Saturday, January 25, 2020

Leetcode solution 1292: Maximum Side Length of a Square with Sum Less than or Equal to Threshold

Problem Statement 

Given a m x n matrix mat and an integer threshold. Return the maximum side-length of a square with a sum less than or equal to threshold or return 0 if there is no such square.

Example 1:

Input: mat = [[1,1,3,2,4,3,2],[1,1,3,2,4,3,2],[1,1,3,2,4,3,2]], threshold = 4
Output: 2
Explanation: The maximum side length of square with sum less than 4 is 2 as shown.
Example 2:
Input: mat = [[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2]], threshold = 1
Output: 0
Example 3:
Input: mat = [[1,1,1,1],[1,0,0,0],[1,0,0,0],[1,0,0,0]], threshold = 6
Output: 3
Example 4:
Input: mat = [[18,70],[61,1],[25,85],[14,40],[11,96],[97,96],[63,45]], threshold = 40184
Output: 2

  • 1 <= m, n <= 300
  • m == mat.length
  • n == mat[i].length
  • 0 <= mat[i][j] <= 10000
  • 0 <= threshold <= 10^5
 Problem link

Video Tutorial

You can find the detailed video tutorial here

Thought Process

There is an absolute brute force way is to calculate the sum of each square in every iteration. There are previous similar problems that we can have a prefix sum matrix, similar to the rolling sum idea on one dimensional array.

A prefix sum matrix is a matrix that for each cell[i, j] represents the sum in original matrix from matrix[0, 0] to matrix [i, j] (inclusive)

The advantage is later we can get any rectangle sum simply performing below

sum from [i, j] to [i + len][j + len] = prefixSum[i + len][j + len] - prefixSum[i - 1][j + len] - prefixSum[i + len][j - 1] + prefixSum[i - 1][j - 1]
Two more optimizations we can do
  • We can apply binary search instead of a linear one to find the max length
  • We can also keep a record of the sum while building up the prefix sum matrix. The reason this works is if it has a square say 3x3 that meets the requirement, then it would definitely have a square 2x2 meet the requirement. So we can start from 1x1, 2x2 till we reach the max. Pretty smart! Kudos to @drunkpiano


Prefix sum implementation

Time Complexity: O(M*N*min(M,N)) where M is row N is col, essentially O(N^3)
Space Complexity: O(M*N) since we need


1 comment:

  1. Thanks. It was very helpful. Your diagrams really helped in understanding. I've a little doubt. In that Brute Force Solution, why are returning len+1 instead of len?

    > Thank you for your reply and that's a good catch. You can think of Len as the delta distance, as a matter of fact, I would probably rename length as delta, and star the for loop with
    for (int len = Math.min(row, col) ; len >= 1; len--) -> for (int len = Math.min(row, col) - 1; len >= 1; len--)


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