数组所有对之间的差平方和

给定大小为N的数组arr [] ，任务是计算所有可能的对的差的平方和。

例子：

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

天真的方法：最简单的方法是生成所有可能的对，并计算每对对的差平方，然后将其继续添加到变量中，例如sum 。最后，打印sum的值。

时间复杂度： O(N ² )
辅助空间： O(1)

高效的方法：最佳的想法是基于以下方式重新排列表达式：

∑_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.

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
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
 
    # Prthe 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

 
// 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
 

输出：

56

时间复杂度： O(N)
辅助空间： O(1)