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 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
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 <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; } |
Java
// Java program to count the pairs in array // such that one element is reverse of another class Beginner { // 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
<script> // Javascript program to count the pairs in array // such that one element is reverse of another // Function to reverse the digits // of the number function 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 other function 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 code var 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 * log10M), where N is the size of the given array and M is the maximum element in the array.
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 <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; } |
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<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
# 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
<script> // Javascript program to count the // pairs in array such that // one element is reverse of // another // Function to reverse the digits // of the number function 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 // other function 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)