What is the time complexity of Squares of a Sorted Array?

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

What data structures are used to solve Squares of a Sorted Array?

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

What are the key insights for solving Squares of a Sorted Array?

Insight 1: Since the input array is sorted, the largest squares will be at either the beginning or the end of the array. Negative numbers with large absolute values become large positive squares. Insight 2: A two-pointer approach can efficiently determine the largest square at each step without needing to explicitly sort the squared array afterwards.

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Squares of a Sorted Array - LeetCode 977 Solution

Q: How to solve Squares of a Sorted Array?

The solution employs a two-pointer approach. One pointer starts at the beginning of the input array (`left`), and the other starts at the end (`right`). We compare the absolute values of the numbers pointed to by these pointers. The number with the larger absolute value will have a larger square. This square is then placed at the end of the `result` array, and the corresponding pointer is moved inwards. This process continues until the `left` and `right` pointers meet or cross each other.

Q: Which companies ask LeetCode 977. Squares of a Sorted Array in interviews?

This problem is frequently asked at Agoda, CrowdStrike, Deutsche Bank, Instacart, Ozon and 4 more companies.

Squares of a Sorted Array - Complete Solution Guide

Squares of a Sorted Array is LeetCode problem 977, 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

Given an integer array nums sorted in non-decreasing order, return an array of the squares of each number sorted in non-decreasing order . Example 1: Input: nums = [-4,-1,0,3,10] Output: [0,1,9,16,100] Explanation: After squaring, the array becomes [16,1,0,9,100]. After sorting, it becomes [0,1,9,16,100]. Example 2: Input: nums = [-7,-3,2,3,11] Output: [4,9,9,49,121] Constraints: 1 <= nums.length <= 10 4 -10 4 <= nums[i] <= 10 4 nums is sorted in non-decreasing order. Follow up: Squaring each el

Detailed Explanation

The problem asks us to take a sorted array of integers (in non-decreasing order) and return a new array containing the squares of each number, also sorted in non-decreasing order. The input is a list of integers, `nums`, and the output is a list of the squares of these integers, sorted. The key constraint is that the original input array is already sorted. This allows for an optimized O(n) solution, rather than the trivial O(n log n) solution of simply squaring each element and then sorting the resulting array.

Solution Approach

The solution employs a two-pointer approach. One pointer starts at the beginning of the input array (`left`), and the other starts at the end (`right`). We compare the absolute values of the numbers pointed to by these pointers. The number with the larger absolute value will have a larger square. This square is then placed at the end of the `result` array, and the corresponding pointer is moved inwards. This process continues until the `left` and `right` pointers meet or cross each other.

Step-by-Step Algorithm

Step 1: Initialize an empty result array of the same size as the input array, filled with default values (e.g., 0 in Python/Java/C++).
Step 2: Initialize two pointers, `left` to 0 and `right` to `n-1`, where `n` is the length of the array.
Step 3: Initialize an index pointer, `index`, also to `n-1`. This pointer will traverse the `result` array from right to left.
Step 4: While `left <= right` (meaning the pointers haven't crossed):
Step 5: Compare the absolute values of `nums[left]` and `nums[right]`.
Step 6: If `abs(nums[left]) > abs(nums[right])`, then `nums[left]` squared is larger. Place `nums[left] * nums[left]` into `result[index]`, increment `left`, and decrement `index`.
Step 7: Otherwise, `nums[right]` squared is larger or equal. Place `nums[right] * nums[right]` into `result[index]`, decrement `right`, and decrement `index`.
Step 8: After the loop completes, the `result` array will contain the sorted squares of the numbers from the input array. Return the `result` array.

Key Insights

Insight 1: Since the input array is sorted, the largest squares will be at either the beginning or the end of the array. Negative numbers with large absolute values become large positive squares.
Insight 2: A two-pointer approach can efficiently determine the largest square at each step without needing to explicitly sort the squared array afterwards.
Insight 3: The squared values need to be inserted into a new array from right to left, ensuring the result remains sorted in non-decreasing order.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Two Pointers, Sorting.

Companies

Asked at: Agoda, CrowdStrike, Deutsche Bank, Instacart, Ozon, PayPal, Tinkoff, Walmart Labs, Whatnot.

