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

• Difficulty Level : Medium
• Last Updated : 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:
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, they return or print the count of such pairs.

Below is the implementation of the above approach:

## C++

 `// C++ program to count the pairs in array``// such that one element is reverse of another``#include ``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);``    ``cout << countReverse(a, n);``    ``return` `0;``}`

## Java

 `// 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

 `# 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`

## C#

 `// 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`

## Javascript

 ``

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 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++

 `// C++ program to count the pairs in array``// such that one element is reverse of another``#include ``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);``    ``cout << countReverse(a, n) << ``'\n'``;``    ``return` `0;``}`

## Java

 `// 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 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

 `# 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`

## C#

 `// 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`

## Javascript

 ``

Output

`2`

Time Complexity: O(N)

Auxiliary Space: O(N)

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