LeetCode #1725 — EASY

Number Of Rectangles That Can Form The Largest Square

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

The Problem

Problem Statement

You are given an array rectangles where rectangles[i] = [l_i, w_i] represents the i^th rectangle of length l_i and width w_i.

You can cut the i^th rectangle to form a square with a side length of k if both k <= l_i and k <= w_i. For example, if you have a rectangle [4,6], you can cut it to get a square with a side length of at most 4.

Let maxLen be the side length of the largest square you can obtain from any of the given rectangles.

Return the number of rectangles that can make a square with a side length of maxLen.

Example 1:

Input: rectangles = [[5,8],[3,9],[5,12],[16,5]]
Output: 3
Explanation: The largest squares you can get from each rectangle are of lengths [5,3,5,5].
The largest possible square is of length 5, and you can get it out of 3 rectangles.

Example 2:

Input: rectangles = [[2,3],[3,7],[4,3],[3,7]]
Output: 3

Constraints:

1 <= rectangles.length <= 1000
rectangles[i].length == 2
1 <= l_i, w_i <= 10⁹
l_i != w_i

Step 01

Brute Force Baseline

Problem summary: You are given an array rectangles where rectangles[i] = [li, wi] represents the ith rectangle of length li and width wi. You can cut the ith rectangle to form a square with a side length of k if both k <= li and k <= wi. For example, if you have a rectangle [4,6], you can cut it to get a square with a side length of at most 4. Let maxLen be the side length of the largest square you can obtain from any of the given rectangles. Return the number of rectangles that can make a square with a side length of maxLen.

Baseline thinking

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

Pattern signal: Array

Example 1

[[5,8],[3,9],[5,12],[16,5]]

Example 2

[[2,3],[3,7],[4,3],[3,7]]

Step 02

Core Insight

What unlocks the optimal approach

What is the length of the largest square the can be cut out of some rectangle? It'll be equal to min(rectangle.length, rectangle.width). Replace each rectangle with this value.
Calculate maxSize by iterating over the given rectangles and maximizing the answer with their values denoted in the first hint.
Then iterate again on the rectangles and calculate the number whose values = maxSize.

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 #1725: Number Of Rectangles That Can Form The Largest Square
class Solution {
    public int countGoodRectangles(int[][] rectangles) {
        int ans = 0, mx = 0;
        for (var e : rectangles) {
            int x = Math.min(e[0], e[1]);
            if (mx < x) {
                mx = x;
                ans = 1;
            } else if (mx == x) {
                ++ans;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1725: Number Of Rectangles That Can Form The Largest Square
func countGoodRectangles(rectangles [][]int) (ans int) {
	mx := 0
	for _, e := range rectangles {
		x := min(e[0], e[1])
		if mx < x {
			mx = x
			ans = 1
		} else if mx == x {
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #1725: Number Of Rectangles That Can Form The Largest Square
class Solution:
    def countGoodRectangles(self, rectangles: List[List[int]]) -> int:
        ans = mx = 0
        for l, w in rectangles:
            x = min(l, w)
            if mx < x:
                ans = 1
                mx = x
            elif mx == x:
                ans += 1
        return ans

// Accepted solution for LeetCode #1725: Number Of Rectangles That Can Form The Largest Square
struct Solution;

use std::collections::HashMap;

impl Solution {
    fn count_good_rectangles(rectangles: Vec<Vec<i32>>) -> i32 {
        let mut count: HashMap<i32, usize> = HashMap::new();
        let mut max = 0;
        for rect in rectangles {
            let min = rect[0].min(rect[1]);
            *count.entry(min).or_default() += 1;
            max = max.max(min);
        }
        count[&max] as i32
    }
}

#[test]
fn test() {
    let rectangles = vec_vec_i32![[5, 8], [3, 9], [5, 12], [16, 5]];
    let res = 3;
    assert_eq!(Solution::count_good_rectangles(rectangles), res);
    let rectangles = vec_vec_i32![[2, 3], [3, 7], [4, 3], [3, 7]];
    let res = 3;
    assert_eq!(Solution::count_good_rectangles(rectangles), res);
}

// Accepted solution for LeetCode #1725: Number Of Rectangles That Can Form The Largest Square
function countGoodRectangles(rectangles: number[][]): number {
    let [ans, mx] = [0, 0];
    for (const [l, w] of rectangles) {
        const x = Math.min(l, w);
        if (mx < x) {
            mx = x;
            ans = 1;
        } else if (mx === x) {
            ++ans;
        }
    }
    return ans;
}

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)

Space

O(1)

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.