Creating Array of Absolute Position Differences for Equal Elements

Given an array arr[] of size N. The task is to create a new array result[] of the same size, where result[i] = ∑ |i – j| for all j such that arr[i] = arr[j].

Examples:

Input: arr = {2, 1, 4, 1, 2, 4, 4}
Output: {4, 2, 7, 2, 4, 4, 5}
Explanation:
=> 0th Index: Another 2 is found at index 4. |0 – 4| = 4
=> 1st Index: Another 1 is found at index 3. |1 – 3| = 2
=> 2nd Index: Two more 4s are found at indices 5 and 6. |2 – 5| + |2 – 6| = 7
=> 3rd Index: Another 1 is found at index 1. |3 – 1| = 2
=> 4th Index: Another 2 is found at index 0. |4 – 0| = 4
=> 5th Index: Two more 4s are found at indices 2 and 6. |5 – 2| + |5 – 6| = 4
=> 6th Index: Two more 4s are found at indices 2 and 5. |6 – 2| + |6 – 5| = 5

Input: arr = {1, 10, 510, 10}
Output: {0, 5, 0, 3, 4}

Constructing an Element Summation Array using Hashing:

To compute result[i], we can apply the following approach:

• Compute leftFreq, the frequency of the same elements to the left of index i.
• Calculate leftSum, the summation of indices where the same element occurs to the left of index i.
• Compute rightSum, the summation of indices where the same element occurs to the right of index i.
• Calculate rightFreq, the frequency of the same elements to the right of index i.
• Compute result[i] using the formula: result[i] = ((leftFreq * i) – leftSum) + (rightSum – (rightFreq * i))

Illustration:

Consider the input arr = {3, 2, 3, 3, 2, 3}

Now, If have to calculate the result for the index i = 3 which is element 3.

• To calculate the absolute value of the required result from the left would be something like => (i – 0) + (i – 2) , which is equals to (2 * i – (0 + 2)).
• To generalise these, we can generate a formula to do the same: (frequency of similar element to the left* i) – (Sum of occurances of such indices).
• To calculate the absolute value of required result from the right would be something like => (5 – i)
• To generalise these, we can generate a formula to do the same: (Sum of occurances of such indices) – (frequency of similar element to the right* i)
• We can conclude from the above that: The result for the ith index would be:
• result[i] = ((leftFreq * i) – leftSum) + (rightSum – (rightFreq * i)).

Step-by-step approach:

• Initialize a map “unmap” of <key, pair<total frequency, total sum>>.
• Initialize another map unmap2 of <key, pair< frequency till any index i, sum till ith index>>.
• Iterate over the given array and update the unmap.
• Create an array result[] for storing the answer
• Iterate over the given array
  • Update the result[i] = ((leftFreq * i) – leftSum) + (rightSum – (rightFreq * i)).
  • Update the unmap2 by increasing the current element’s frequency and the sum of all occurrences of similar elements till the ith index.
• Finally, return the result[] as the required answer.

Below is the implementation of the above approach:

C++

// C++ code to implement the approach

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

// Function to find the resultant array
vector<long long> solve(vector<long long>& arr)
{
    int n = arr.size();

    // Declare a unordered map unmap for
    // storing the <key, <Total frequency,
    // total sum>> Declare a unordered map
    // unmap2 for storing the <key,
    // <frequency till any index i, sum
    // till ith index>>
    unordered_map<int, pair<long long, long long> > unmap,
        unmap2; // < key, <freq, sum>>

    // Iterate over the array and
    // update the unmap
    for (int i = 0; i < n; i++) {
        unmap[arr[i]] = { unmap[arr[i]].first + 1,
                        unmap[arr[i]].second + i };
    }

    // Declare an array result for
    // storing the result
    vector<long long> result(n);

