Sum of squares of differences between all pairs of an array

Given an array arr[] of size N, the task is to compute the sum of squares of differences of all possible pairs.

Examples:

Input: arr[] = {2, 8, 4}
Output: 56
Explanation: 
Sum of squared differences of all possible pairs = (2 – 8)² + (2 – 4)² + (8 – 4)² = 56

Input: arr[] = {-5, 8, 9, -4, -3}
Output: 950

Naive Approach: The simplest approach is to generate all possible pairs and calculate the square of their differences of each pair and keep adding it to a variable, say sum. Finally, print the value of sum.

Time Complexity: O(N²)
Auxiliary Space: O(1)

Efficient Approach: The optimal idea is based on rearranging the expression in the following manner:

∑_i=2 ⁿ ∑_j=1^i-1 (A_i-A_j)²

Since A_i – A_i  = 0, the above expression can be rearranged as 1/2*(∑_i=1ⁿ ∑_j=1ⁿ (A_i-A_j)²) which can be further be simplified using the identity:

(A – B)² = A² + B² – 2 * A * B 

As 1/2*(∑_i=1ⁿ ∑_j=1ⁿ (A_i² + A_j² – 2*A_i*A_j)) = 1/2 * (2*n*∑_i=1ⁿ A_i²  – 2*∑_i=1ⁿ (A_j * ∑_j=1ⁿ A_j))

The final expression is n*∑_i=1ⁿ A_i² – (∑_i=1ⁿA_i)² which can be computed by linearly iterating over the array.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

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

// Function to calculate sum of squares
// of differences of all possible pairs
void sumOfSquaredDifferences(int arr[],
                             int N)
{
    // Stores the final sum
    int ans = 0;

    // Stores temporary values
    int sumA = 0, sumB = 0;

    // Traverse the array
    for (int i = 0; i < N; i++) {
        sumA += (arr[i] * arr[i]);
        sumB += arr[i];
    }

    sumA = N * sumA;
    sumB = (sumB * sumB);

    // Final sum
    ans = sumA - sumB;

    // Print the answer
    cout << ans;
}

// Driver Code
int main()
{
    // Given array
    int arr[] = { 2, 8, 4 };

    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function call to find sum of square
    // of differences of all possible pairs
    sumOfSquaredDifferences(arr, N);

    return 0;
}

Java

// Java program for the above approach
class GFG{

// Function to calculate sum of squares
// of differences of all possible pairs
static void sumOfSquaredDifferences(int arr[],
                                    int N)
{

    // Stores the final sum
    int ans = 0;

    // Stores temporary values
    int sumA = 0, sumB = 0;

    // Traverse the array
    for(int i = 0; i < N; i++)
    {
        sumA += (arr[i] * arr[i]);
        sumB += arr[i];
    }

    sumA = N * sumA;
    sumB = (sumB * sumB);

    // Final sum
    ans = sumA - sumB;

    // Print the answer
    System.out.println(ans);
}

// Driver Code
public static void main (String[] args)
{

    // Given array
    int arr[] = { 2, 8, 4 };

    // Size of the array
    int N = arr.length;

    // Function call to find sum of square
    // of differences of all possible pairs
    sumOfSquaredDifferences(arr, N);
}
}

// This code is contributed by AnkThon

Python3

# Python3 program for the above approach

# Function to calculate sum of squares
# of differences of all possible pairs
def sumOfSquaredDifferences(arr, N):

    # Stores the final sum
    ans = 0

    # Stores temporary values
    sumA, sumB = 0, 0

    # Traverse the array
    for i in range(N):
        sumA += (arr[i] * arr[i])
        sumB += arr[i]

    sumA = N * sumA
    sumB = (sumB * sumB)

    # Final sum
    ans = sumA - sumB

    # Print the answer
    print(ans)

# Driver Code
if __name__ == '__main__':

    # Given array
    arr = [2, 8, 4]

    # Size of the array
    N = len(arr)

    # Function call to find sum of square
    # of differences of all possible pairs
    sumOfSquaredDifferences(arr, N)

# This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;

class GFG{

// Function to calculate sum of squares
// of differences of all possible pairs
static void sumOfSquaredDifferences(int []arr,
                                    int N)
{

    // Stores the final sum
    int ans = 0;

    // Stores temporary values
    int sumA = 0, sumB = 0;

    // Traverse the array
    for(int i = 0; i < N; i++)
    {
        sumA += (arr[i] * arr[i]);
        sumB += arr[i];
    }

    sumA = N * sumA;
    sumB = (sumB * sumB);

    // Final sum
    ans = sumA - sumB;

    // Print the answer
    Console.WriteLine(ans);
}

// Driver Code
public static void Main(string[] args)
{

    // Given array
    int []arr = { 2, 8, 4 };

    // Size of the array
    int N = arr.Length;

    // Function call to find sum of square
    // of differences of all possible pairs
    sumOfSquaredDifferences(arr, N);
}
}

// This code is contributed by AnkThon

Javascript

<script>

// Javascript program for the above approach

// Function to calculate sum of squares
// of differences of all possible pairs
function sumOfSquaredDifferences(arr, N)
{

    // Stores the final sum
    let ans = 0;

    // Stores temporary values
    let sumA = 0, sumB = 0;

    // Traverse the array
    for(let i = 0; i < N; i++)
    {
        sumA += (arr[i] * arr[i]);
        sumB += arr[i];
    }

    sumA = N * sumA;
    sumB = (sumB * sumB);

    // Final sum
    ans = sumA - sumB;

    // Print the answer
    document.write(ans);
}

// Driver Code

// Given array
let arr = [ 2, 8, 4 ];

// Size of the array
let N = arr.length;

// Function call to find sum of square
// of differences of all possible pairs
sumOfSquaredDifferences(arr, N);

// This code is contributed by subhammahato348

</script>

Output:

56

Time Complexity: O(N)
Auxiliary Space: O(1)