Sum of absolute differences of indices of occurrences of each array element

Last Updated : 23 Jul, 2025

Given an array arr[] consisting of N integers, the task for each array element arr[i] is to print the sum of |i - j| for all possible indices j such that arr[i] = arr[j].

Examples:

Input: arr[] = {1, 3, 1, 1, 2}
Output: 5 0 3 4 0
Explanation:
For arr[0], sum = |0 - 0| + |0 - 2| + |0 - 3| = 5.
For arr[1], sum = |1 - 1| = 0.
For arr[2], sum = |2 - 0| + |2 - 2| + |2 - 3| = 3.
For arr[3], sum = |3 - 0| + |3 - 2| + |3 - 3| = 4.
For arr[4], sum = |4 - 4| = 0.
Therefore, the required output is 5 0 3 4 0.

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

Naive approach: The simplest approach is to traverse the given array and for each element arr[i] ( 0 ? i ? N ), traverse the array and check if arr[i] is same as arr[j] ( 0 ? j ? N ). If found to be true, add abs(i - j) to the sum print the sum obtained for each array element.

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find sum of differences
// of indices of occurrences of each
// unique array element
void sum(int arr[], int n)
{
    // Traverse over the array
    for (int i = 0; i < n; i++) {
        int sum = 0;

        // Check for every other elements of the array
        for (int j = 0; j < n; j++) {

            // Add into sum if such pair found
            if (arr[i] == arr[j]) {
                sum += abs(i - j);
            }
        }

        // Print the sum
        cout << sum << " ";
    }
}

// Driver Code
int main()
{

    // Given array
    int arr[] = { 1, 3, 1, 1, 2 };

    // Given size
    int n = sizeof(arr) / sizeof(arr[0]);

    // Function call
    sum(arr, n);

    return 0;
}

Java

import java.util.Arrays;

public class Main {
    // Function to find sum of differences
    // of indices of occurrences of each
    // unique array element
    public static void sum(int[] arr, int n) {
        // Traverse over the array
        for (int i = 0; i < n; i++) {
            int sum = 0;

            // Check for every other elements of the array
            for (int j = 0; j < n; j++) {

                // Add into sum if such pair found
                if (arr[i] == arr[j]) {
                    sum += Math.abs(i - j);
                }
            }

            // Print the sum
            System.out.print(sum + " ");
        }
    }

    // Driver Code
    public static void main(String[] args) {
        // Given array
        int[] arr = {1, 3, 1, 1, 2};

        // Given size
        int n = arr.length;

        // Function call
        sum(arr, n);
    }
}

Python3

def sum(arr, n):
    # Traverse over the array
    for i in range(n):
        s = 0

        # Check for every other elements of the array
        for j in range(n):

            # Add into sum if such pair found
            if arr[i] == arr[j]:
                s += abs(i - j)

        # Print the sum
        print(s, end=" ")

# Driver code
if __name__ == '__main__':
    # Given array
    arr = [1, 3, 1, 1, 2]

    # Given size
    n = len(arr)

    # Function call
    sum(arr, n)

JavaScript

// JavaScript program for the above approach

// Function to find sum of differences
// of indices of occurrences of each
// unique array element
function sum(arr, n) {
    // Traverse over the array
    for (let i = 0; i < n; i++) {
        let sum = 0;

        // Check for every other elements of the array
        for (let j = 0; j < n; j++) {

            // Add into sum if such pair found
            if (arr[i] === arr[j]) {
                sum += Math.abs(i - j);
            }
        }

        // Print the sum
        console.log(sum + " ");
    }
}

// Driver Code
(function main() {
    // Given array
    let arr = [1, 3, 1, 1, 2];

    // Given size
    let n = arr.length;

    // Function call
    sum(arr, n);
})();

using System;

public class GFG
{
    // Function to find sum of differences
    // of indices of occurrences of each
    // unique array element
    public static void Sum(int[] arr, int n)
    {
        // Traverse over the array
        for (int i = 0; i < n; i++)
        {
            int sum = 0;

            // Check for every other elements of the array
            for (int j = 0; j < n; j++)
            {
                // Add into sum if such pair found
                if (arr[i] == arr[j])
                {
                    sum += Math.Abs(i - j);
                }
            }

            // Print the sum
            Console.Write(sum + " ");
        }
    }

    public static void Main()
    {
        // Given array
        int[] arr = { 1, 3, 1, 1, 2 };

        // Given size
        int n = arr.Length;

        // Function call
        Sum(arr, n);
    }
}

Output

5 0 3 4 0

Time Complexity: O(N²) where N is the size of the given array.
Auxiliary Space: O(1)

Efficient Approach: The idea is to use Map data structure to optimize the above approach. Follow the steps below to solve the problem:

Initialize a Map to store the vector of indices for repetitions of each unique element present in the array.
Traverse the given array from i = 0 to N - 1 and for each array element arr[i], initialize the sum with 0 and traverse the vector map[arr[i]] which stores the indices of the occurrences of the element arr[i].
For each value j present in the vector, increment the sum by abs(i - j).
After traversing the vector, store the sum for the element at index i and print the sum.

