Count pairs such that difference between them and indices are different

Given an array arr[] of size N, the task is to count all pair of indices (i, j) such that i < j and j - i != arr[j] - arr[i].

Examples:

Input: arr[] = {4, 1, 3, 3}
Output: 5
Explanation:
The pair (0, 1) is a valid pair since 1 - 0 != 1 - 4.
The pair (0, 2) is a valid pair since 2 - 0 != 3 - 4, 2 != -1.
The pair (0, 3) is a valid pair since 3 - 0 != 3 - 4, 3 != -1.
The pair (1, 2) is a valid pair since 2 - 1 != 3 - 1, 1 != 2.
The pair (2, 3) is a valid pair since 3 - 2 != 3 - 3, 1 != 0.
There are a total of 5 valid pairs, so we return 5.

Input: arr = {1, 2, 3, 4, 5}
Output: 0
Explanation: There are no valid pairs.

An approach using Hashing:

As per the valid pair is concerned the value of  j - i != arr[j] - arr[i], This can also be written as j - arr[j] != i - arr[i]. 

We'll use map to keep the count of element that has value (i - arr[i]) occurred till i^th index. If we are at i^th index then the valid pair for i^th element would be just (Total number of pairs till i^th index - count of element that has value (i - arr[i]) till i^th index).

Follow the steps below to implement the above idea:

• Initialize a map for storing the value of (i - arr[i])
• Initialize a variable result for storing the count of all valid pairs
• Iterate over the given array
  • Store the count of all the previous element that has value (i - arr[i]) into a variable count, as this will act as all invalid pair with the element at the i^th index. 
  • Add all valid pairs by removing the invalid pairs into the result (i.e, result += i - count).
  • Increment the count of (i - arr[i]) into map for every i^th index.
• Return the result.

Below is the implementation of the above approach.

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the number of valid pairs
long long validPairs(vector<int>& arr)
{
    // Initialize a map for storing the
    // value of (i - arr[i])
    unordered_map<int, int> unmap;

    // Initialize a variable result for
    // storing the count of all valid
    // pairs
    long long result = 0;

    // Iterate over the given array
    for (int i = 0; i < arr.size(); i++) {

        // Calculate number of time value
        // (i - arr[i]) has occurred till
        // ith index All the previous
        // element that has value
        // (i - arr[i]) will be act as
        // invalid Pair with element at
        // ith index.
        long long count = unmap[i - arr[i]];

        // Add all valid pair by remove
        // the invalid pairs
        result += i - count;

        // Store the value of (i - arr[i])
        // into map
        unmap[i - arr[i]]++;
    }

    // Return the result.
    return result;
}

// Driver's code
int main()
{
    vector<int> arr = { 4, 1, 3, 3 };

    // Function Call
    cout << validPairs(arr);

    return 0;
}

// Java code to implement the approach
import java.io.*;
import java.util.*;

class GFG {

  // Function to find the number of valid pairs
  static long validPairs(int[] arr)
  {

    // Initialize a map for storing the
    // value of (i - arr[i])
    HashMap<Integer, Integer> unmap = new HashMap<>();

    // Initialize a variable result for
    // storing the count of all valid
    // pairs
    long result = 0;

    // Iterate over the given array
    for (int i = 0; i < arr.length; i++) {
      long count = 0;

      // Calculate number of time value
      // (i - arr[i]) has occurred till
      // ith index All the previous
      // element that has value
      // (i - arr[i]) will be act as
      // invalid Pair with element at
      // ith index.
      if (unmap.containsKey(i - arr[i])) {
        count = unmap.get(i - arr[i]);
      }

      // Add all valid pair by remove
      // the invalid pairs
      result += i - count;

      // Store the value of (i - arr[i])
      // into map
      if (unmap.containsKey(i - arr[i])) {
        unmap.put(i - arr[i],
                  unmap.get(i - arr[i]) + 1);
      }
      else {
        unmap.put(i - arr[i], 1);
      }
    }
    // Return the result.
    return result;
  }

  public static void main(String[] args)
  {
    int[] arr = { 4, 1, 3, 3 };

