【前綴和】B001_LC_元素和小於等於閾值的正方形的最大邊長（PS + 子矩陣 / 二分優化）

一、Problem

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.

Constraints:

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

二、Solution

方法一：前綴和 +求子矩陣和

class Solution {
	public int maxSideLength(int[][] mat, int thre) {
		int R = mat.length, C = mat[0].length, mi = Math.min(R, C), s[][] = new int[R+1][C+1];
		for (int i = 1; i <= R; i++)
		for (int j = 1; j <= C; j++) {
			s[i][j] = mat[i-1][j-1] + s[i-1][j] + s[i][j-1] - s[i-1][j-1];
		}
		int max = 0;
		for (int k = 1; k <= mi; k++)	//枚舉子矩陣的邊長
		for (int i = 1; i <= R; i++)
		for (int j = 1; j <= C; j++) {
			if (i-k < 0 || j-k < 0)
				continue;
			int t = s[i][j] - s[i-k][j] - s[i][j-k] + s[i-k][j-k];
			if (t <= thre && k > max) {
				max = k;
			}
		}
		return max;
	}
}

複雜度分析

• 時間複雜度：$O(min(R, C) × R × C)$，
• 空間複雜度：$O( R × C)$，

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方法二：二分優化

由於邊長 k 越大，子矩陣的總和就越大，所以邊長 k 具備單調性質，可以用二分來優化掉枚舉 $[1, min(R, C)]$ 這一層循環。

class Solution {
	int R, C, mi, s[][];
	boolean ck(int k, int thre) {
		for (int i = 1; i <= R; i++)
		for (int j = 1; j <= C; j++) {
			if (i-k < 0 || j-k < 0)
				continue;
			int t = s[i][j] - s[i-k][j] - s[i][j-k] + s[i-k][j-k];
			if (t <= thre)
				return true;
		}
		return false;
	}
	public int maxSideLength(int[][] mat, int thre) {
		R = mat.length; C = mat[0].length; s = new int[R+1][C+1];
		for (int i = 1; i <= R; i++)
		for (int j = 1; j <= C; j++) {
			s[i][j] = mat[i-1][j-1] + s[i-1][j] + s[i][j-1] - s[i-1][j-1];
		}
		int max = 0;
		int l = 1, r = Math.min(R, C), res = 0;

		while (l <= r) {
			int k = l + r >>> 1;
			if (ck(k, thre)) {l = k + 1; res = k;}
			else 			 r = k - 1;
		}
		return res;
	}
}

複雜度分析

• 時間複雜度：$O(logMin(R, C) × R × C)$，
• 空間複雜度：$O( R × C)$，