Given an array arr[] consisting of N positive integers, the task is to find the count of unique pairs (i, j) such that the sum of arr[i] and the reverse(arr[j]) is same as the sum of reverse(arr[i]) and arr[j].
Examples:
Input: arr[] = {2, 15, 11, 7}
Output: 3
Explanation:
The pairs are (0, 2), (0, 3) and (2, 3).
- (0, 2): arr[0] + reverse(arr[2]) (= 2 + 11 = 211) and reverse(arr[0]) + arr[2](= 2 + 11 = 211).
- (0, 3): arr[0] + reverse(arr[3]) (= 2 + 7 = 27) and reverse(arr[0]) + arr[3](= 2 + 7 = 27).
- (2, 3): arr[2] + reverse(arr[3]) (= 11 + 7 = 117) and reverse(arr[2]) + arr[3](= 11 + 7 = 117).
Input: A[] = {22, 115, 7, 313, 17, 23, 22}
Output: 6
Naive Approach: The simplest approach is to generate all possible pairs of the given array and if any pair of elements satisfy the given conditions then count these pairs. After completing the above stepsm, print the value of count as the result.
Time Complexity: O(N2*log M), where M is the maximum element in A[]
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized by using Hashing technique and rewriting the equation as:
A[i] + reverse(A[j]) = reverse(A[i]) + A[j]
=> A[i] – reverse(A[i]) = A[j] – reverse(A[j])
Now, the idea is to count the frequency of (A[i] – reverse(A[i])) for every element arr[i] and then count possible number of valid pairs satisfying the given condition. Follow the steps below to solve the problem:
- Maintain a Hashmap, say u_map to store the frequency count of A[i] – reverse(A[i]) for any index i.
- Initialize a variable pairs to store the number of pairs that satisfy the given condition.
- Traverse the given array A[] using the variable i and perform the following operations:
- Store the frequency of A[i] – reverse(A[i]) in val.
- Increment pairs by val.
- Update the frequency of val in u_map.
- After completing the above steps, print the value of pairs as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find the // reverse of the number n int reverse( int n) { int temp = n, rev = 0, r; // Iterate until temp is 0 while (temp) { r = temp % 10; rev = rev * 10 + r; temp /= 10; } // Return the reversed number return rev; } // Function to count number of unique // pairs (i, j) from the array A[] // which satisfies the given condition void countPairs( int A[], int N) { // Store the frequency of keys // as A[i] - reverse(A[i]) unordered_map< int , int > u_map; // Stores count of desired pairs int pairs = 0; // Iterate the array A[] for ( int i = 0; i < N; i++) { int val = A[i] - reverse(A[i]); // Add frequency of val // to the required answer pairs += u_map[val]; // Increment frequency of val u_map[val]++; } // Print the number of pairs cout << pairs; } // Driver Code int main() { int arr[] = { 2, 15, 11, 7 }; int N = sizeof (arr) / sizeof (arr[0]); // Function Call countPairs(arr, N); return 0; } |
Java
// Java program for the above approach import java.io.*; import java.lang.*; import java.util.*; class GFG{ // Function to find the // reverse of the number n static int reverse( int n) { int temp = n, rev = 0 , r; // Iterate until temp is 0 while (temp > 0 ) { r = temp % 10 ; rev = rev * 10 + r; temp /= 10 ; } // Return the reversed number return rev; } // Function to count number of unique // pairs (i, j) from the array A[] // which satisfies the given condition static void countPairs( int A[], int N) { // Store the frequency of keys // as A[i] - reverse(A[i]) HashMap<Integer, Integer> map = new HashMap<>(); // Stores count of desired pairs int pairs = 0 ; // Iterate the array A[] for ( int i = 0 ; i < N; i++) { int val = A[i] - reverse(A[i]); // Add frequency of val // to the required answer pairs += map.getOrDefault(val, 0 ); // Increment frequency of val map.put(val, map.getOrDefault(val, 0 ) + 1 ); } // Print the number of pairs System.out.println(pairs); } // Driver Code public static void main(String[] args) { int arr[] = { 2 , 15 , 11 , 7 }; int N = arr.length; // Function Call countPairs(arr, N); } } // This code is contributed by Kingash |
Python3
# Python3 program for the above approach from collections import defaultdict # Function to find the # reverse of the number n def reverse(n): temp = n rev = 0 # Iterate until temp is 0 while (temp): r = temp % 10 rev = rev * 10 + r temp / / = 10 # Return the reversed number return rev # Function to count number of unique # pairs (i, j) from the array A[] # which satisfies the given condition def countPairs(A, N): # Store the frequency of keys # as A[i] - reverse(A[i]) u_map = defaultdict( int ) # Stores count of desired pairs pairs = 0 # Iterate the array A[] for i in range (N): val = A[i] - reverse(A[i]) # Add frequency of val # to the required answer pairs + = u_map[val] # Increment frequency of val u_map[val] + = 1 # Print the number of pairs print (pairs) # Driver Code if __name__ = = "__main__" : arr = [ 2 , 15 , 11 , 7 ] N = len (arr) # Function Call countPairs(arr, N) # This code is contributed by chitranayal. |
3
Time Complexity: O(N*log M), where M is the largest element in the array.
Auxiliary Space: O(1)
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