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Kth highest XOR of diagonal elements from a Matrix

  • Last Updated : 19 Jul, 2021

Given a square matrix mat[][] of size N * N, the task is to calculate XOR value of every diagonal elements and find the Kth maximum XOR value obtained. 
Note: There are 2 * N – 1 diagonals in the matrix. Starting point of ith diagonal is (min(N, i), (1 + max(i – N, 0))).

Examples:

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Input: mat[][] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, K = 3
Output: 6
Explanation:  
XOR of 1st diagonal = 1.
XOR of 2nd diagonal = 4 ^ 2 = 6.
XOR of 3rd diagonal = 7 ^ 5 ^ 3 = 1.
XOR of 4th diagonal = 8 ^ 6 = 14.
XOR of 5th diagonal = 9.



Input: mat[][] = {{1, 2}, {4, 5}}, K = 2
Output: 6
Explanation: 
XOR of 1st diagonal =1.
XOR of 2nd diagonal = 4 ^ 2 = 6.
XOR of 3rd diagonal = 5.

Approach: Follow the steps below to solve the problem

  • Traverse the matrix diagonally.
  • For every ith diagonal, starting point is (min(N, i), (1 + max(i – N, 0))).
  • Store XOR of each diagonal in a vector.
  • Sort the vector.
  • Print the Kth maximum value obtained.

Below is the implementation of the above approach:

C++




// C++ Program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find K-th maximum XOR
// of any diagonal in the matrix
void findXOR(vector<vector<int> > mat, int K)
{
    // Number or rows
    int N = mat.size();
 
    // Number of columns
    int M = mat[0].size();
 
    // Store XOR of diagonals
    vector<int> digXOR;
 
    // Traverse each diagonal
    for (int l = 1; l <= (N + M - 1); l++) {
 
        // Starting column of diagonal
        int s_col = max(0, l - N);
 
        // Count total elements in the diagonal
        int count = min({ l, (M - s_col), N });
 
        // Store XOR of current diagonal
        int currXOR = 0;
 
        for (int j = 0; j < count; j++) {
            currXOR
                = (currXOR
                   ^ mat[min(N, l) - j - 1][s_col + j]);
        }
 
        // Push XOR of current diagonal
        digXOR.push_back(currXOR);
    }
 
    // Sort XOR values of diagonals
    sort(digXOR.begin(), digXOR.end());
 
    // Print the K-th Maximum XOR
    cout << digXOR[N + M - 1 - K];
}
 
// Driver Code
int main()
{
    vector<vector<int> > mat
        = { { 1, 2, 3 },
            { 4, 5, 6 },
            { 7, 8, 9 } };
 
    int K = 3;
 
    findXOR(mat, K);
    return 0;
}

Java




// Java Program to implement
// the above approach
import java.util.*;
class GFG
{
 
// Function to find K-th maximum XOR
// of any diagonal in the matrix
static void findXOR(int[][] mat, int K)
{
   
    // Number or rows
    int N = mat.length;
 
    // Number of columns
    int M = mat[0].length;
 
    // Store XOR of diagonals
    ArrayList<Integer> digXOR
            = new ArrayList<Integer>();
 
    // Traverse each diagonal
    for (int l = 1; l <= (N + M - 1); l++) {
 
        // Starting column of diagonal
        int s_col = Math.max(0, l - N);
 
        // Count total elements in the diagonal
        int count = Math.min( l, Math.min((M - s_col), N));
 
        // Store XOR of current diagonal
        int currXOR = 0;
 
        for (int j = 0; j < count; j++) {
            currXOR
                = (currXOR
                   ^ mat[Math.min(N, l) - j - 1][s_col + j]);
        }
 
        // Push XOR of current diagonal
        digXOR.add(currXOR);
    }
 
    // Sort XOR values of diagonals
    Collections.sort(digXOR);
 
    // Print the K-th Maximum XOR
    System.out.print(digXOR.get(N + M - 1 - K));
}
 
// Driver Code
public static void main(String[] args)
{
    int[][] mat
        = { { 1, 2, 3 },
            { 4, 5, 6 },
            { 7, 8, 9 } };
 
    int K = 3;
 
    findXOR(mat, K);
}
}
 
// This code is contributed by code_hunt.

