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Program for addition of two matrices
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Mirror of matrix across diagonal

Last Updated : 12 Feb, 2025
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Given a 2-D array of order N x N, print a matrix that is the mirror of the given tree across the diagonal. We need to print the result in a way: swap the values of the triangle above the diagonal with the values of the triangle below it like a mirror image swap. Print the 2-D array obtained in a matrix layout.

Examples:  

Input : int mat[][] = {{1 2 4 }
                       {5 9 0}
                       { 3 1 7}}
Output :  1 5 3 
          2 9 1
          4 0 7

Input : mat[][] = {{1  2  3  4 }
                   {5  6  7  8 }
                   {9  10 11 12}
                   {13 14 15 16} }
Output : 1 5 9 13 
         2 6 10 14  
         3 7 11 15 
         4 8 12 16

A simple solution to this problem involves extra space. We traverse all right diagonal (right-to-left) one by one. During the traversal of the diagonal, first, we push all the elements into the stack and then we traverse it again and replace every element of the diagonal with the stack element. 

Below is the implementation of the above idea. 

C++ 
// Simple CPP program to find mirror of // matrix across diagonal. #include <bits/stdc++.h> using namespace std;  const int MAX = 100;  void imageSwap(vector<vector<int>> &mat, int n) {     // for diagonal which start from at      // first row of matrix     int row = 0;      // traverse all top right diagonal     for (int j = 0; j < n; j++) {          // here we use stack for reversing         // the element of diagonal         stack<int> s;         int i = row, k = j;         while (i < n && k >= 0)              s.push(mat[i++][k--]);                  // push all element back to matrix          // in reverse order         i = row, k = j;         while (i < n && k >= 0) {             mat[i++][k--] = s.top();             s.pop();         }     }      // do the same process for all the     // diagonal which start from last     // column     int column = n - 1;     for (int j = 1; j < n; j++) {          // here we use stack for reversing          // the elements of diagonal         stack<int> s;         int i = j, k = column;         while (i < n && k >= 0)              s.push(mat[i++][k--]);                  // push all element back to matrix          // in reverse order         i = j;         k = column;         while (i < n && k >= 0) {             mat[i++][k--] = s.top();             s.pop();         }     } }  // Utility function to print a matrix void printMatrix(vector<vector<int>> &mat, int n) {     for (int i = 0; i < n; i++) {         for (int j = 0; j < n; j++)             cout << mat[i][j] << " ";         cout << endl;     } }  // driver program to test above function int main() {     vector<vector<int>> mat = { { 1, 2, 3, 4 },                                { 5, 6, 7, 8 },                                { 9, 10, 11, 12 },                                { 13, 14, 15, 16 } };     int n = 4;     imageSwap(mat, n);     printMatrix(mat, n);     return 0; }
Java 
// Simple Java program to find mirror of // matrix across diagonal.  import java.util.*;  class GFG  {      static int MAX = 100;      static void imageSwap(int mat[][], int n)      {         // for diagonal which start from at          // first row of matrix         int row = 0;          // traverse all top right diagonal         for (int j = 0; j < n; j++)          {              // here we use stack for reversing             // the element of diagonal             Stack<Integer> s = new Stack<>();             int i = row, k = j;             while (i < n && k >= 0)             {                 s.push(mat[i++][k--]);             }              // push all element back to matrix              // in reverse order             i = row;             k = j;             while (i < n && k >= 0)             {                 mat[i++][k--] = s.peek();                 s.pop();             }         }          // do the same process for all the         // diagonal which start from last         // column         int column = n - 1;         for (int j = 1; j < n; j++)         {              // here we use stack for reversing              // the elements of diagonal             Stack<Integer> s = new Stack<>();             int i = j, k = column;             while (i < n && k >= 0)             {                 s.push(mat[i++][k--]);             }              // push all element back to matrix              // in reverse order             i = j;             k = column;             while (i < n && k >= 0)             {                 mat[i++][k--] = s.peek();                 s.pop();             }         }     }      // Utility function to print a matrix     static void printMatrix(int mat[][], int n)      {         for (int i = 0; i < n; i++)          {             for (int j = 0; j < n; j++)             {                 System.out.print(mat[i][j] + " ");             }             System.out.println("");         }     }      // Driver program to test above function     public static void main(String[] args)     {          int mat[][] = {{1, 2, 3, 4},         {5, 6, 7, 8},         {9, 10, 11, 12},         {13, 14, 15, 16}};         int n = 4;         imageSwap(mat, n);         printMatrix(mat, n);     } }  // This code contributed by Rajput-Ji
Python 
