LeetCode-Max Sum of Rectangle No Larger Than K

LiBlog發表於2016-07-20
LeetCode

Given a non-empty 2D matrix matrix and an integer k, find the max sum of a rectangle in the matrix such that its sum is no larger than k.

Example:

Given matrix = [
  [1,  0, 1],
  [0, -2, 3]
]
k = 2

The answer is 2. Because the sum of rectangle [[0, 1], [-2, 3]] is 2 and 2 is the max number no larger than k (k = 2).

Note:

  1. The rectangle inside the matrix must have an area > 0.
  2. What if the number of rows is much larger than the number of columns?

Credits:
Special thanks to @fujiaozhu for adding this problem and creating all test cases.

Analysis:

Take the following analysis from https://discuss.leetcode.com/topic/48854/java-binary-search-solution-time-complexity-min-m-n-2-max-m-n-log-max-m-n

 

/* first  consider the situation matrix is 1D
    we can save every sum of 0~i(0<=i<len) and binary search previous sum to find 
    possible result for every index, time complexity is O(NlogN).
    so in 2D matrix, we can sum up all values from row i to row j and create a 1D array 
    to use 1D array solution.
    If col number is less than row number, we can sum up all values from col i to col j 
    then use 1D array solution.
*/


Solution:
 1 public class Solution {
 2     public int maxSumSubmatrix(int[][] matrix, int k) {
 3         if (matrix.length == 0 || matrix[0].length == 0)
 4             return Integer.MIN_VALUE;
 5 
 6         int res = Integer.MIN_VALUE;
 7         int row = matrix.length;
 8         int col = matrix[0].length;
 9         int m = Math.min(row, col);
10         int n = Math.max(row, col);
11         boolean moreCol = (col > row);
12 
13         for (int i = 0; i < m; i++) {
14             int[] sum = new int[n];
15             for (int j = i; j >= 0; j--) {
16                 int val = 0;
17                 TreeSet<Integer> sumSet = new TreeSet<Integer>();
18                 sumSet.add(0);
19                 for (int l = 0; l < n; l++) {
20                     sum[l] = sum[l] + ((moreCol) ? matrix[j][l] : matrix[l][j]);
21                     val += sum[l];
22                     Integer other = sumSet.ceiling(val - k);
23                     if (other != null) {
24                         res = Math.max(res, val - other);
25                     }
26                     sumSet.add(val);
27                 }
28             }
29         }
30 
31         return res;
32     }
33 }

 

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