Check if two elements of a matrix are on the same diagonal or not
Given a matrix mat[][], and two integers X and Y, the task is to check if X and Y are on the same diagonal of the given matrix or not.
Examples:
Input: mat[][]= {{1, 2}. {3, 4}}, X = 1, Y = 4
Output: Yes
Explanation:
Both X and Y lie on the same diagonal.Input: mat[][]= {{1, 2}. {3, 4}}, X = 2, Y = 4
Output: No
Approach: The key observation to solve the problem is that the two elements of the matrix are on the same diagonal only if the sum of the indices or the difference of the indices of the elements are equal. Follow the steps below to solve the problem:
- Traverse the matrix and find the indices of the elements of the matrix.
- Check if the elements are on the same diagonal.
Let the indices of the elements of the matrix are (P, Q) and (X, Y), then the condition that both the elements are on the same diagonal is given by the following equation:
P – Q == X – Y or P + Q == X + Y
- If the above condition is satisfied, print “Yes”. Otherwise, print “No”.
Below is the implementation of the above approach:
C++
// C++ program for // the above approach #include <bits/stdc++.h> using namespace std; // Function to check if two // integers are on the same // diagonal of the matrix void checkSameDiag( int li[6][5], int x, int y, int m, int n) { // Storing indexes of x in I, J int I = 0, J = 0; // Storing Indexes of y in P, Q int P = 0, Q = 0; for ( int i = 0; i < m; i++) { for ( int j = 0; j < n; j++) { if (li[i][j] == x) { I = i; J = j; } if (li[i][j] == y) { P = i; Q = j; } } } // Condition to check if the // both the elements are in // same diagonal of a matrix if (P - Q == I - J || P + Q == I + J) { cout << "YES" ; } else cout << "NO" ; } // Driver code int main() { // Dimensions of Matrix int m = 6; int n = 5; // Given Matrix int li[6][5] = {{32, 94, 99, 26, 82}, {51, 69, 52, 63, 17}, {90, 36, 88, 55, 33}, {93, 42, 73, 39, 28}, {81, 31, 83, 53, 10}, {12, 29, 85, 80, 87}}; // Elements to be checked int x = 42; int y = 80; // Function call checkSameDiag(li, x, y, m, n); return 0; } //This code is contributed by 29AjayKumar |
Java
// Java implementation of the // above approach class GFG{ // Function to check if two // integers are on the same // diagonal of the matrix static void checkSameDiag( int li[][], int x, int y, int m, int n) { // Storing indexes of x in I, J int I = 0 , J = 0 ; // Storing Indexes of y in P, Q int P = 0 , Q = 0 ; for ( int i = 0 ; i < m; i++) { for ( int j = 0 ; j < n; j++) { if (li[i][j] == x) { I = i; J = j; } if (li[i][j] == y) { P = i; Q = j; } } } // Condition to check if the // both the elements are in // same diagonal of a matrix if (P - Q == I - J || P + Q == I + J) { System.out.println( "YES" ); } else System.out.println( "NO" ); } // Driver Code public static void main(String[] args) { // Dimensions of Matrix int m = 6 ; int n = 5 ; // Given Matrix int [][] li = { { 32 , 94 , 99 , 26 , 82 }, { 51 , 69 , 52 , 63 , 17 }, { 90 , 36 , 88 , 55 , 33 }, { 93 , 42 , 73 , 39 , 28 }, { 81 , 31 , 83 , 53 , 10 }, { 12 , 29 , 85 , 80 , 87 } }; // Elements to be checked int x = 42 ; int y = 80 ; // Function call checkSameDiag(li, x, y, m, n); } } // This code is contributed by Amit Katiyar |
Python3
# Python3 implementation of the # above approach # Function to check if two # integers are on the same # diagonal of the matrix def checkSameDiag(x, y): # Storing indexes of x in I, J # Storing Indexes of y in P, Q for i in range (m): for j in range (n): if li[i][j] = = x: I, J = i, j if li[i][j] = = y: P, Q = i, j # Condition to check if the # both the elements are in # same diagonal of a matrix if P - Q = = I - J or P + Q = = I + J: print ( "YES" ) else : print ( "NO" ) # Driver Code if __name__ = = "__main__" : # Dimensions of Matrix m, n = 6 , 5 # Given Matrix li = [[ 32 , 94 , 99 , 26 , 82 ], [ 51 , 69 , 52 , 63 , 17 ], [ 90 , 36 , 88 , 55 , 33 ], [ 93 , 42 , 73 , 39 , 28 ], [ 81 , 31 , 83 , 53 , 10 ], [ 12 , 29 , 85 , 80 , 87 ]] # elements to be checked x, y = 42 , 80 # Function Call checkSameDiag(x, y) |
C#
// C# implementation of the // above approach using System; class GFG{ // Function to check if two // integers are on the same // diagonal of the matrix static void checkSameDiag( int [,]li, int x, int y, int m, int n) { // Storing indexes of x in I, J int I = 0, J = 0; // Storing Indexes of y in P, Q int P = 0, Q = 0; for ( int i = 0; i < m; i++) { for ( int j = 0; j < n; j++) { if (li[i, j] == x) { I = i; J = j; } if (li[i, j] == y) { P = i; Q = j; } } } // Condition to check if the // both the elements are in // same diagonal of a matrix if (P - Q == I - J || P + Q == I + J) { Console.WriteLine( "YES" ); } else Console.WriteLine( "NO" ); } // Driver Code public static void Main(String[] args) { // Dimensions of Matrix int m = 6; int n = 5; // Given Matrix int [,]li = {{32, 94, 99, 26, 82}, {51, 69, 52, 63, 17}, {90, 36, 88, 55, 33}, {93, 42, 73, 39, 28}, {81, 31, 83, 53, 10}, {12, 29, 85, 80, 87}}; // Elements to be checked int x = 42; int y = 80; // Function call checkSameDiag(li, x, y, m, n); } } // This code is contributed by 29AjayKumar |
Javascript
<script> // JavaScript program for // the above approach // Function to check if two // integers are on the same // diagonal of the matrix function checkSameDiag(li, x, y, m, n) { // Storing indexes of x in I, J let I = 0, J = 0; // Storing Indexes of y in P, Q let P = 0, Q = 0; for (let i = 0; i < m; i++) { for (let j = 0; j < n; j++) { if (li[i][j] == x) { I = i; J = j; } if (li[i][j] == y) { P = i; Q = j; } } } // Condition to check if the // both the elements are in // same diagonal of a matrix if (P - Q == I - J || P + Q == I + J) { document.write( "YES" ); } else document.write( "NO" ); } // Driver code // Dimensions of Matrix let m = 6; let n = 5; // Given Matrix let li = [[ 32, 94, 99, 26, 82 ], [ 51, 69, 52, 63, 17 ], [ 90, 36, 88, 55, 33 ], [ 93, 42, 73, 39, 28 ], [ 81, 31, 83, 53, 10 ], [ 12, 29, 85, 80, 87 ]]; // Elements to be checked let x = 42; let y = 80; // Function call checkSameDiag(li, x, y, m, n); // This code is contributed by splevel62. </script> |
Output:
YES
Time Complexity: O(N * M)
Auxiliary Space: O(1)
Contact Us