# Simple Python3 program to find mirror of # matrix across diagonal. MAX = 100  def imageSwap(mat, n):          # for diagonal which start from at      # first row of matrix     row = 0          # traverse all top right diagonal     for j in range(n):                  # here we use stack for reversing         # the element of diagonal         s = []         i = row         k = j         while (i < n and k >= 0):             s.append(mat[i][k])             i += 1             k -= 1                      # push all element back to matrix          # in reverse order         i = row         k = j         while (i < n and k >= 0):             mat[i][k] = s[-1]             k -= 1             i += 1             s.pop()                  # do the same process for all the     # diagonal which start from last     # column     column = n - 1     for j in range(1, n):                   # here we use stack for reversing          # the elements of diagonal         s = []         i = j         k = column         while (i < n and k >= 0):             s.append(mat[i][k])             i += 1             k -= 1                      # push all element back to matrix          # in reverse order         i = j         k = column         while (i < n and k >= 0):             mat[i][k] = s[-1]             i += 1             k -= 1             s.pop()  # Utility function to print a matrix def printMatrix(mat, n):     for i in range(n):         for j in range(n):             print(mat[i][j], end=" ")         print()          # Driver code mat = [[1, 2, 3, 4],[5, 6, 7, 8],         [9, 10, 11, 12],[13, 14, 15, 16]] n = 4 imageSwap(mat, n) printMatrix(mat, n)  # This code is contributed by shubhamsingh10
C# 
// Simple C# program to find mirror of // matrix across diagonal. using System; using System.Collections.Generic;  class GFG  {      static int MAX = 100;      static void imageSwap(int [,]mat, int n)      {         // for diagonal which start from at          // first row of matrix         int row = 0;          // traverse all top right diagonal         for (int j = 0; j < n; j++)          {              // here we use stack for reversing             // the element of diagonal             Stack<int> s = new Stack<int>();             int i = row, k = j;             while (i < n && k >= 0)             {                 s.Push(mat[i++,k--]);             }              // push all element back to matrix              // in reverse order             i = row;             k = j;             while (i < n && k >= 0)             {                 mat[i++,k--] = s.Peek();                 s.Pop();             }         }          // do the same process for all the         // diagonal which start from last         // column         int column = n - 1;         for (int j = 1; j < n; j++)         {              // here we use stack for reversing              // the elements of diagonal             Stack<int> s = new Stack<int>();             int i = j, k = column;             while (i < n && k >= 0)             {                 s.Push(mat[i++,k--]);             }              // push all element back to matrix              // in reverse order             i = j;             k = column;             while (i < n && k >= 0)             {                 mat[i++,k--] = s.Peek();                 s.Pop();             }         }     }      // Utility function to print a matrix     static void printMatrix(int [,]mat, int n)      {         for (int i = 0; i < n; i++)          {             for (int j = 0; j < n; j++)             {                 Console.Write(mat[i,j] + " ");             }             Console.WriteLine("");         }     }      // Driver code     public static void Main(String[] args)     {          int [,]mat = {{1, 2, 3, 4},                     {5, 6, 7, 8},                     {9, 10, 11, 12},                     {13, 14, 15, 16}};         int n = 4;         imageSwap(mat, n);         printMatrix(mat, n);     } }  /* This code contributed by PrinciRaj1992 */
JavaScript 
<script>         // Simple Javascript program to find mirror of matrix across diagonal.          let MAX = 100;       function imageSwap(mat, n)     {         // for diagonal which start from at         // first row of matrix         let row = 0;           // traverse all top right diagonal         for (let j = 0; j < n; j++)         {               // here we use stack for reversing             // the element of diagonal             let s = [];             let i = row, k = j;             while (i < n && k >= 0)             {                 s.push(mat[i++][k--]);             }               // push all element back to matrix             // in reverse order             i = row;             k = j;             while (i < n && k >= 0)             {                 mat[i++][k--] = s[s.length - 1];                 s.pop();             }         }           // do the same process for all the         // diagonal which start from last         // column         let column = n - 1;         for (let j = 1; j < n; j++)         {               // here we use stack for reversing             // the elements of diagonal             let s = [];             let i = j, k = column;             while (i < n && k >= 0)             {                 s.push(mat[i++][k--]);             }               // push all element back to matrix             // in reverse order             i = j;             k = column;             while (i < n && k >= 0)             {                 mat[i++][k--] = s[s.length - 1];                 s.pop();             }         }     }       // Utility function to print a matrix     function printMatrix(mat, n)     {         for (let i = 0; i < n; i++)         {             for (let j = 0; j < n; j++)             {                 document.write(mat[i][j] + " ");             }             document.write("</br>");         }     }          let mat = [[1, 2, 3, 4],                [5, 6, 7, 8],                [9, 10, 11, 12],                [13, 14, 15, 16]];     let n = 4;     imageSwap(mat, n);     printMatrix(mat, n);      </script>