    // Iterate over the given array
    for (int i = 0; i < n; i++) {
        auto curr = unmap2[arr[i]];

        long long leftSum = curr.second;
        long long leftFreq = curr.first;

        long long rightSum
            = unmap[arr[i]].second - leftSum - i;
        long long rightFreq
            = unmap[arr[i]].first - leftFreq - 1;

        // Update the result[i] with value mentioned
        result[i] = ((leftFreq * i) - leftSum)
                    + (rightSum - (rightFreq * i));

        // Update the unmap2 by increasing
        // the current elements frequency
        // and sum of all occurance of
        // similar element till ith index.
        unmap2[arr[i]] = { unmap2[arr[i]].first + 1,
                        unmap2[arr[i]].second + i };
    }

    // Finally return result.
    return result;
}

// Driver code
int main()
{
    vector<long long> arr = { 2, 1, 4, 1, 2, 4, 4 };

    // Function call
    vector<long long> result = solve(arr);

    for (auto i : result) {
        cout << i << " ";
    }

    return 0;
}

Java

import java.util.HashMap;
import java.util.Map;
import java.util.Vector;

public class Main {
    // Function to find the resultant array
    static Vector<Long> solve(Vector<Long> arr)
    {
        int n = arr.size();

        // Declare a map for storing the <key, <Total
        // frequency, total sum>>.
        Map<Long, Pair<Long, Long> > unmap
            = new HashMap<>();
        // Declare a map for storing the <key, <frequency
        // till any index i, sum till ith index>>.
        Map<Long, Pair<Long, Long> > unmap2
            = new HashMap<>();

        // Iterate over the array and update the unmap
        for (int i = 0; i < n; i++) {
            long key = arr.get(i);
            Pair<Long, Long> value = unmap.getOrDefault(
                key, new Pair<>(0L, 0L));
            unmap.put(key, new Pair<>(value.first + 1,
                                      value.second + i));
        }

        // Declare an array result for storing the result
        Vector<Long> result = new Vector<>(n);

        // Iterate over the given array
        for (int i = 0; i < n; i++) {
            long key = arr.get(i);
            Pair<Long, Long> curr = unmap2.getOrDefault(
                key, new Pair<>(0L, 0L));

            long leftSum = curr.second;
            long leftFreq = curr.first;

            Pair<Long, Long> value = unmap.get(key);
            long rightSum = value.second - leftSum - i;
            long rightFreq = value.first - leftFreq - 1;

            // Update the result[i] with value mentioned
            result.add(((leftFreq * i) - leftSum)
                       + (rightSum - (rightFreq * i)));

            // Update the unmap2 by increasing the current
            // elements frequency and sum of all occurrences
            // of similar elements till ith index.
            unmap2.put(key, new Pair<>(curr.first + 1,
                                       curr.second + i));
        }

        // Finally return result.
        return result;
    }

    // Driver code
    public static void main(String[] args)
    {
        Vector<Long> arr = new Vector<>();
        arr.add(2L);
        arr.add(1L);
        arr.add(4L);
        arr.add(1L);
        arr.add(2L);
        arr.add(4L);
        arr.add(4L);

        // Function call
        Vector<Long> result = solve(arr);

        // Print the result
        for (long i : result) {
            System.out.print(i + " ");
        }
    }

    static class Pair<T, U> {
        T first;
        U second;

        Pair(T first, U second)
        {
            this.first = first;
            this.second = second;
        }
    }
}

Python3

from typing import List, Tuple

def solve(arr: List[int]) -> List[int]:
    n = len(arr)

    # Declare a dictionary for storing the <key, <Total frequency, total sum>>.
    unmap = {}
    # Declare a dictionary for storing the <key, <frequency till any index i, sum till ith index>>.
    unmap2 = {}

    # Iterate over the array and update the unmap
    for i in range(n):
        key = arr[i]
        value = unmap.get(key, (0, 0))
        unmap[key] = (value[0] + 1, value[1] + i)

