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1738. Find Kth Largest XOR Coordinate Value
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.lengthn == matrix[i].length1 <= m, n <= 10000 <= matrix[i][j] <= 1061 <= 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.
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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]; } }; -
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] -
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]; }