LeetCode #1051 — EASY

Height Checker

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

The Problem

Problem Statement

A school is trying to take an annual photo of all the students. The students are asked to stand in a single file line in non-decreasing order by height. Let this ordering be represented by the integer array expected where expected[i] is the expected height of the i^th student in line.

You are given an integer array heights representing the current order that the students are standing in. Each heights[i] is the height of the i^th student in line (0-indexed).

Return the number of indices where heights[i] != expected[i].

Example 1:

Input: heights = [1,1,4,2,1,3]
Output: 3
Explanation: 
heights:  [1,1,4,2,1,3]
expected: [1,1,1,2,3,4]
Indices 2, 4, and 5 do not match.

Example 2:

Input: heights = [5,1,2,3,4]
Output: 5
Explanation:
heights:  [5,1,2,3,4]
expected: [1,2,3,4,5]
All indices do not match.

Example 3:

Input: heights = [1,2,3,4,5]
Output: 0
Explanation:
heights:  [1,2,3,4,5]
expected: [1,2,3,4,5]
All indices match.

Constraints:

1 <= heights.length <= 100
1 <= heights[i] <= 100

Step 01

Brute Force Baseline

Problem summary: A school is trying to take an annual photo of all the students. The students are asked to stand in a single file line in non-decreasing order by height. Let this ordering be represented by the integer array expected where expected[i] is the expected height of the ith student in line. You are given an integer array heights representing the current order that the students are standing in. Each heights[i] is the height of the ith student in line (0-indexed). Return the number of indices where heights[i] != expected[i].

Baseline thinking

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

Pattern signal: Array

Example 1

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

Example 2

[5,1,2,3,4]

Example 3

[1,2,3,4,5]

Step 02

Core Insight

What unlocks the optimal approach

Build the correct order of heights by sorting another array, then compare the two arrays.

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 #1051: Height Checker
class Solution {
    public int heightChecker(int[] heights) {
        int[] expected = heights.clone();
        Arrays.sort(expected);
        int ans = 0;
        for (int i = 0; i < heights.length; ++i) {
            if (heights[i] != expected[i]) {
                ++ans;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1051: Height Checker
func heightChecker(heights []int) (ans int) {
	expected := slices.Clone(heights)
	sort.Ints(expected)
	for i, v := range heights {
		if v != expected[i] {
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #1051: Height Checker
class Solution:
    def heightChecker(self, heights: List[int]) -> int:
        expected = sorted(heights)
        return sum(a != b for a, b in zip(heights, expected))

// Accepted solution for LeetCode #1051: Height Checker
struct Solution;

impl Solution {
    fn height_checker(heights: Vec<i32>) -> i32 {
        let mut sorted = heights.to_vec();
        sorted.sort_unstable();
        heights
            .iter()
            .zip(sorted.iter())
            .fold(0, |sum, (a, b)| if a != b { sum + 1 } else { sum })
    }
}

#[test]
fn test() {
    let heights = vec![1, 1, 4, 2, 1, 3];
    assert_eq!(Solution::height_checker(heights), 3);
}

// Accepted solution for LeetCode #1051: Height Checker
function heightChecker(heights: number[]): number {
    const expected = [...heights].sort((a, b) => a - b);
    return heights.reduce((acc, cur, i) => acc + (cur !== expected[i] ? 1 : 0), 0);
}

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 × log n)

Space

O(n)

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.