Squares of a Sorted Array | LeetCode 977

Easy

Array Two Pointers Sorting

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

Solution Code

Available in:

class Solution {
    public int[] sortedSquares(int[] nums) {
        int n = nums.length;
        int[] result = new int[n];
        int left = 0;
        int right = n - 1;
        int index = n - 1;

        while (left <= right) {
            int leftSquare = nums[left] * nums[left];
            int rightSquare = nums[right] * nums[right];

            if (leftSquare > rightSquare) {
                result[index] = leftSquare;
                left++;
            } else {
                result[index] = rightSquare;
                right--;
            }
            index--;
        }

        return result;
    }
}

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Problem Info

Description

Problem 977: Squares of a Sorted Array - Detailed explanation and solution approaches.

Topics

Array Two Pointers Sorting

Companies

AgodaCrowdStrikeDeutsche BankInstacartOzonPayPalTinkoffWalmart Labs

Stats

Difficulty:Easy

Languages:4

Companies:9

Problem #:977

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

8 steps

1
Step 1: Initialize an empty result array of the same size as the input array, filled with default values (e.g., 0 in Python/Java/C++).
2
Step 2: Initialize two pointers, `left` to 0 and `right` to `n-1`, where `n` is the length of the array.
3
Step 3: Initialize an index pointer, `index`, also to `n-1`. This pointer will traverse the `result` array from right to left.
4
Step 4: While `left <= right` (meaning the pointers haven't crossed):
5
Step 5: Compare the absolute values of `nums[left]` and `nums[right]`.
6
Step 6: If `abs(nums[left]) > abs(nums[right])`, then `nums[left]` squared is larger. Place `nums[left] * nums[left]` into `result[index]`, increment `left`, and decrement `index`.
7
Step 7: Otherwise, `nums[right]` squared is larger or equal. Place `nums[right] * nums[right]` into `result[index]`, decrement `right`, and decrement `index`.
8
Step 8: After the loop completes, the `result` array will contain the sorted squares of the numbers from the input array. Return the `result` array.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because we iterate through the input array once using two pointers. Each comparison and assignment takes constant time. The space complexity is O(n) because we create a new array `result` of the same size as the input array to store the squared and sorted values. This new array is necessary to fulfill the problem requirements and to store the output.

📖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

Naive Approach: Squaring and Sorting

This involves squaring each element of the array and then using a standard sorting algorithm (e.g., merge sort, quicksort) to sort the squared array.

Time: O(n log n)Space: O(1) or O(n), depending on the sorting algorithm used. In-place sorts have O(1) space, while some sorts might create a copy of the array, thus O(n)

Trade-offs: Simpler to implement, but less efficient than the two-pointer approach, especially for large input arrays. Use only if you don't need the performance optimization, or the input sizes are very small.

✨

Key Insights

1Insight 1: Since the input array is sorted, the largest squares will be at either the beginning or the end of the array. Negative numbers with large absolute values become large positive squares.
2Insight 2: A two-pointer approach can efficiently determine the largest square at each step without needing to explicitly sort the squared array afterwards.
3Insight 3: The squared values need to be inserted into a new array from right to left, ensuring the result remains sorted in non-decreasing order.

📚

Key Terms Explained

For Students

Array

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

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Sorting

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

Edge Cases

Edge case 1: Array with all negative numbers: The algorithm correctly handles this by iterating from both ends and placing the squares in the correct order. For example, `nums = [-5, -4, -3, -2, -1]` will be correctly processed.
Edge case 2: Array with all positive numbers: Similarly, the algorithm handles this by iterating from both ends. For example, `nums = [1, 2, 3, 4, 5]` will be correctly processed.
Edge case 3: Array with mixed positive and negative numbers, including zero: For example, `nums = [-3, -2, -1, 0, 1, 2, 3]`. The algorithm correctly places the square of zero in its correct position, and correctly handles the other numbers as well.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not realizing the input array is sorted and attempting a more complex sorting algorithm. The problem explicitly states the array is sorted, which is a hint for an optimized solution.

✗Common mistake 2: Not handling the edge case of negative numbers correctly. Negative numbers, when squared, become positive and can be larger than the squares of positive numbers in the array.

✗Common mistake 3: Forgetting to create a new array to store the results and trying to modify the original array in place. The problem specifies returning a *new* array.

✅

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: Data analysis and preprocessing: When analyzing datasets containing sorted numerical values, squaring the values and sorting them can be a common preprocessing step for certain machine learning algorithms or statistical analyses.

🏦Real-world scenario 2: Image processing: Calculating the magnitude of gradients in images often involves squaring values. If the gradient values are sorted or can be sorted efficiently, this algorithm can be applied for optimization.

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ 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: August 17, 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

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

@InterviewProJul 14

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

@CodeNinjaJul 7

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 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.

How do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

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 "Squares of a Sorted Array"?

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 two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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