Question

1738. Find Kth Largest XOR Coordinate Value

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Description

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.

 

Constraints:

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

Solutions

Solution 1: Two-dimensional Prefix XOR + Sorting or Quick Selection

We define a two-dimensional prefix XOR array $s$, where $s[i][j]$ represents the XOR result of the elements in the first $i$ rows and the first $j$ columns of the matrix, i.e.,

$$ s[i][j] = \bigoplus_{0 \leq x \leq i, 0 \leq y \leq j} matrix[x][y] $$

And $s[i][j]$ can be calculated from the three elements $s[i - 1][j]$, $s[i][j - 1]$ and $s[i - 1][j - 1]$, i.e.,

$$ s[i][j] = s[i - 1][j] \oplus s[i][j - 1] \oplus s[i - 1][j - 1] \oplus matrix[i - 1][j - 1] $$

We traverse the matrix, calculate all $s[i][j]$, then sort them, and finally return the $k$th largest element. If you don't want to use sorting, you can also use the quick selection algorithm, which can optimize the time complexity.

The time complexity is $O(m \times n \times \log (m \times n))$ or $O(m \times n)$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the number of rows and columns of the matrix, respectively.

class Solution:
  def kthLargestValue(self, matrix: List[List[int]], k: int) -> int:
    m, n = len(matrix), len(matrix[0])
    s = [[0] * (n + 1) for _ in range(m + 1)]
    ans = []
    for i in range(m):
      for j in range(n):
        s[i + 1][j + 1] = s[i + 1][j] ^ s[i][j + 1] ^ s[i][j] ^ matrix[i][j]
        ans.append(s[i + 1][j + 1])
    return nlargest(k, ans)[-1]

class Solution {
  public int kthLargestValue(int[][] matrix, int k) {
    int m = matrix.length, n = matrix[0].length;
    int[][] s = new int[m + 1][n + 1];
    List<Integer> ans = new ArrayList<>();
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        s[i + 1][j + 1] = s[i + 1][j] ^ s[i][j + 1] ^ s[i][j] ^ matrix[i][j];
        ans.add(s[i + 1][j + 1]);
      }
    }
    Collections.sort(ans);
    return ans.get(ans.size() - k);
  }
}

class Solution {
public:
  int kthLargestValue(vector<vector<int>>& matrix, int k) {
    int m = matrix.size(), n = matrix[0].size();
    vector<vector<int>> s(m + 1, vector<int>(n + 1));
    vector<int> ans;
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        s[i + 1][j + 1] = s[i + 1][j] ^ s[i][j + 1] ^ s[i][j] ^ matrix[i][j];
        ans.push_back(s[i + 1][j + 1]);
      }
    }
    sort(ans.begin(), ans.end());
    return ans[ans.size() - k];
  }
};

func kthLargestValue(matrix [][]int, k int) int {
  m, n := len(matrix), len(matrix[0])
  s := make([][]int, m+1)
  for i := range s {
    s[i] = make([]int, n+1)
  }
  var ans []int
  for i := 0; i < m; i++ {
    for j := 0; j < n; j++ {
      s[i+1][j+1] = s[i+1][j] ^ s[i][j+1] ^ s[i][j] ^ matrix[i][j]
      ans = append(ans, s[i+1][j+1])
    }
  }
  sort.Ints(ans)
  return ans[len(ans)-k]
}

function kthLargestValue(matrix: number[][], k: number): number {
  const m: number = matrix.length;
  const n: number = matrix[0].length;
  const s = Array.from({ length: m + 1 }, () => Array.from({ length: n + 1 }, () => 0));
  const ans: number[] = [];
  for (let i = 0; i < m; ++i) {
    for (let j = 0; j < n; ++j) {
      s[i + 1][j + 1] = s[i + 1][j] ^ s[i][j + 1] ^ s[i][j] ^ matrix[i][j];
      ans.push(s[i + 1][j + 1]);
    }
  }
  ans.sort((a, b) => b - a);
  return ans[k - 1];
}