The value of coordinate (a, b) of the matrix is the XOR of all matrix[i][j] where 0 <= i <= a < m and 0 <= j <= b < n (0-indexed).

Find the kth largest value (1-indexed) of all the coordinates of matrix.

Example 1:

Input: matrix = [[5,2],[1,6]], k = 1
Output: 7
Explanation: The value of coordinate (0,1) is 5 XOR 2 = 7, which is the largest value.

Example 2:

Input: matrix = [[5,2],[1,6]], k = 2
Output: 5
Explanation: The value of coordinate (0,0) is 5 = 5, which is the 2nd largest value.

Example 3:

Input: matrix = [[5,2],[1,6]], k = 3
Output: 4
Explanation: The value of coordinate (1,0) is 5 XOR 1 = 4, which is the 3rd largest value.

Example 4:

Input: matrix = [[5,2],[1,6]], k = 4
Output: 0
Explanation: The value of coordinate (1,1) is 5 XOR 2 XOR 1 XOR 6 = 0, which is the 4th largest value.

Constraints:

• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 1000
• 0 <= matrix[i][j] <= 106
• 1 <= k <= m * n

Solution: DP

Similar to 花花酱 LeetCode 304. Range Sum Query 2D – Immutable

xor[i][j] = matrix[i][j] ^ xor[i – 1][j – 1] ^ xor[i – 1][j] ^ xor[i][j- 1]

Time complexity: O(mn)
Space complexity: O(mn)

// Author: Huahua
class Solution {
public:
  int kthLargestValue(vector<vector<int>>& matrix, int k) {    
    const int m = matrix.size(), n = matrix[0].size();
    vector<int> v;
    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        v.push_back(matrix[i][j] 
                      ^= (i ? matrix[i - 1][j] : 0) 
                       ^ (j ? matrix[i][j - 1] : 0) 
                       ^ (i * j ? matrix[i - 1][j - 1] : 0));
    nth_element(begin(v), begin(v) + k - 1, end(v), greater<int>());    
    return v[k - 1];
  }
};

The post 花花酱 LeetCode 1738. Find Kth Largest XOR Coordinate Value appeared first on Huahua's Tech Road.