Count pairs in array such that one element is reverse of another

Given an array arr[], the task is to count the pairs in the array such that elements are reverse of each other.

Examples:

Input: arr[] = { 16, 61, 12, 21, 25 }
Output: 2
Explanation:
The 2 pairs such that one number is the reverse of the other are {16, 61} and {12, 21}.

Input: arr[] = {10, 11, 12}
Output: 0

Approach: The idea is to use nested loops to get all the possible pairs of numbers in the array. Then, for each pair, check whether an element is a reverse of another. If it is, then increase the required count by one. When all the pairs have been checked, return or print the count of such pair.



Below is the implementation of the above approach:

C++

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// C++ program to count the pairs in array
// such that one element is reverse of another
#include <bits/stdc++.h>
using namespace std;
  
// Function to reverse the digits
// of the number
int reverse(int num)
{
    int rev_num = 0;
  
    // Loop to iterate till the number is
    // greater than 0
    while (num > 0) {
  
        // Extract the last digit and keep
        // multiplying it by 10 to get the
        // reverse of the number
        rev_num = rev_num * 10 + num % 10;
        num = num / 10;
    }
    return rev_num;
}
  
// Function to find the pairs from the 
// such that one number is reverse of
// the other
int countReverse(int arr[], int n)
{
    int res = 0;
  
    // Iterate through all pairs
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
  
            // Increment count if one is
            // the reverse of other
            if (reverse(arr[i]) == arr[j]) {
                res++;
            }
  
    return res;
}
  
// Driver code
int main()
{
    int a[] = { 16, 61, 12, 21, 25 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << countReverse(a, n);
    return 0;
}

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Java

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// Java program to count the pairs in array
// such that one element is reverse of another
  
class Geeks {
    // Function to reverse the digits
    // of the number
    static int reverse(int num) {
        int rev_num = 0;
  
        // Loop to iterate till the number is
        // greater than 0
        while (num > 0) {
  
            // Extract the last digit and keep
            // multiplying it by 10 to get the
            // reverse of the number
            rev_num = rev_num * 10 + num % 10;
            num = num / 10;
        }
        return rev_num;
    }
  
    // Function to find the pairs from the
    // such that one number is reverse of
    // the other
    static int countReverse(int arr[], int n) {
        int res = 0;
  
        // Iterate through all pairs
        for (int i = 0; i < n; i++)
            for (int j = i + 1; j < n; j++)
  
                // Increment count if one is
                // the reverse of other
                if (reverse(arr[i]) == arr[j]) {
                    res++;
                }
  
        return res;
    }
  
    // Driver code
    public static void main(String[] args) {
        int a[] = { 16, 61, 12, 21, 25 };
        int n = a.length;
        System.out.print(countReverse(a, n));
    }
}
  
// This code is contributed by Rajnis09

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C#

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// C# program to count the pairs in array
// such that one element is reverse of another
using System;
   
class GFG 
{
   
// Function to reverse the digits
// of the number
static int reverse(int num)
{
    int rev_num = 0;
   
    // Loop to iterate till the number is
    // greater than 0
    while (num > 0) {
   
        // Extract the last digit and keep
        // multiplying it by 10 to get the
        // reverse of the number
        rev_num = rev_num * 10 + num % 10;
        num = num / 10;
    }
    return rev_num;
}
   
// Function to find the pairs from the 
// such that one number is reverse of
// the other
static int countReverse(int []arr, int n)
{
    int res = 0;
   
    // Iterate through all pairs
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
   
            // Increment count if one is
            // the reverse of other
            if (reverse(arr[i]) == arr[j]) {
                res++;
            }
   
    return res;
}
   
// Driver code 
public static void Main(String []arr) 
{
    int []a = { 16, 61, 12, 21, 25 };
    int n = a.Length;
    Console.Write(countReverse(a, n));
}
}
  
// This code is contributed by shivanisinghss2110

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Python3

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# Python3 program to count the pairs in array
# such that one element is reverse of another
  
# Function to reverse the digits
# of the number
def reverse(num):
    rev_num = 0
  
    # Loop to iterate till the number is
    # greater than 0
    while (num > 0):
  
        # Extract the last digit and keep
        # multiplying it by 10 to get the
        # reverse of the number
        rev_num = rev_num * 10 + num % 10
        num = num // 10
  
    return rev_num
  
# Function to find the pairs from the 
# such that one number is reverse of
# the other
def countReverse(arr,n):
    res = 0
  
    # Iterate through all pairs
    for i in range(n):
        for j in range(i + 1, n):
  
            # Increment count if one is
            # the reverse of other
            if (reverse(arr[i]) == arr[j]):
                res += 1
  
    return res
  
# Driver code
if __name__ == '__main__':
    a =  [16, 61, 12, 21, 25]
    n =  len(a)
    print(countReverse(a, n))
  
# This code is contributed by Surendra_Gangwar

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Output:

2

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