What is the brute force approach and complexity for Squares of a Sorted Array?

Square every number, then sort the result with time complexity O(N log N) and space complexity O(N).

What is the optimal approach and complexity for Squares of a Sorted Array?

Bigger square of the two ends goes into the last empty slot with optimal time complexity O(N) and space complexity O(N).

How do you identify the pattern for Squares of a Sorted Array?

Look for key signals in the prompt: "sorted array (with negatives)" indicates The extreme values sit at the two ends → look at both ends.; "squares" indicates Big negative and big positive both give big squares, so the two ends compete for "biggest".; "sorted output, O(n)" indicates No sorting allowed → build the output in order, one end at a time..

Squares of a Sorted Array

Easy·LeetCode #977 Pro OnlyMedium

LeetCode

Problem Statement

You are given an integer array nums sorted from smallest to largest. It can contain negative numbers.

Return a new array with the square of every number, also sorted from smallest to largest.

Follow-up: squaring and then sorting is easy. Can you do it in O(n) time?

Input

An integer array nums, sorted in non-decreasing order.

Output

An array of the squares of nums, sorted in non-decreasing order.

Example 1

Input: nums = [-4, -1, 0, 3, 10]

Output: [0, 1, 9, 16, 100]

The squares are [16, 1, 0, 9, 100]. Sorted, that's [0, 1, 9, 16, 100].

Example 2

Input: nums = [-7, -3, 2, 3, 11]

Output: [4, 9, 9, 49, 121]

The squares are [49, 9, 4, 9, 121]. Sorted, that's [4, 9, 9, 49, 121].

Constraints

• 1 ≤ nums.length ≤ 10^4
• −10^4 ≤ nums[i] ≤ 10^4
• nums is sorted in non-decreasing order

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“sorted array (with negatives)”

→ The extreme values sit at the two ends → look at both ends.

“squares”

→ Big negative and big positive both give big squares, so the two ends compete for "biggest".

“sorted output, O(n)”

→ No sorting allowed → build the output in order, one end at a time.

Approach Complexity Ladder

Brute Force — square, then sortO(N log N)O(N)

Square every number, then sort the result

Optimal — two ends compete, fill from the backO(N)O(N)

Bigger square of the two ends goes into the last empty slot

Brute Force Intuition

Square each number into a new array, then sort it.

Recognition rule for this sub-pattern:

Start at both ends of a sorted (or symmetric) structure. Each step, move the side that cannot do better.

Remember this · Two Pointers

Trigger

Sorted array, pairs/triplets with a target, comparing from both ends, compacting an array in place.

Think

If I compare the elements at two positions, can I safely discard one side?

Key question

Which pointer moves, and why can the skipped elements never be the answer?

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