1738

1738. Find Kth Largest XOR Coordinate Value (Medium)

You are given a 2D matrix of size m x n, consisting of non-negative integers. You are also given an integer k.

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

Related Topics:
Array

Solution 1.

// OJ: https://leetcode.com/problems/find-kth-largest-xor-coordinate-value/
// Author: github.com/lzl124631x
// Time: O(MNlog(MN))
// Space: O(MN)
class Solution {
public:
    int kthLargestValue(vector<vector<int>>& A, int k) {
        vector<int> v;
        int M = A.size(), N = A[0].size();
        for (int i = 0; i < M; ++i) {
            int val = 0;
            for (int j = 0; j < N; ++j) {
                val ^= A[i][j];
                A[i][j] = val;
                if (i - 1 >= 0) A[i][j] ^= A[i - 1][j];
                v.push_back(A[i][j]);
            }
        }
        sort(begin(v), end(v), greater<>());
        return v[k - 1];
    }
};

Or use Heap

// OJ: https://leetcode.com/problems/find-kth-largest-xor-coordinate-value/
// Author: github.com/lzl124631x
// Time: O(MNlog(K))
// Space: O(K)
class Solution {
public:
    int kthLargestValue(vector<vector<int>>& A, int k) {
        priority_queue<int, vector<int>, greater<>> pq;
        int M= A.size(), N = A[0].size();
        for (int i = 0; i < M; ++i) {
            int val = 0;
            for (int j = 0; j < N; ++j) {
                val ^= A[i][j];
                A[i][j] = val;
                if (i - 1 >= 0)  A[i][j] ^= A[i - 1][j];
                if (pq.size() < k || A[i][j] > pq.top()) pq.push(A[i][j]);
                if (pq.size() > k) pq.pop();
            }
        }
        return pq.top();
    }
};