Given a matrix arr[][] of size N * N, the task is to find the number of palindromic rows.
Examples:
Input: arr[][] = {{1, 3, 1},
{2, 2, 3},
{2, 1, 2}}
Output: 2
Explanation: First and third row forms a palindrome i.e 1 3 1 and 2 1 2.
Therefore, count of palindromic rows is 2.Input: arr[][] = {{2, 2, 3, 2},
{1, 3, 3, 1},
{4, 2, 2, 4},
{5, 6, 6, 5}}
Output: 3
Approach: The task can be solved using a two-pointer approach. Follow the steps mentioned below:
- Iterate over each row of the matrix.
- For each row:
- Use two pointers to point the starting of the row and the end of the row.
- If values of both the pointer are same increment the starting pointer and decrement the end pointer.
- Keep on doing this until both the pointers point to the same element or they have different values.
- If they have different values the row is not a palindrome. Stop the iteration and return false. Else continue for the next row.
- After all the rows are traversed and everyone is palindrome return true.
Below is the implementation of the approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;const int MAX = 100;// Function to count the number of// palindromic rowsint countPalindrome( vector<vector<int> >& arr, int N){ int count = 0; for (int i = 0; i < N; i++) { int j = 0, k = N - 1; bool t = true; while (j < k) { if (arr[i][j] != arr[i][k]) { t = false; break; } j++; k--; } if (t) count++; } return count;}// Driver Codeint main(){ int N = 3; vector<vector<int> > arr = { { 1, 3, 1 }, { 2, 2, 3 }, { 2, 1, 2 } }; cout << countPalindrome(arr, N); return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.lang.*;import java.util.*;class Solution { // Function to count the number of // palindromic rows static int countPalindrome(int arr[][], int N) { int count = 0; for (int i = 0; i < N; i++) { int j = 0, k = N - 1; boolean t = true; while (j < k) { if (arr[i][j] != arr[i][k]) { t = false; break; } j++; k--; } if (t) count++; } return count; } // Driver code public static void main(String[] args) { int N = 3; int arr[][] = { { 1, 3, 1 }, { 2, 2, 3 }, { 2, 1, 2 } }; System.out.println( countPalindrome(arr, N)); }} |
Python3
# Python program for the above approachMAX = 100;# Function to count the number of# palindromic rowsdef countPalindrome(arr, N): count = 0; for i in range(N): j = 0 k = N - 1 t = True while (j < k): if (arr[i][j] != arr[i][k]): t = False break; j += 1 k -= 1 if (t): count += 1 return count;# Driver CodeN = 3;arr = [[1, 3, 1], [2, 2, 3], [2, 1, 2]];print(countPalindrome(arr, N));# This code is contributed by gfgking |
C#
// C# program for the above approachusing System;public class GFG{ // Function to count the number of // palindromic rows static int countPalindrome(int[,] arr, int N) { int count = 0; for (int i = 0; i < N; i++) { int j = 0, k = N - 1; bool t = true; while (j < k) { if (arr[i, j] != arr[i, k]) { t = false; break; } j++; k--; } if (t) count++; } return count; } // Driver code static public void Main (){ int N = 3; int[,] arr = new int[3, 3] { { 1, 3, 1 }, { 2, 2, 3 }, { 2, 1, 2 } }; Console.WriteLine( countPalindrome(arr, N)); }}// This code is contributed by hrithikgarg03188. |
Javascript
<script> // JavaScript program for the above approach const MAX = 100; // Function to count the number of // palindromic rows const countPalindrome = (arr, N) => { let count = 0; for (let i = 0; i < N; i++) { let j = 0, k = N - 1; let t = true; while (j < k) { if (arr[i][j] != arr[i][k]) { t = false; break; } j++; k--; } if (t) count++; } return count; } // Driver Code let N = 3; let arr = [[1, 3, 1], [2, 2, 3], [2, 1, 2]]; document.write(countPalindrome(arr, N));// This code is contributed by rakeshsahni</script> |
Output
2
Time Complexity: O(N*N)
Auxiliary Space: O(1)
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