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 sum of differences
// of indices of occurrences of each
// unique array element
void sum(int arr[], int n)
{
    // Stores indices of each
    // array element
    map<int, vector<int> > mp;

    // Store the indices
    for (int i = 0; i < n; i++) {
        mp[arr[i]].push_back(i);
    }

    // Stores the sums
    int ans[n];

    // Traverse the array
    for (int i = 0; i < n; i++) {

        // Find sum for each element
        int sum = 0;

        // Iterate over the Map
        for (auto it : mp[arr[i]]) {

            // Calculate sum of
            // occurrences of arr[i]
            sum += abs(it - i);
        
        }

        // Store sum for
        // current element
        ans[i] = sum;
    }

    // Print answer for each element
    for (int i = 0; i < n; i++) {
        cout << ans[i] << " ";
    }
    return;
}

// Driver Code
int main()
{

    // Given array
    int arr[] = { 1, 3, 1, 1, 2 };

    // Given size
    int n = sizeof(arr)
            / sizeof(arr[0]);

    // Function call
    sum(arr, n);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find sum of differences
// of indices of occurrences of each
// unique array element
static void sum(int arr[], int n)
{
    
    // Stores indices of each
    // array element
    HashMap<Integer, Vector<Integer>> mp = new HashMap<>();

    // Store the indices
    for(int i = 0; i < n; i++) 
    {
        Vector<Integer> v = new Vector<>();
        v.add(i);
        
        if (mp.containsKey(arr[i]))
            v.addAll(mp.get(arr[i]));
        
        mp.put(arr[i], v);
    }

    // Stores the sums
    int []ans = new int[n];

    // Traverse the array
    for(int i = 0; i < n; i++) 
    {
        
        // Find sum for each element
        int sum = 0;
        
        // Iterate over the Map
        for(int it : mp.get(arr[i]))
        {
            
            // Calculate sum of
            // occurrences of arr[i]
            sum += Math.abs(it - i);
        }

        // Store sum for
        // current element
        ans[i] = sum;
    }

    // Print answer for each element
    for(int i = 0; i < n; i++)
    {
        System.out.print(ans[i] + " ");
    }
    return;
}

// Driver Code
public static void main(String[] args)
{
    
    // Given array
    int arr[] = { 1, 3, 1, 1, 2 };

    // Given size
    int n = arr.length;

    // Function call
    sum(arr, n);
}
}

// This code is contributed by Princi Singh

Python3

# Python3 program for the above approach 
from collections import defaultdict

# Function to find sum of differences 
# of indices of occurrences of each
# unique array element
def sum_i(arr, n):
  
    # Stores indices of each
    # array element
    mp = defaultdict(lambda : [])

    # Store the indices
    for i in range(n):
        mp[arr[i]].append(i)

    # Stores the sums
    ans = [0] * n

    # Traverse the array
    for i in range(n):

        # Find sum for each element
        sum = 0

        # Iterate over the Map
        for it in mp[arr[i]]:

            # Calculate sum of
            # occurrences of arr[i]
            sum += abs(it - i)

            # Store sum for
            # current element
            ans[i] = sum

    # Print answer for each element
    for i in range(n):
        print(ans[i], end = " ")

# Driver code
if __name__ == '__main__':
  
    # Given array
    arr = [ 1, 3, 1, 1, 2 ]

    # Given size
    n = len(arr)

    # Function Call
    sum_i(arr, n)

# This code is contributed by Shivam Singh

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to find sum of differences
// of indices of occurrences of each
// unique array element
static void sum(int []arr, int n)
{
    
    // Stores indices of each
    // array element
    Dictionary<int, 
          List<int>> mp = new Dictionary<int, 
                                    List<int>>();

    // Store the indices
    for(int i = 0; i < n; i++) 
    {
        List<int> v = new List<int>();
        v.Add(i);
        
        if (mp.ContainsKey(arr[i]))
            v.AddRange(mp[arr[i]]);
        
        mp[arr[i]]= v;
    }

    // Stores the sums
    int []ans = new int[n];

    // Traverse the array
    for(int i = 0; i < n; i++) 
    {
        
        // Find sum for each element
        int sum = 0;
        
        // Iterate over the Map
        foreach(int it in mp[arr[i]])
        {
            
            // Calculate sum of
            // occurrences of arr[i]
            sum += Math.Abs(it - i);
        }

        // Store sum for
        // current element
        ans[i] = sum;
    }

    // Print answer for each element
    for(int i = 0; i < n; i++)
    {
        Console.Write(ans[i] + " ");
    }
    return;
}

// Driver Code
public static void Main(String[] args)
{
    
    // Given array
    int []arr = { 1, 3, 1, 1, 2 };

    // Given size
    int n = arr.Length;

    // Function call
    sum(arr, n);
}
}

// This code is contributed by shikhasingrajput

JavaScript

<script>
 
// JavaScript program for the above approach

// Function to find sum of differences
// of indices of occurrences of each
// unique array element
function sum(arr, n)
{
    // Stores indices of each
    // array element
    var mp = new Map();

    // Store the indices
    for (var i = 0; i < n; i++) {
        if(mp.has(arr[i]))
        {
            var tmp = mp.get(arr[i]);
            tmp.push(i);
            mp.set(arr[i], tmp);
        }
        else
        {
            mp.set(arr[i], [i]);
        }
    }

    // Stores the sums
    var ans = Array(n);

    // Traverse the array
    for (var i = 0; i < n; i++) {

        // Find sum for each element
        var sum = 0;

        // Iterate over the Map
        mp.get(arr[i]).forEach(it => {

            // Calculate sum of
            // occurrences of arr[i]
            sum += Math.abs(it - i);
        
        });

        // Store sum for
        // current element
        ans[i] = sum;
    }

    // Print answer for each element
    for (var i = 0; i < n; i++) {
        document.write( ans[i] + " ");
    }
    return;
}

// Driver Code

// Given array
var arr = [1, 3, 1, 1, 2];

// Given size
var n = arr.length;

// Function call
sum(arr, n);

</script>

Output:

5 0 3 4 0

Time Complexity: O(N * L) where N is the size of the given array and L is the maximum frequency of any array element.
Auxiliary Space: O(N)

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