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Count of pairs in Array with difference equal to the difference with digits reversed

  • Last Updated : 15 Nov, 2021

Given an array arr[] of N integers, the task is to find the number of pairs of array elements (arr[i], arr[j]) such that the difference between the pairs is equal to the difference when the digits of both the numbers are reversed. 

Examples:

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Input: arr[] = {42, 11, 1, 97}
Output: 2
Explanation:
The valid pairs of array elements are (42, 97), (11, 1) as:
1. 42 – 97 = 24 – 79 = (-55)
2. 11 – 1   = 11 – 1 = (10)



Input: arr[] = {1, 2, 3, 4}
Output: 6

 

Approach: The given problem can be solved by using Hashing which is based on the following observations:

A valid pair (i, j) will follow the equation as

=> arr[i] – arr[j] = rev(arr[i]) – rev(arr[j])
=> arr[i] – rev(arr[i]) = arr[j] – rev(arr[j])

Follow the below steps to solve the problem:

  • Now, create a function reverseDigits, which will take an integer as an argument and reverse the digits of that integer.
  • Store the frequency of values arr[i] – rev(arr[i]) in an unordered map, say mp.
  • For each key(= difference) of frequency X the number of pairs that can be formed is given by \binom{X}{2} = \frac{X * (X - 1)}{2}          .
  • The total count of pairs is given by the sum of the value of the above expression for each frequency stored in the map mp.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to reverse the digits
// of an integer
int reverseDigits(int n)
{
    // Convert the given number
    // to a string
    string s = to_string(n);
 
    // Reverse the string
    reverse(s.begin(), s.end());
 
    // Return the value of the string
    return stoi(s);
}
int countValidPairs(vector<int> arr)
{
    // Stores resultant count of pairs
    long long res = 0;
 
    // Stores the frequencies of
    // differences
    unordered_map<int, int> mp;
    for (int i = 0; i < arr.size(); i++) {
        mp[arr[i] - reverseDigits(arr[i])]++;
    }
 
    // Traverse the map and count pairs
    // formed for all frequency values
    for (auto i : mp) {
        long long int t = i.second;
        res += t * (t - 1) / 2;
    }
 
    // Return the resultant count
    return res;
}
 
// Driver Code
int main()
{
    vector<int> arr = { 1, 2, 3, 4 };
    cout << countValidPairs(arr);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.HashMap;
 
class GFG {
 
    // Function to reverse the digits
    // of an integer
    public static int reverseDigits(int n)
    {
       
        // Convert the given number
        // to a string
        String s = String.valueOf(n);
 
        // Reverse the string
        s = new StringBuffer(s).reverse().toString();
 
        // Return the value of the string
        return Integer.parseInt(s);
    }
 
    public static int countValidPairs(int[] arr)
    {
       
        // Stores resultant count of pairs
        int res = 0;
 
        // Stores the frequencies of
        // differences
        HashMap<Integer, Integer> mp = new HashMap<Integer, Integer>();
        for (int i = 0; i < arr.length; i++) {
            if (mp.containsKey(arr[i] - reverseDigits(arr[i]))) {
                mp.put(arr[i] - reverseDigits(arr[i]), mp.get(arr[i] - reverseDigits(arr[i])) + 1);
            } else {
                mp.put(arr[i] - reverseDigits(arr[i]), 1);
            }
        }
 
        // Traverse the map and count pairs
        // formed for all frequency values
        for (int i : mp.keySet()) {
            int t = mp.get(i);
            res += t * (t - 1) / 2;
        }
 
        // Return the resultant count
        return res;
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int[] arr = { 1, 2, 3, 4 };
        System.out.println(countValidPairs(arr));
    }
 
}
 
// This code is contributed by saurabh_jaiswal.

Python3




# python program for the above approach
 
# Function to reverse the digits
# of an integer
def reverseDigits(n):
 
    # Convert the given number
    # to a string
    s = str(n)
 
    # Reverse the string
    s = "".join(reversed(s))
 
    # Return the value of the string
    return int(s)
 
def countValidPairs(arr):
 
    # Stores resultant count of pairs
    res = 0
 
    # Stores the frequencies of
    # differences
    mp = {}
 
    for i in range(0, len(arr)):
        if not arr[i] - reverseDigits(arr[i]) in mp:
            mp[arr[i] - reverseDigits(arr[i])] = 1
        else:
            mp[arr[i] - reverseDigits(arr[i])] += 1
 
        # Traverse the map and count pairs
        # formed for all frequency values
    for i in mp:
        t = mp[i]
        res += (t * (t - 1)) // 2
 
        # Return the resultant count
    return res
 
# Driver Code
if __name__ == "__main__":
 
    arr = [1, 2, 3, 4]
    print(countValidPairs(arr))
 
    # This code is contributed by rakeshsahni

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG {
 
    // Function to reverse the digits
    // of an integer
    public static int reverseDigits(int n)
    {
 
        // Convert the given number
        // to a string
        string s = n.ToString();
 
        // Reverse the string
        char[] arr = s.ToCharArray();
        Array.Reverse(arr);
        string st = new string(arr);
 
        // Return the value of the string
        return Int32.Parse(st);
    }
 
    public static int countValidPairs(int[] arr)
    {
 
        // Stores resultant count of pairs
        int res = 0;
 
        // Stores the frequencies of
        // differences
        Dictionary<int, int> mp
            = new Dictionary<int, int>();
        for (int i = 0; i < arr.Length; i++) {
            if (mp.ContainsKey(arr[i]
                               - reverseDigits(arr[i]))) {
                mp[arr[i] - reverseDigits(arr[i])]
                    = mp[arr[i] - reverseDigits(arr[i])]
                      + 1;
            }
            else {
                mp[arr[i] - reverseDigits(arr[i])] = 1;
            }
        }
 
        // Traverse the map and count pairs
        // formed for all frequency values
        foreach(int i in mp.Keys)
        {
            int t = mp[i];
            res += t * (t - 1) / 2;
        }
 
        // Return the resultant count
        return res;
    }
 
    // Driver Code
    public static void Main()
    {
        int[] arr = { 1, 2, 3, 4 };
        Console.WriteLine(countValidPairs(arr));
    }
}
 
// This code is contributed by ukasp.

Javascript




<script>
// Javascript program for the above approach
 
// Function to reverse the digits
// of an integer
function reverseDigits(n)
{
 
  // Convert the given number
  // to a string
  let s = new String(n);
 
  // Reverse the string
  s = s.split("").reverse().join("");
 
  // Return the value of the string
  return parseInt(s);
}
function countValidPairs(arr)
{
 
  // Stores resultant count of pairs
  let res = 0;
 
  // Stores the frequencies of
  // differences
  let mp = new Map();
  for (let i = 0; i < arr.length; i++) {
    let temp = arr[i] - reverseDigits(arr[i]);
    if (mp.has(temp)) {
      mp.set(temp, mp.get(temp) + 1);
    } else {
      mp.set(temp, 1);
    }
  }
 
  // Traverse the map and count pairs
  // formed for all frequency values
  for (i of mp) {
    let t = i[1];
    res += (t * (t - 1)) / 2;
  }
 
  // Return the resultant count
  return res;
}
 
// Driver Code
let arr = [1, 2, 3, 4];
document.write(countValidPairs(arr));
 
// This code is contributed by gfgking.
</script>
Output: 
6

 

Time Complexity: O(N)
Auxiliary Space: O(N)




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