Python3




# Python3 Program to implement
# the above approach
 
# Function to find K-th maximum XOR
# of any diagonal in the matrix
def findXOR(mat, K):
   
    # Number or rows
    N = len(mat)
 
    # Number of columns
    M = len(mat[0])
 
    # Store XOR of diagonals
    digXOR = []
 
    # Traverse each diagonal
    for l in range(1, N + M, 1):
       
        # Starting column of diagonal
        s_col = max(0, l - N)
 
        # Count total elements in the diagonal
        count = min([l, (M - s_col), N])
 
        # Store XOR of current diagonal
        currXOR = 0
        for j in range(count):
            currXOR = (currXOR ^ mat[min(N, l) - j - 1][s_col + j])
 
        # Push XOR of current diagonal
        digXOR.append(currXOR)
 
    # Sort XOR values of diagonals
    digXOR.sort(reverse=False)
 
    # Print the K-th Maximum XOR
    print(digXOR[N + M - 1 - K])
 
# Driver Code
if __name__ == '__main__':
    mat = [[1, 2, 3],
           [4, 5, 6],
           [7, 8, 9]]
    K = 3
    findXOR(mat, K)
 
    # This code is contributed by SURENDRA_GANGWAR.

C#




// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find K-th maximum XOR
// of any diagonal in the matrix
static void findXOR(int[,]mat, int K)
{
     
    // Number or rows
    int N = mat.GetLength(0);
 
    // Number of columns
    int M = mat.GetLength(1);
 
    // Store XOR of diagonals
    List<int> digXOR = new List<int>();
 
    // Traverse each diagonal
    for(int l = 1; l <= (N + M - 1); l++)
    {
         
        // Starting column of diagonal
        int s_col = Math.Max(0, l - N);
 
        // Count total elements in the diagonal
        int count = Math.Min(l, Math.Min((M - s_col), N));
 
        // Store XOR of current diagonal
        int currXOR = 0;
 
        for(int j = 0; j < count; j++)
        {
            currXOR = (currXOR ^ mat[Math.Min(N, l) - j - 1,
                                              s_col + j]);
        }
 
        // Push XOR of current diagonal
        digXOR.Add(currXOR);
    }
 
    // Sort XOR values of diagonals
    digXOR.Sort();
 
    // Print the K-th Maximum XOR
    Console.Write(digXOR[N + M - 1 - K]);
}
 
// Driver Code
public static void Main(String[] args)
{
    int[,] mat = { { 1, 2, 3 },
                   { 4, 5, 6 },
                   { 7, 8, 9 } };
 
    int K = 3;
 
    findXOR(mat, K);
}
}
 
// This code is contributed by shikhasingrajput

Javascript




<script>
// Javascript Program to implement
// the above approach
 
// Function to find K-th maximum XOR
// of any diagonal in the matrix
function findXOR(mat, K)
{
 
    // Number or rows
    let N = mat.length;
 
    // Number of columns
    let M = mat[0].length;
 
    // Store XOR of diagonals
    let digXOR = [];
 
    // Traverse each diagonal
    for (let l = 1; l <= (N + M - 1); l++) {
 
        // Starting column of diagonal
        let s_col = Math.max(0, l - N);
 
        // Count total elements in the diagonal
        let count = Math.min(l, (M - s_col), N);
 
        // Store XOR of current diagonal
        let currXOR = 0;
 
        for (let j = 0; j < count; j++) {
            currXOR
                = (currXOR
                    ^ mat[Math.min(N, l) - j - 1][s_col + j]);
        }
 
        // Push XOR of current diagonal
        digXOR.push(currXOR);
    }
 
    // Sort XOR values of diagonals
    digXOR.sort((a, b) => a - b);
 
    // Print the K-th Maximum XOR
    document.write(digXOR[N + M - 1 - K]);
}
 
// Driver Code
let mat
    = [[1, 2, 3],
    [4, 5, 6],
    [7, 8, 9]];
 
let K = 3;
findXOR(mat, K);
 
// This code is contributed by gfgking.
</script>
Output: 
6

 

Time Complexity: O(N * M)
Space Complexity: O(N * M)




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