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

03 Jun, 2021
Given an array arr[], the task is to count the pairs in the array such that the elements are the 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, they return or print the count of such pairs. 
Below is the implementation of the above approach: 
C++
Java
Python3
C#
Javascript
// 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 numberint 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 otherint 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 codeint main(){    int a[] = { 16, 61, 12, 21, 25 };    int n = sizeof(a) / sizeof(a[0]);    cout << countReverse(a, n);    return 0;}
// 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
# Python3 program to count the pairs in array# such that one element is reverse of another # Function to reverse the digits# of the numberdef 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 otherdef 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 codeif __name__ == '__main__':    a =  [16, 61, 12, 21, 25]    n =  len(a)    print(countReverse(a, n)) # This code is contributed by Surendra_Gangwar
// C# program to count the pairs in array// such that one element is reverse of anotherusing System;  class GFG{  // Function to reverse the digits// of the numberstatic 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 otherstatic 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 codepublic 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
<script> // Javascript program to count the pairs in array// such that one element is reverse of another // Function to reverse the digits// of the numberfunction reverse(num){    var 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 = parseInt(num / 10);    }    return rev_num;} // Function to find the pairs from the array// such that one number is reverse of// the otherfunction countReverse(arr, n){    var res = 0;     // Iterate through all pairs    for (var i = 0; i < n; i++)        for (var 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 codevar a = [ 16, 61, 12, 21, 25 ];var n = a.length;document.write( countReverse(a, n)); // This code is contributed by noob2000.</script>
2
 
Time Complexity: O(N2)
Auxiliary Space: O(1)
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 increase 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++
Java
Python3
C#
Javascript
// 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 numberint 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 otherint 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 codeint main() {    int a[] = { 16, 61, 12, 21, 25 };    int n = sizeof(a) / sizeof(a[0]);    cout << countReverse(a, n) << '\n';    return 0;}
// Java program to count the// pairs in array such that// one element is reverse of// anotherimport java.util.*;class GFG{     // Function to reverse the digits// of the numberpublic 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// otherpublic 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
# Python3 program to count# the pairs in array such# that one element is reverse# of anotherfrom collections import defaultdict # Function to reverse# the digits of the numberdef 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 otherdef 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 codeif __name__ == "__main__":       a = [16, 61, 12, 21, 25]    n = len(a)    print (countReverse(a, n))     # This code is contributed by Chitranayal
// C# program to count the// pairs in array such that// one element is reverse of// anotherusing System;using System.Collections.Generic;   class GFG{     // Function to reverse the digits// of the numberstatic 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// otherstatic 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
<script>// Javascript program to count the// pairs in array such that// one element is reverse of// another // Function to reverse the digits// of the numberfunction reverse(num){    let 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 = Math.floor(num / 10);      }      return rev_num;} // Function to find the pairs// from the array such that// one number is reverse of the// otherfunction countReverse(arr,n){     let freq = new Map();        // Iterate over every element      // in the array and increase      // the frequency of the element      // in hash map      for(let i = 0; i < n; ++i)      {        if(freq.has(arr[i]))        {          freq.set(arr[i],          freq.get(arr[i]) + 1);        }        else        {          freq.set(arr[i], 1);        }      }       let res = 0;         // Iterate over every element      // in the array      for (let 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.has(arr[i]))        {              freq.set(arr[i],              freq.get(arr[i]) - 1);        }        else        {              freq.set(arr[i], -1);        }          // Increment the count        // by the frequency of        // reverse of the number        if(freq.has(reverse(arr[i])))        {              res += freq.get(reverse(arr[i]));        }      }      return res;} // Driver code  let a=[16, 61, 12, 21, 25];let n = a.length;document.write(countReverse(a, n));  // This code is contributed by avanitrachhadiya2155</script>
2
Time Complexity: O(N)
Auxiliary Space: O(N)
SURENDRA_GANGWAR
Rajnis09
shivanisinghss2110
abhijeet_ar
ukasp
divyeshrabadiya07
divyesh072019
noob2000
avanitrachhadiya2155
Arrays
Mathematical
Arrays
Mathematical
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.