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Sum of absolute differences of indices of occurrences of each array element
  • Last Updated : 25 Nov, 2020

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.

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

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



  1. Initialize a Map to store the vector of indices for repetitions of each unique element present in the array.
  2. 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].
  3. For each value j present in the vector, increment the sum by abs(i – j).
  4. 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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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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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