Output: 

1 5 9 13 
2 6 10 14 
3 7 11 15 
4 8 12 16

Time complexity : O(n2)
Auxiliary Space: O(n), as stack is used

An efficient solution to this problem is that if we observe an output matrix, then we notice that we just have to swap (mat[i][j] to mat[j][i]). 
Below is the implementation of the above idea. 

Implementation:

C++ 
// Efficient CPP program to find mirror of // matrix across diagonal. #include <bits/stdc++.h> using namespace std;  const int MAX = 100;  void imageSwap(vector<vector<int>>&mat, int n) {     // traverse a matrix and swap      // mat[i][j] with mat[j][i]     for (int i = 0; i < n; i++)         for (int j = 0; j <= i; j++)              mat[i][j] = mat[i][j] + mat[j][i] -                         (mat[j][i] = mat[i][j]);        }  // Utility function to print a matrix void printMatrix(vector<vector<int>> &mat, int n) {     for (int i = 0; i < n; i++) {         for (int j = 0; j < n; j++)             cout << mat[i][j] << " ";         cout << endl;     } }  // driver program to test above function int main() {  vector<vector<int>> mat = { { 1, 2, 3, 4 },                              { 5, 6, 7, 8 },                              { 9, 10, 11, 12 },                              { 13, 14, 15, 16 } };     int n = 4;     imageSwap(mat, n);     printMatrix(mat, n);     return 0; }
Java 
// Efficient Java program to find mirror of // matrix across diagonal. import java.io.*;  class GFG {          static int MAX = 100;          static void imageSwap(int mat[][], int n)     {                  // traverse a matrix and swap          // mat[i][j] with mat[j][i]         for (int i = 0; i < n; i++)             for (int j = 0; j <= i; j++)                  mat[i][j] = mat[i][j] + mat[j][i]                        - (mat[j][i] = mat[i][j]);          }          // Utility function to print a matrix     static void printMatrix(int mat[][], int n)     {         for (int i = 0; i < n; i++) {             for (int j = 0; j < n; j++)                 System.out.print( mat[i][j] + " ");             System.out.println();         }     }          // driver program to test above function     public static void main (String[] args)      {         int mat[][] = { { 1, 2, 3, 4 },                         { 5, 6, 7, 8 },                         { 9, 10, 11, 12 },                         { 13, 14, 15, 16 } };         int n = 4;         imageSwap(mat, n);         printMatrix(mat, n);     } }  // This code is contributed by anuj_67.
Python 
# Efficient Python3 program to find mirror of # matrix across diagonal. from builtins import range MAX = 100;  def imageSwap(mat, n):      # traverse a matrix and swap     # mat[i][j] with mat[j][i]     for i in range(n):         for j in range(i + 1):             t = mat[i][j];             mat[i][j] = mat[j][i]             mat[j][i] = t  # Utility function to print a matrix def printMatrix(mat, n):     for i in range(n):         for j in range(n):             print(mat[i][j], end=" ");         print();  # Driver code if __name__ == '__main__':     mat = [1, 2, 3, 4], \         [5, 6, 7, 8], \         [9, 10, 11, 12], \         [13, 14, 15, 16];     n = 4;     imageSwap(mat, n);     printMatrix(mat, n);  # This code is contributed by Rajput-Ji
C# 