    // Function call
    System.out.print(validPairs(arr));
  }
}

// This code is contributed by lokesh

# Python code to implement the approach

# Function to find the number of valid pairs
def validPairs(arr):
    # Initialize a map for storing the
    # value of (i - arr[i])
    unmap={}
    
    # Initialize a variable result for
    # storing the count of all valid
    # pairs
    result=0
    
    # Iterate over the given array
    for i in range(len(arr)):
        # Calculate number of time value
        # (i - arr[i]) has occurred till
        # ith index All the previous
        # element that has value
        # (i - arr[i]) will be act as
        # invalid Pair with element at
        # ith index.
        count=0
        if(i-arr[i] in unmap):
            count=unmap[i-arr[i]]
        
        # Add all valid pair by remove
        # the invalid pairs
        result=result+i-count
        
        # Store the value of (i - arr[i])
        # into map
        if(i-arr[i] in unmap):
            unmap[i-arr[i]]=unmap[i-arr[i]]+1
        else:
            unmap[i-arr[i]]=1
    
    # Return the result.
    return result
    
# Driver code
arr=[4,1,3,3]

# Function Call
print(validPairs(arr))


# This code is contributed Pushpesh Raj.

// C# code to implement the above approach
using System;
using System.Collections.Generic;

public class GFG
{
  // Function to find the number of valid pairs
  static long validPairs(int[] arr)
  {

    // Initialize a map for storing the
    // value of (i - arr[i])
    Dictionary<int,
    int> unmap = new Dictionary<int,int>();

    // Initialize a variable result for
    // storing the count of all valid
    // pairs
    long result = 0;

    // Iterate over the given array
    for (int i = 0; i < arr.Length; i++) {
      long count = 0;

      // Calculate number of time value
      // (i - arr[i]) has occurred till
      // ith index All the previous
      // element that has value
      // (i - arr[i]) will be act as
      // invalid Pair with element at
      // ith index.
      if (unmap.ContainsKey(i - arr[i])) {
        count = unmap[(i - arr[i])];
      }

      // Add all valid pair by remove
      // the invalid pairs
      result += i - count;

      // Store the value of (i - arr[i])
      // into map
      if (unmap.ContainsKey(i - arr[i])) {
        unmap[i - arr[i]]++;
      }
      else {
        unmap[i - arr[i]] = 1;
      }
    }
    
    // Return the result.
    return result;
  }

  // Driver function
  public static void Main(string[] args)
  {
    int[] arr = { 4, 1, 3, 3 };

    // Function call
    Console.WriteLine(validPairs(arr));
  }

}

// This code is contributed by sanjoy_62.

// JavaScript code for the above approach

// Function to find the number of valid pairs
function validPairs(arr)
{
    // Initialize a map for storing the
    // value of (i - arr[i])
    let unmap = new Map();

    // Initialize a variable result for
    // storing the count of all valid
    // pairs
    let result = 0;

    // Iterate over the given array
    for (let i = 0; i < arr.length; i++) {

        // Calculate number of time value
        // (i - arr[i]) has occurred till
        // ith index All the previous
        // element that has value
        // (i - arr[i]) will be act as
        // invalid Pair with element at
        // ith index.
        let count = 0;
        if (unmap.has(i - arr[i]))
            count = unmap.get(i - arr[i]);

        // Add all valid pair by remove
        // the invalid pairs
        result += i - count;

        // Store the value of (i - arr[i])
        // into map
        if (unmap.has(i - arr[i]))
            unmap.set(i - arr[i],
                      unmap.get(i - arr[i]) + 1);
        else
            unmap.set(i - arr[i], 1);
    }

    // Return the result.
    return result;
}

// Driver's code

let arr = [ 4, 1, 3, 3 ];

// Function Call
console.log(validPairs(arr));

// This code is contributed by Potta Lokesh

5

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