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In this video, we solve LeetCode 1292: Maximum Side Length of a Square with Sum Less than or Equal to Threshold. Code: https://leetcode.com/problems/maximum... Upsolve Leetcode Contest: • Leetcode Contests Greedy & Heaps: • Greedy & Heaps Two pointers: • Two pointers Sliding Window: • Sliding Window Maths & Geometry: • Maths & Geometry Stack: • Stack Set & Map: • Set & Map Bit manipulation: • Bit Manipulation Backtracking: • Backtracking Linked List: • Linked List Binary Search: • Binary Search Graph: • Graph Dynamic Progamming: • Dynamic Programming You are given a matrix and a threshold. The task is to find the maximum possible side length of a square submatrix whose sum is less than or equal to the given threshold. I explain this problem using THREE progressive approaches, starting from brute force and ending with the optimal solution. -------------------------------------------------- 1️⃣ Brute Force Approach -------------------------------------------------- We try all possible square sizes. For every cell, we treat it as the top-left corner. We manually calculate the sum of each square. If the sum is less than or equal to the threshold, we update the answer. This approach is easy to understand but very slow due to repeated sum calculations. Time Complexity: O(n³ × m) -------------------------------------------------- 2️⃣ Optimized Approach -------------------------------------------------- In the optimized approach, we improve the brute force solution in TWO ways: 🔹 Row-wise Prefix Sum: We build prefix sums for each row. This allows faster calculation of row sums inside a square. 🔹 Search Optimization: Instead of checking all square sizes blindly, we optimize the search for the maximum valid square size. This significantly reduces unnecessary computations. Time Complexity: ~O(n³) -------------------------------------------------- 3️⃣ Optimal Approach (Full 2D Prefix Sum) -------------------------------------------------- This is the most efficient solution. We build a 2D prefix sum array. prefix[i][j] represents the sum of the complete top area and left area up to (i-1, j-1). Using this, we can compute the sum of any square in O(1) time. Formula used: sum = prefix[x2][y2] - prefix[x1][y2] - prefix[x2][y1] + prefix[x1][y1] This allows us to efficiently find the maximum square size. Time Complexity: O(n × m) Space Complexity: O(n × m) -------------------------------------------------- LeetCode Problem Link: https://leetcode.com/problems/maximum... Timestamps: 00:00 Introduction 00:50 Problem statement 01:35 Brute Force 07:30 Optimized Approach 25:55 Optimal Approach #leetcode #leetcode1292 #dailyleetcode #prefixsum #2dprefixsum #rowsum #binarysearch #optimization #bruteforce #optimal_solution #matrix #submatrix #squarematrix #datastructures #algorithms #dsa #competitiveprogramming #codinginterview #java #faang #google #amazon #microsoft #placementprep #studyplacement #problem_solving #codingpractice