// Efficient C# program to find mirror of // matrix across diagonal. using System; class GFG {          //static int MAX = 100;          static void imageSwap(int [,]mat, int n)     {                  // traverse a matrix and swap          // mat[i][j] with mat[j][i]         for (int i = 0; i < n; i++)             for (int j = 0; j <= i; j++)                  mat[i,j] = mat[i,j] + mat[j,i]                     - (mat[j,i] = mat[i,j]);          }          // Utility function to print a matrix     static void printMatrix(int [,]mat, int n)     {         for (int i = 0; i < n; i++) {             for (int j = 0; j < n; j++)                 Console.Write( mat[i,j] + " ");             Console.WriteLine();         }     }          // driver program to test above function     public static void Main ()      {         int [,]mat = { { 1, 2, 3, 4 },                         { 5, 6, 7, 8 },                         { 9, 10, 11, 12 },                         { 13, 14, 15, 16 } };         int n = 4;         imageSwap(mat, n);         printMatrix(mat, n);     } }  // This code is contributed by anuj_67.
JavaScript 
<script>     // Efficient Javascript program to find mirror of     // matrix across diagonal.          let MAX = 100;            function imageSwap(mat, n)     {                    // traverse a matrix and swap          // mat[i][j] with mat[j][i]         for (let i = 0; i < n; i++)             for (let j = 0; j <= i; j++)                  mat[i][j] = mat[i][j] + mat[j][i]                        - (mat[j][i] = mat[i][j]);          }            // Utility function to print a matrix     function printMatrix(mat, n)     {         for (let i = 0; i < n; i++) {             for (let j = 0; j < n; j++)                 document.write(mat[i][j] + " ");             document.write("</br>");         }     }          let mat = [ [ 1, 2, 3, 4 ],                 [ 5, 6, 7, 8 ],                 [ 9, 10, 11, 12 ],                 [ 13, 14, 15, 16 ] ];     let n = 4;     imageSwap(mat, n);     printMatrix(mat, n);        </script>
PHP 
<?php // Efficient PHP program to find mirror  // of matrix across diagonal.  function imageSwap(&$mat, $n) {     // traverse a matrix and swap      // mat[i][j] with mat[j][i]     for ($i = 0; $i < $n; $i++)         for ($j = 0; $j <= $i; $j++)              $mat[$i][$j] = $mat[$i][$j] + $mat[$j][$i] -                            ($mat[$j][$i] = $mat[$i][$j]);  }  // Utility function to print a matrix function printMatrix(&$mat, $n) {     for ($i = 0; $i < $n; $i++)     {         for ($j = 0; $j < $n; $j++)         {             echo ($mat[$i][$j]);             echo (" ");         }         echo ("\n");     } }  // Driver Code $mat = array(array(1, 2, 3, 4),              array(5, 6, 7, 8),              array(9, 10, 11, 12),              array(13, 14, 15, 16)); $n = 4; imageSwap($mat, $n); printMatrix($mat, $n);  // This code is contributed  // by Shivi_Aggarwal ?>

