to solve Find the Peaks

A peak is a local condition, so a single sweep that never touches the two ends answers the whole question.

Approach

  1. Run i from 1 up to n - 2 inclusive.
  2. Record i when mountain[i] > mountain[i - 1] and mountain[i] > mountain[i + 1].
  3. Return the recorded indices; the loop order already makes them increasing.

Why it works

Restricting the loop to the interior is the statement's own rule that the endpoints cannot qualify, and testing both neighbours is the definition verbatim, so nothing is missed and nothing extra is reported.

Complexity

  • Time — O(n)
  • Space — O(1) beyond the output

Pitfalls

  • Using >= turns a flat run such as [2,4,4] into a false peak.
  • Looping from 0 or to n - 1 reads out of bounds — or, worse, silently reads a default value.

Reference solution

Python

from typing import List

def findPeaks(mountain: List[int]) -> List[int]:
    out = []
    for i in range(1, len(mountain) - 1):
        if mountain[i] > mountain[i - 1] and mountain[i] > mountain[i + 1]:
            out.append(i)
    return out

JavaScript

var findPeaks = function(mountain) {
    var out = [];
    for (var i = 1; i + 1 < mountain.length; i++) {
        if (mountain[i] > mountain[i - 1] && mountain[i] > mountain[i + 1]) out.push(i);
    }
    return out;
};

Also on the editorial tab: C, C#, C++, Go, Java, Kotlin, PHP, Ruby, Rust, Swift, TypeScript.

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