    # Declare a list result for storing the result
    result = []

    # Iterate over the given array
    for i in range(n):
        key = arr[i]
        curr = unmap2.get(key, (0, 0))

        leftSum, leftFreq = curr[1], curr[0]

        value = unmap[key]
        rightSum = value[1] - leftSum - i
        rightFreq = value[0] - leftFreq - 1

        # Update the result[i] with value mentioned
        result.append(((leftFreq * i) - leftSum) + (rightSum - (rightFreq * i)))

        # Update the unmap2 by increasing the current
        # elements frequency and sum of all occurrences
        # of similar elements till ith index.
        unmap2[key] = (curr[0] + 1, curr[1] + i)

    # Finally return result.
    return result

# Driver code
if __name__ == "__main__":
    arr = [2, 1, 4, 1, 2, 4, 4]

    # Function call
    result = solve(arr)

    # Print the result
    print(*result)

C#

using System;
using System.Collections.Generic;

class GFG
{
    public static List<long> Solve(List<long> arr)
    {
        int n = arr.Count;
        var unmap = new Dictionary<long, Tuple<long, long>>();
        var unmap2 = new Dictionary<long, Tuple<long, long>>();

        for (int i = 0; i < n; i++)
        {
            long key = arr[i];
            if (!unmap.TryGetValue(key, out var value))
            {
                value = new Tuple<long, long>(0, 0);
            }
            unmap[key] = new Tuple<long, long>(value.Item1 + 1, value.Item2 + i);
        }

        List<long> result = new List<long>();

        for (int i = 0; i < n; i++)
        {
            long key = arr[i];
            if (!unmap2.TryGetValue(key, out var curr))
            {
                curr = new Tuple<long, long>(0, 0);
            }

            long leftSum = curr.Item2;
            long leftFreq = curr.Item1;

            var value = unmap[key];
            long rightSum = value.Item2 - leftSum - i;
            long rightFreq = value.Item1 - leftFreq - 1;

            result.Add(((leftFreq * i) - leftSum) + (rightSum - (rightFreq * i)));

            unmap2[key] = new Tuple<long, long>(curr.Item1 + 1, curr.Item2 + i);
        }

        return result;
    }

    static void Main(string[] args)
    {
        List<long> arr = new List<long> { 2, 1, 4, 1, 2, 4, 4 };
        List<long> result = Solve(arr);
        foreach (var item in result)
        {
            Console.Write(item + " ");
        }
    }
}

Javascript

// Function to find the resultant array
function solve(arr) {
    const n = arr.length;

    // Declare a Map for storing the <key, <Total frequency, total sum>>
    const unmap = new Map();
    // Declare a Map for storing the <key, <frequency till any index i, sum till ith index>>
    const unmap2 = new Map();

    // Iterate over the array and update the unmap
    for (let i = 0; i < n; i++) {
        const key = arr[i];
        if (!unmap.has(key)) {
            unmap.set(key, [0, 0]);
        }
        const [freq, sum] = unmap.get(key);
        unmap.set(key, [freq + 1, sum + i]);
    }

    // Declare an array result for storing the result
    const result = [];

    // Iterate over the given array
    for (let i = 0; i < n; i++) {
        const key = arr[i];
        const curr = unmap2.get(key) || [0, 0];

        const leftSum = curr[1];
        const leftFreq = curr[0];

        const [freq, sum] = unmap.get(key);
        const rightSum = sum - leftSum - i;
        const rightFreq = freq - leftFreq - 1;

        // Update the result[i] with value mentioned
        result.push((leftFreq * i) - leftSum + rightSum - (rightFreq * i));

        // Update the unmap2 by increasing the current element's frequency
        // and sum of all occurrences of a similar element till ith index.
        unmap2.set(key, [curr[0] + 1, curr[1] + i]);
    }

    // Finally return result.
    return result;
}

// Driver code
const arr = [2, 1, 4, 1, 2, 4, 4];

// Function call
const result = solve(arr);

console.log(result.join(' '));

// This code is contributed by rambabuguphka

Output

4 2 7 2 4 4 5 

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