Output: 

1 5 9 13 
2 6 10 14 
3 7 11 15 
4 8 12 16

Time complexity : O(n2)
Auxiliary Space: O(1)


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    Given two m x n matrices m1 and m2, the task is to subtract m2 from m1 and return res.Input: m1 = {{1, 2}, {3, 4}}, m2 = {{4, 3}, {2, 1}}Output: {{-3, -1}, {1, 3}}Input: m1 = {{3, 3, 3}, {3, 3, 3}}, m1 = {{2, 2, 2}, {1, 1, 1}},Output: {{1, 1, 1}, {2, 2, 2}},We traverse both matrices element by eleme
    5 min read

    Intermediate problems on Matrix

    Program for Conway's Game Of Life | Set 1
    Given a Binary Matrix mat[][] of order m*n. A cell with a value of zero is a Dead Cell, while a cell with a value of one is a Live Cell. The state of cells in a matrix mat[][] is known as Generation. The task is to find the next generation of cells based on the following rules:Any live cell with few
    12 min read
    Program to multiply two matrices
    Given two matrices, the task is to multiply them. Matrices can either be square or rectangular:Examples: (Square Matrix Multiplication)Input: m1[m][n] = { {1, 1}, {2, 2} }m2[n][p] = { {1, 1}, {2, 2} }Output: res[m][p] = { {3, 3}, {6, 6} }(Rectangular Matrix Multiplication)Input: m1[3][2] = { {1, 1},
    7 min read
    Rotate an Image 90 Degree Counterclockwise
    Given an image represented by m x n matrix, rotate the image by 90 degrees in counterclockwise direction. Please note the dimensions of the result matrix are going to n x m for an m x n input matrix.Input: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Output: 4 8 12 16 3 7 11 15 2 6 10 14 1 5 9 13Input: 1
    6 min read
    Check if all rows of a matrix are circular rotations of each other
    Given a matrix of n*n size, the task is to find whether all rows are circular rotations of each other or not. Examples: Input: mat[][] = 1, 2, 3 3, 1, 2 2, 3, 1 Output: Yes All rows are rotated permutation of each other. Input: mat[3][3] = 1, 2, 3 3, 2, 1 1, 3, 2 Output: No Explanation : As 3, 2, 1
    8 min read
    Largest Cross Bordered Square
    Given a matrix mat[][] of size n x n where every element is either 'O' or 'X', the task is to find the size of the largest square subgrid that is completely surrounded by 'X', i.e. the largest square where all its border cells are 'X'. Examples: Input: mat[][] = [ ['X', 'X'], ['X', 'X'] ]Output: 2Ex
    15 min read
    Count zeros in a row wise and column wise sorted matrix
    Given a n x n binary matrix (elements in matrix can be either 1 or 0) where each row and column of the matrix is sorted in ascending order, count number of 0s present in it.Examples: Input: [0, 0, 0, 0, 1][0, 0, 0, 1, 1][0, 1, 1, 1, 1][1, 1, 1, 1, 1][1, 1, 1, 1, 1]Output: 8Input: [0, 0][0, 0]Output:
    6 min read
    Queries in a Matrix
    Given two integers m and n, that describes the order m*n of a matrix mat[][], initially filled with integers from 1 to m*n sequentially in a row-major order. Also, there is a 2d array query[][] consisting of q queries, which contains three integers each, where the first integer t describes the type
    15 min read
    Find pairs with given sum such that elements of pair are in different rows
    Given a matrix of distinct values and a sum. The task is to find all the pairs in a given matrix whose summation is equal to the given sum. Each element of a pair must be from different rows i.e; the pair must not lie in the same row.Examples: Input : mat[][] = {{1, 3, 2, 4}, {5, 8, 7, 6}, {9, 10, 1
    15+ min read
    Find all permuted rows of a given row in a matrix
    Given a matrix mat[][] of order m*n, and an index ind. The task is to find all the rows in the matrix mat[][] which are permutations of rows at index ind.Note: All the elements of a row are distinct.Examples: Input: mat[][] = [[3, 1, 4, 2], [1, 6, 9, 3], [1, 2, 3, 4], [4, 3, 2, 1]] ind = 3 Output: 0
    9 min read
    Find number of transformation to make two Matrix Equal
    Given two matrices a and b of size n*m. The task is to find the required number of transformation steps so that both matrices become equal. Print -1 if this is not possible. The transformation step is as follows: Select any one matrix out of two matrices. Choose either row/column of the selected mat
    8 min read
    Inplace (Fixed space) M x N size matrix transpose
    Given an M x N matrix, transpose the matrix without auxiliary memory.It is easy to transpose matrix using an auxiliary array. If the matrix is symmetric in size, we can transpose the matrix inplace by mirroring the 2D array across it's diagonal (try yourself). How to transpose an arbitrary size matr
    15+ min read
    Minimum flip required to make Binary Matrix symmetric
    Given a Binary Matrix mat[][] of size n x n, consisting of 1s and 0s. The task is to find the minimum flips required to make the matrix symmetric along the main diagonal.Examples : Input: mat[][] = [[0, 0, 1], [1, 1, 1], [1, 0, 0]];Output: 2Value of mat[1][0] is not equal to mat[0][1].Value of mat[2
    8 min read
    Magic Square of Odd Order
    Given a positive integer n, your task is to generate a magic square of order n * n. A magic square of order n is an n * n grid filled with the numbers 1 through n² so that every row, every column, and both main diagonals each add up to the same total, called the magic constant (or magic sum) M. Beca
    15+ min read

