Find the sum of all possible pairs in an array of N elements
Given an array arr[] of N integers, the task is to find the sum of all the pairs possible from the given array. Note that,
- (arr[i], arr[i]) is also considered as a valid pair.
- (arr[i], arr[j]) and (arr[j], arr[i]) are considered as two different pairs.
Examples:
Input: arr[] = {1, 2}
Output: 12
All valid pairs are (1, 1), (1, 2), (2, 1) and (2, 2).
1 + 1 + 1 + 2 + 2 + 1 + 2 + 2 = 12Input: arr[] = {1, 2, 3, 1, 4}
Output: 110
Naive approach: Find all the possible pairs and calculate the sum of the elements of each pair.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the sum of the elements // of all possible pairs from the array int sumPairs(int arr[], int n) { // To store the required sum int sum = 0; // Nested loop for all possible pairs for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { // Add the sum of the elements // of the current pair sum += (arr[i] + arr[j]); } } return sum; } // Driver code int main() { int arr[] = { 1, 2, 3 }; int n = sizeof(arr) / sizeof(arr[0]); cout << sumPairs(arr, n); return 0; } |
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Java
// Java implementation of the approach import java.util.*; class GFG { // Function to return the sum of the elements // of all possible pairs from the array static int sumPairs(int arr[], int n) { // To store the required sum int sum = 0; // Nested loop for all possible pairs for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { // Add the sum of the elements // of the current pair sum += (arr[i] + arr[j]); } } return sum; } // Driver code public static void main(String[] args) { int arr[] = {1, 2, 3}; int n = arr.length; System.out.println(sumPairs(arr, n)); } } // This code is contributed by PrinciRaj1992 |
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Python3
# Python3 implementation of the approach # Function to return the summ of the elements # of all possible pairs from the array def summPairs(arr, n): # To store the required summ summ = 0 # Nested loop for all possible pairs for i in range(n): for j in range(n): # Add the summ of the elements # of the current pair summ += (arr[i] + arr[j]) return summ # Driver code arr = [1, 2, 3] n = len(arr) print(summPairs(arr, n)) # This code is contributed by Mohit Kumar |
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C#
// C# implementation of the approach using System; class GFG { // Function to return the sum of the elements // of all possible pairs from the array static int sumPairs(int []arr, int n) { // To store the required sum int sum = 0; // Nested loop for all possible pairs for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { // Add the sum of the elements // of the current pair sum += (arr[i] + arr[j]); } } return sum; } // Driver code public static void Main(String[] args) { int []arr = {1, 2, 3}; int n = arr.Length; Console.WriteLine(sumPairs(arr, n)); } } // This code is contributed by PrinciRaj1992 |
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Output:
36
Time Complexity: O(N2)
Efficient approach: It can be observed that each element appears exactly (2 * N) times as one of the elements of the pair (x, y). Exactly N times as x and exactly N times as y.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the sum of the elements // of all possible pairs from the array int sumPairs(int arr[], int n) { // To store the required sum int sum = 0; // For every element of the array for (int i = 0; i < n; i++) { // It appears (2 * n) times sum = sum + (arr[i] * (2 * n)); } return sum; } // Driver code int main() { int arr[] = { 1, 2, 3 }; int n = sizeof(arr) / sizeof(arr[0]); cout << sumPairs(arr, n); return 0; } |
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Java
// Java implementation of the approach import java.util.*; class GFG { // Function to return the sum of the elements // of all possible pairs from the array static int sumPairs(int arr[], int n) { // To store the required sum int sum = 0; // For every element of the array for (int i = 0; i < n; i++) { // It appears (2 * n) times sum = sum + (arr[i] * (2 * n)); } return sum; } // Driver code static public void main(String []arg) { int arr[] = { 1, 2, 3 }; int n = arr.length; System.out.println(sumPairs(arr, n)); } } // This code is contributed by 29AjayKumar |
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Python3
# Python3 implementation of the approach # Function to return the sum of the elements # of all possible pairs from the array def sumPairs(arr, n) : # To store the required sum sum = 0; # For every element of the array for i in range(n) : # It appears (2 * n) times sum = sum + (arr[i] * (2 * n)); return sum; # Driver code if __name__ == "__main__" : arr = [ 1, 2, 3 ]; n = len(arr); print(sumPairs(arr, n)); # This code is contributed by AnkitRai01 |
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C#
// C# implementation of the approach using System; class GFG { // Function to return the sum of the elements // of all possible pairs from the array static int sumPairs(int []arr, int n) { // To store the required sum int sum = 0; // For every element of the array for (int i = 0; i < n; i++) { // It appears (2 * n) times sum = sum + (arr[i] * (2 * n)); } return sum; } // Driver code static public void Main(String []arg) { int []arr = { 1, 2, 3 }; int n = arr.Length; Console.WriteLine(sumPairs(arr, n)); } } // This code contributed by Rajput-Ji |
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Output:
36
Time Complexity: O(N)
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