Given an array arr[] of N elements, the task is to find the product of absolute differences of all pairs in the given array.
Examples:
Input: arr[] = {1, 2, 3, 4}
Output: 12
Explanation:
Product of |2-1| * |3-1| * |4-1| * |3-2| * |4-2| * |4-3| = 12
Input: arr[] = {1, 8, 9, 15, 16}
Output: 27659520
Approach: The idea is to generate every possible pairs of the given array arr[] and find the product of the absolute difference of all the pairs.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to return the product of // abs diff of all pairs (x, y) int getProduct( int a[], int n)
{ // To store product
int p = 1;
// Iterate all possible pairs
for ( int i = 0; i < n; i++) {
for ( int j = i + 1; j < n; j++) {
// Find the product
p *= abs (a[i] - a[j]);
}
}
// Return product
return p;
} // Driver Code int main()
{ // Given array arr[]
int arr[] = { 1, 2, 3, 4 };
int N = sizeof (arr) / sizeof (arr[0]);
// Function Call
cout << getProduct(arr, N);
return 0;
} |
Java
// Java program for the above approach import java.util.*;
class GFG{
// Function to return the product of // abs diff of all pairs (x, y) static int getProduct( int a[], int n)
{ // To store product
int p = 1 ;
// Iterate all possible pairs
for ( int i = 0 ; i < n; i++)
{
for ( int j = i + 1 ; j < n; j++)
{
// Find the product
p *= Math.abs(a[i] - a[j]);
}
}
// Return product
return p;
} // Driver Code public static void main(String[] args)
{ // Given array arr[]
int arr[] = { 1 , 2 , 3 , 4 };
int N = arr.length;
// Function call
System.out.println(getProduct(arr, N));
} } // This code is contributed by Ritik Bansal |
Python3
# Python3 program for # the above approach # Function to return the product of # abs diff of all pairs (x, y) def getProduct(a, n):
# To store product
p = 1
# Iterate all possible pairs
for i in range (n):
for j in range (i + 1 , n):
# Find the product
p * = abs (a[i] - a[j])
# Return product
return p
# Driver Code if __name__ = = "__main__" :
# Given array arr[]
arr = [ 1 , 2 , 3 , 4 ]
N = len (arr)
# Function Call
print (getProduct(arr, N))
# This code is contributed by Chitranayal |
C#
// C# program for the above approach using System;
class GFG{
// Function to return the product of // abs diff of all pairs (x, y) static int getProduct( int []a, int n)
{ // To store product
int p = 1;
// Iterate all possible pairs
for ( int i = 0; i < n; i++)
{
for ( int j = i + 1; j < n; j++)
{
// Find the product
p *= Math.Abs(a[i] - a[j]);
}
}
// Return product
return p;
} // Driver Code public static void Main( string [] args)
{ // Given array arr[]
int []arr = { 1, 2, 3, 4 };
int N = arr.Length;
// Function call
Console.Write(getProduct(arr, N));
} } // This code is contributed by rutvik_56 |
Javascript
<script> // Javascript program for the above approach // Function to return the product of // abs diff of all pairs (x, y) function getProduct( a, n)
{ // To store product
var p = 1;
// Iterate all possible pairs
for ( var i = 0; i < n; i++) {
for ( var j = i + 1; j < n; j++) {
// Find the product
p *= Math.abs(a[i] - a[j]);
}
}
// Return product
return p;
} // Driver Code // Given array arr[] var arr = [ 1, 2, 3, 4 ];
var N = arr.length;
// Function Call document.write( getProduct(arr, N)); // This code is contributed by itsok. </script> |
Output:
12
Time Complexity: O(N2)
Auxiliary Space: O(1)
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