    Hard problems on Matrix

    Number of Islands
    Given an n x m grid of 'W' (Water) and 'L' (Land), the task is to count the number of islands. An island is a group of adjacent 'L' cells connected horizontally, vertically, or diagonally, and it is surrounded by water or the grid boundary. The goal is to determine how many distinct islands exist in
    15+ min read
    A Boolean Matrix Question
    Given a boolean matrix mat where each cell contains either 0 or 1, the task is to modify it such that if a matrix cell matrix[i][j] is 1 then all the cells in its ith row and jth column will become 1.Examples:Input: [[1, 0], [0, 0]]Output: [[1, 1], [1, 0]]Input: [[1, 0, 0, 1], [0, 0, 1, 0], [0, 0, 0
    15+ min read
    Matrix Chain Multiplication
    Given the dimension of a sequence of matrices in an array arr[], where the dimension of the ith matrix is (arr[i-1] * arr[i]), the task is to find the most efficient way to multiply these matrices together such that the total number of element multiplications is minimum. When two matrices of size m*
    15+ min read
    Maximum size rectangle binary sub-matrix with all 1s
    Given a 2d binary matrix mat[][], the task is to find the maximum size rectangle binary-sub-matrix with all 1's. Examples: Input: mat = [ [0, 1, 1, 0], [1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 0, 0] ]Output : 8Explanation : The largest rectangle with only 1's is from (1, 0) to (2, 3) which is[1, 1, 1, 1][
    15 min read
    Construct Ancestor Matrix from a Given Binary Tree
    Given a Binary Tree where all values are from 0 to n-1. Construct an ancestor matrix mat[n][n] where the ancestor matrix is defined as below. mat[i][j] = 1 if i is ancestor of jmat[i][j] = 0, otherwiseExamples: Input: Output: {{0 1 1} {0 0 0} {0 0 0}}Input: Output: {{0 0 0 0 0 0} {1 0 0 0 1 0} {0 0
    15+ min read
    K'th element in spiral form of matrix
    Given a matrix of size n * m. You have to find the kth element which will obtain while traversing the matrix spirally starting from the top-left corner of the matrix.Examples:Input: mat[][] = [ [1, 2, 3], [4, 5, 6], [7, 8, 9] ], k = 4Output: 6Explanation: Spiral traversal of matrix: {1, 2, 3, 6, 9,
    13 min read
    Largest Plus or '+' formed by all ones in a binary square matrix
    Given an n × n binary matrix mat consisting of 0s and 1s. Your task is to find the size of the largest ‘+’ shape that can be formed using only 1s. A ‘+’ shape consists of a center cell with four arms extending in all four directions (up, down, left, and right) while remaining within the matrix bound
    10 min read
    Shortest path in a Binary Maze
    Given an M x N matrix where each element can either be 0 or 1. We need to find the shortest path between a given source cell to a destination cell. The path can only be created out of a cell if its value is 1.Note: You can move into an adjacent cell in one of the four directions, Up, Down, Left, and
    15+ min read
    Maximum sum square sub-matrix of given size
    Given a 2d array mat[][] of order n * n, and an integer k. Your task is to find a submatrix of order k * k, such that sum of all the elements in the submatrix is maximum possible.Note: Matrix mat[][] contains zero, positive and negative integers.Examples:Input: k = 3mat[][] = [ [ 1, 2, -1, 4 ] [ -8,
    15+ min read
    Validity of a given Tic-Tac-Toe board configuration
    A Tic-Tac-Toe board is given after some moves are played. Find out if the given board is valid, i.e., is it possible to reach this board position after some moves or not.Note that every arbitrary filled grid of 9 spaces isn't valid e.g. a grid filled with 3 X and 6 O isn't valid situation because ea
    15+ min read
    Minimum Initial Points to Reach Destination
    Given a m*n grid with each cell consisting of positive, negative, or no points i.e., zero points. From a cell (i, j) we can move to (i+1, j) or (i, j+1) and we can move to a cell only if we have positive points ( > 0 ) when we move to that cell. Whenever we pass through a cell, points in that cel
    15+ min read
    Program for Sudoku Generator
    Given an integer k, the task is to generate a 9 x 9 Sudoku grid having k empty cells while following the below set of rules:In all 9 submatrices 3x3, the elements should be 1-9, without repetition.In all rows, there should be elements between 1-9, without repetition.In all columns, there should be e
    15+ min read
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