What is the time complexity of Height Checker?

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Height Checker?

This problem uses Array, Sorting, Counting Sort. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Height Checker?

**Constructing the 'Ground Truth':** The crucial first step is to establish what the perfectly sorted lineup *should* look like. This is done by creating a sorted copy of the input `heights` array. Without this `expected` array, comparing individual `heights[i]` values wouldn't have a meaningful reference point. It explicitly defines the target state for comparison. **Preserving Original Order for Comparison:** When sorting, it's essential to create a *new*, sorted array (like `expected`) rather than sorting the `heights` array in-place. This ensures that the original `heights` array, representing the current student arrangement, remains intact. We need both the current (unsorted) and the ideal (sorted) arrangements available simultaneously to perform the direct index-by-index comparison.

Which companies ask LeetCode 1051. Height Checker in interviews?

This problem is frequently asked at Salesforce.

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Height Checker - LeetCode 1051 Solution

Height Checker - Complete Solution Guide

Height Checker is LeetCode problem 1051, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 h

Detailed Explanation

Imagine you're trying to line up a group of students for a photo, where everyone needs to be ordered from shortest to tallest. You've got the students standing in some arbitrary current arrangement. This problem asks you to figure out how many students are *not* in the correct spot they *should* be in if the line were perfectly sorted by height. We're interested in counting how many *positions* are occupied by a student whose height doesn't match the height of the student who *should* be at that position in the ideal, sorted line.

Solution Approach

The most intuitive way to tackle this 'spot the difference' challenge is to first construct the 'ideal' sorted line. The provided solution achieves this by taking the input `heights` array and creating a new array, `expected`, which is simply a sorted version of `heights`. This `expected` array now represents the non-decreasing order of heights that the students *should* be in. Once we have both `heights` (the current arrangement) and `expected` (the ideal arrangement), the solution iterates through both arrays simultaneously using an index `i`. For each position `i`, it compares `heights[i]` with `expected[i]`. If these two values are different, it means the student currently at position `i` is not the correct height for that position in a perfectly sorted line. A counter, `count`, is incremented for each such mismatch. Finally, after checking every position, the total `count` of mismatches is returned. This direct, element-by-element comparison against a pre-computed ideal state is highly effective because it precisely isolates every instance where the current reality deviates from the desired order.

Step-by-Step Algorithm

Step 1: Create a copy of the `heights` array (or sort in place, depending on the language).
Step 2: Sort the copied array (this becomes the `expected` array).
Step 3: Initialize a counter `count` to 0.
Step 4: Iterate through the `heights` array and compare each element `heights[i]` with the corresponding element `expected[i]`.
Step 5: If `heights[i]` != `expected[i]`, increment `count`.
Step 6: After iterating through the entire array, return the value of `count`.

Key Insights

**Constructing the 'Ground Truth':** The crucial first step is to establish what the perfectly sorted lineup *should* look like. This is done by creating a sorted copy of the input `heights` array. Without this `expected` array, comparing individual `heights[i]` values wouldn't have a meaningful reference point. It explicitly defines the target state for comparison.
**Preserving Original Order for Comparison:** When sorting, it's essential to create a *new*, sorted array (like `expected`) rather than sorting the `heights` array in-place. This ensures that the original `heights` array, representing the current student arrangement, remains intact. We need both the current (unsorted) and the ideal (sorted) arrangements available simultaneously to perform the direct index-by-index comparison.
**Direct Index-to-Index Mismatch Tally:** The problem simplifies to a straightforward `if (heights[i] != expected[i]) count++` loop. There's no need for complex graph traversals, dynamic programming, or intricate permutation logic. Once the `expected` array is available, the problem boils down to a single pass, O(N) scan, making it remarkably efficient after the initial sort.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Sorting, Counting Sort.

Companies

Asked at: Salesforce.

Height Checker | LeetCode 1051

Easy

Array Sorting Counting Sort

Time: O(n log n)Space: O(n)

Solution Code

Available in:

import java.util.Arrays;

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

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Problem Info

Description

Problem 1051: Height Checker - Detailed explanation and solution approaches.

Topics

Array Sorting Counting Sort

Companies

Salesforce

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1051

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Create a copy of the `heights` array (or sort in place, depending on the language).
2
Step 2: Sort the copied array (this becomes the `expected` array).
3
Step 3: Initialize a counter `count` to 0.
4
Step 4: Iterate through the `heights` array and compare each element `heights[i]` with the corresponding element `expected[i]`.
5
Step 5: If `heights[i]` != `expected[i]`, increment `count`.
6
Step 6: After iterating through the entire array, return the value of `count`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The dominant operation is the sorting of the `heights` array in Step 2, which typically has a time complexity of O(n log n) for efficient algorithms like merge sort or quicksort. Creating a copy of the array adds O(n) space complexity. The comparison loop in Step 4 takes O(n) time, but it's asymptotically less significant than the sorting step. Therefore, the overall time complexity is O(n log n), and the space complexity is O(n) due to the array copy.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Counting Sort

Since the heights are constrained to a relatively small range (1-100), counting sort can be used to sort the array in O(n) time, improving the time complexity to O(n + k), where k is the range of heights (100 in this case).

Time: O(n + k)Space: O(k)

Trade-offs: Counting sort is better when the range of input values is relatively small. It offers linear time complexity but requires extra space proportional to the range of values. For very large ranges, this approach becomes less efficient than comparison-based sorts.

✨

Key Insights

1**Constructing the 'Ground Truth':** The crucial first step is to establish what the perfectly sorted lineup *should* look like. This is done by creating a sorted copy of the input `heights` array. Without this `expected` array, comparing individual `heights[i]` values wouldn't have a meaningful reference point. It explicitly defines the target state for comparison.
2**Preserving Original Order for Comparison:** When sorting, it's essential to create a *new*, sorted array (like `expected`) rather than sorting the `heights` array in-place. This ensures that the original `heights` array, representing the current student arrangement, remains intact. We need both the current (unsorted) and the ideal (sorted) arrangements available simultaneously to perform the direct index-by-index comparison.
3**Direct Index-to-Index Mismatch Tally:** The problem simplifies to a straightforward `if (heights[i] != expected[i]) count++` loop. There's no need for complex graph traversals, dynamic programming, or intricate permutation logic. Once the `expected` array is available, the problem boils down to a single pass, O(N) scan, making it remarkably efficient after the initial sort.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty input array. The solution correctly handles this by implicitly returning 0 when the array is empty.
Edge case 2: Already sorted array. If `heights` is already sorted, the solution correctly returns 0 because no mismatches will be found.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to create a copy of the `heights` array before sorting. Modifying the original array directly will lead to incorrect results.

✗Common mistake 2: Using an inefficient sorting algorithm. The performance will suffer significantly with an O(n^2) sorting algorithm like bubble sort for larger inputs.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Checking the accuracy of a sorted data set. Imagine a system that's supposed to sort student records by ID; this algorithm can quickly check if the sorting was done correctly.

🏦Real-world scenario 2: Quality control in manufacturing. If a machine is supposed to produce items in a specific order (e.g., by size), this algorithm could verify the output.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 30, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Height Checker"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether sorting techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?