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# Count of unique pairs (i, j) in an array such that sum of A[i] and reverse of A[j] is equal to sum of reverse of A[i] and A[j]

• Difficulty Level : Easy
• Last Updated : 28 Dec, 2022

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 the 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 + reverse(arr) (= 2 + 11 = 13) and reverse(arr) + arr(= 2 + 11 = 13).
• (0, 3): arr + reverse(arr) (= 2 + 7 = 9) and reverse(arr) + arr(= 2 + 7 = 9).
• (2, 3): arr + reverse(arr) (= 11 + 7 = 18) and reverse(arr) + arr(= 11 + 7 = 18).

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 steps, 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 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 ` `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);`   `    ``// 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 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.`

## C#

 `// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic; `   `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])` `    ``Dictionary<``int``,` `               ``int``> u_map = ``new` `Dictionary<``int``,` `                                           ``int``>();` `                                           `  `    ``// 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` `        ``if` `(u_map.ContainsKey(val))` `            ``pairs += u_map[val];`   `        ``// Increment frequency of val` `        ``if` `(u_map.ContainsKey(val))` `            ``u_map[val]++;` `        ``else` `            ``u_map.Add(val, 1);` `    ``}`   `    ``// Print the number of pairs` `    ``Console.Write(pairs);` `}`   `// Driver Code` `public` `static` `void` `Main()` `{` `    ``int` `[]arr = { 2, 15, 11, 7 };` `    ``int` `N = arr.Length;`   `    ``// Function Call` `    ``countPairs(arr, N);` `}` `}`   `// This code is contributed by SURENDRA_GANGWAR`

## Javascript

 ``

Output

`3`

Time Complexity: O(N*log10 M), where M is the largest element in the array.
Auxiliary Space: O(N)

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