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Count pairs in array such that one element is reverse of another
  • Last Updated : 26 Nov, 2020

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:
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  

Method 1:

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 pairs. 



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 array
// 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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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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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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Output: 

2

 

Time Complexity: O(N2)

Auxiliary Space: O(1)

 

Method 2: (Using Hash-Map)

We can observe that the most expensive operation here is searching for the reversed element in the array(which takes O(N)). By using a hash map, this can be reduced to O(1). 

Approach: The idea is to store all the elements of the array in the hash map(by increasing the frequency of the present element to tackle the problem of duplicates) and check how many times the reversed element is repeated and increases the count by that frequency. To avoid recounting the number when it is a palindrome or when we visit its reverse, we need to delete the present number from the hash map(this is achieved by decreasing the frequency of that number).

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 array
// such that one number is reverse of
// the other
int countReverse(int arr[], int n)
{
    unordered_map<int, int> freq;
 
    // Iterate over every element in the array
    // and increase the frequency of the element
    // in hash map
    for(int i = 0; i < n; ++i)
        ++freq[arr[i]];
 
    int res = 0;
    // Iterate over every element in the array
    for (int i = 0; i < n; i++){
 
        // remove the current element from
        // the hash map by decreasing the
        // frequency to avoid counting
        // when the number is a palindrome
        // or when we visit its reverse
        --freq[arr[i]];
 
        // Increment the count
        // by the frequency of
        // reverse of the number
        res += freq[reverse(arr[i])];
    }
    return res;
}
 
// Driver code
int main() {
    int a[] = { 16, 61, 12, 21, 25 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << countReverse(a, n) << '\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
import java.util.*;
class GFG{
     
// Function to reverse the digits
// of the number
public 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 array such that
// one number is reverse of the
// other
public static int countReverse(int arr[],
                               int n)
{
  HashMap<Integer,
          Integer> freq =
          new HashMap<>();
 
  // Iterate over every element
  // in the array and increase
  // the frequency of the element
  // in hash map
  for(int i = 0; i < n; ++i)
  {
    if(freq.containsKey(arr[i]))
    {
      freq.replace(arr[i],
      freq.get(arr[i]) + 1);
    }
    else
    {
      freq.put(arr[i], 1);
    }
  }
 
  int res = 0;
   
  // Iterate over every element
  // in the array
  for (int i = 0; i < n; i++)
  {
    // remove the current element from
    // the hash map by decreasing the
    // frequency to avoid counting
    // when the number is a palindrome
    // or when we visit its reverse
    if(freq.containsKey(arr[i]))
    {
      freq.replace(arr[i],
      freq.get(arr[i]) - 1);
    }
    else
    {
      freq.put(arr[i], -1);
    }
 
    // Increment the count
    // by the frequency of
    // reverse of the number
    if(freq.containsKey(reverse(arr[i])))
    {
      res += freq.get(reverse(arr[i]));
    }
  }
  return res;
}
 
// Driver code   
public static void main(String[] args)
{
  int a[] = {16, 61, 12, 21, 25};
  int n = a.length;
  System.out.println(countReverse(a, n));
}
}
 
// This code is contributed by divyeshrabadiya07

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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
from collections import defaultdict
 
# 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 array such that
# one number is reverse of
# the other
def countReverse(arr, n):
 
    freq = defaultdict (int)
 
    # Iterate over every element
    # in the array and increase
    # the frequency of the element
    # in hash map
    for i in range (n):
        freq[arr[i]] += 1
 
    res = 0
     
    # Iterate over every
    # element in the array
    for i in range (n):
 
        # remove the current element from
        # the hash map by decreasing the
        # frequency to avoid counting
        # when the number is a palindrome
        # or when we visit its reverse
        freq[arr[i]] -= 1
 
        # Increment the count
        # by the frequency of
        # reverse of the number
        res += freq[reverse(arr[i])]
    
    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 Chitranayal

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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;
using System.Collections.Generic;  
 
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 array such that
// one number is reverse of the
// other
static int countReverse(int[] arr, int n)
{
    Dictionary<int,
               int> freq = new Dictionary<int,
                                          int>(); 
     
    // Iterate over every element
    // in the array and increase
    // the frequency of the element
    // in hash map
    for(int i = 0; i < n; ++i)
    {
        if (freq.ContainsKey(arr[i]))
        {
            freq[arr[i]]++;
        }
        else
        {
            freq.Add(arr[i], 1);
        }
    }
     
    int res = 0;
     
    // Iterate over every element
    // in the array
    for(int i = 0; i < n; i++)
    {
         
        // Remove the current element from
        // the hash map by decreasing the
        // frequency to avoid counting
        // when the number is a palindrome
        // or when we visit its reverse
        if (freq.ContainsKey(arr[i]))
        {
            freq[arr[i]]--;
        }
        else
        {
            freq.Add(arr[i], -1);
        }
     
        // Increment the count
        // by the frequency of
        // reverse of the number
        if (freq.ContainsKey(reverse(arr[i])))
        {
            res += freq[reverse(arr[i])];
        }
    }
    return res;
}
 
// Driver code   
static void Main()
{
    int[] a = { 16, 61, 12, 21, 25 };
    int n = a.Length;
     
    Console.WriteLine(countReverse(a, n));
}
}
 
// This code is contributed by divyesh072019

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Output

2

Time Complexity: O(N)

Auxiliary Space: O(N)

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