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Find an index such that difference between product of elements before and after it is minimum
  • Last Updated : 07 Dec, 2020

Given an integer arr[], the task is to find an index such that the difference between the product of elements up to that index (including that index) and the product of rest of the elements is minimum. If more than one such index is present, then return the minimum index as the answer. 

Examples:  

Input : arr[] = { 2, 2, 1 } 
Output :
For index 0: abs((2) – (2 * 1)) = 0 
For index 1: abs((2 * 2) – (1)) = 3

Input : arr[] = { 3, 2, 5, 7, 2, 9 } 
Output :

A Simple Solution is to traverse through all elements starting from first to second last element. For every element, find the product of elements till this element (including this element). Then find the product of elements after it. Finally compute the difference. If the difference is minimum so far, update the result.



Better Approach: The problem can be easily solved using a prefix product array prod[] where the prod[i] stores the product of elements from arr[0] to arr[i]. Therefore, the product of rest of the elements can be easily found by dividing the total product of the array by the product up to current index. Now, iterate the product array to find the index with minimum difference. 

Below is the implementation of the above approach: 

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
 
// Function to return the index i such that
// the absolute difference between product
// of elements up to that index and the
// product of rest of the elements
// of the array is minimum
int findIndex(int a[], int n)
{
    // To store the required index
    int res;
 
    ll min_diff = INT_MAX;
 
    // Prefix product array
    ll prod[n];
    prod[0] = a[0];
 
    // Compute the product array
    for (int i = 1; i < n; i++)
        prod[i] = prod[i - 1] * a[i];
 
    // Iterate the product array to find the index
    for (int i = 0; i < n - 1; i++) {
        ll curr_diff = abs((prod[n - 1] / prod[i]) - prod[i]);
 
        if (curr_diff < min_diff) {
            min_diff = curr_diff;
            res = i;
        }
    }
 
    return res;
}
 
// Driver code
int main()
{
    int arr[] = { 3, 2, 5, 7, 2, 9 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << findIndex(arr, N);
 
    return 0;
}

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Java

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// Java implementation of the approach
class GFG{
     
// Function to return the index i such that
// the absolute difference between product
// of elements up to that index and the
// product of rest of the elements
// of the array is minimum
static int findIndex(int a[], int n)
{
    // To store the required index
    int res = 0;
 
    long min_diff = Long.MAX_VALUE;
 
    // Prefix product array
    long prod[] = new long[n];
    prod[0] = a[0];
 
    // Compute the product array
    for (int i = 1; i < n; i++)
        prod[i] = prod[i - 1] * a[i];
 
    // Iterate the product array to find the index
    for (int i = 0; i < n - 1; i++)
    {
        long curr_diff = Math.abs((prod[n - 1] /
                                   prod[i]) - prod[i]);
 
        if (curr_diff < min_diff)
        {
            min_diff = curr_diff;
            res = i;
        }
    }
 
    return res;
}
 
// Driver code
public static void main(String arg[])
{
    int arr[] = { 3, 2, 5, 7, 2, 9 };
    int N = arr.length;
 
    System.out.println(findIndex(arr, N));
}
}
 
// This code is contributed by rutvik_56

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Python3

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# Python3 implementation of the approach
 
# Function to return the index i such that
# the absolute difference between product of
# elements up to that index and the product of
# rest of the elements of the array is minimum
def findIndex(a, n):
  
    # To store the required index
    res, min_diff = None, float('inf')
 
    # Prefix product array
    prod = [None] * n
    prod[0] = a[0]
 
    # Compute the product array
    for i in range(1, n):
        prod[i] = prod[i - 1] * a[i]
 
    # Iterate the product array to find the index
    for i in range(0, n - 1): 
        curr_diff = abs((prod[n - 1] // prod[i]) - prod[i])
 
        if curr_diff < min_diff: 
            min_diff = curr_diff
            res = i
          
    return res
 
# Driver code
if __name__ == "__main__":
  
    arr = [3, 2, 5, 7, 2, 9
    N = len(arr)
 
    print(findIndex(arr, N))
 
# This code is contributed by Rituraj Jain

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C#

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// C# implementation of the approach
using System;
 
class GFG
{
    // Function to return the index i such that 
    // the absolute difference between product 
    // of elements up to that index and the 
    // product of rest of the elements 
    // of the array is minimum 
    static int findIndex(int[] a, int n) 
    
        // To store the required index 
        int res = 0; 
       
        long min_diff = Int64.MaxValue; 
       
        // Prefix product array 
        long[] prod = new long[n]; 
        prod[0] = a[0]; 
       
        // Compute the product array 
        for (int i = 1; i < n; i++) 
            prod[i] = prod[i - 1] * a[i]; 
       
        // Iterate the product array to find the index 
        for (int i = 0; i < n - 1; i++)
        
            long curr_diff = Math.Abs((prod[n - 1] / 
                                       prod[i]) - prod[i]); 
       
            if (curr_diff < min_diff) 
            
                min_diff = curr_diff; 
                res = i; 
            
        
       
        return res; 
    
   
  // Driver code
  static void Main()
  {
        int[] arr = { 3, 2, 5, 7, 2, 9 }; 
        int N = arr.Length; 
       
        Console.WriteLine(findIndex(arr, N)); 
  }
}
 
// This code is contributed by divyeshrabadiya07

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PHP

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<?php
// PHP implementation of the approach
 
// Function to return the index i such that
// the absolute difference between product
// of elements up to that index and the
// product of rest of the elements
// of the array is minimum
function findIndex($a, $n)
{
    $min_diff = PHP_INT_MAX;
 
    // Prefix product array
    $prod = array();
    $prod[0] = $a[0];
 
    // Compute the product array
    for ($i = 1; $i < $n; $i++)
        $prod[$i] = $prod[$i - 1] * $a[$i];
 
    // Iterate the product array to find the index
    for ($i = 0; $i < $n - 1; $i++)
    {
        $curr_diff = abs(($prod[$n - 1] /
                    $prod[$i]) - $prod[$i]);
 
        if ($curr_diff < $min_diff)
        {
            $min_diff = $curr_diff;
            $res = $i;
        }
    }
 
    return $res;
}
 
    // Driver code
    $arr = array( 3, 2, 5, 7, 2, 9 );
    $N = count($arr);
 
    echo findIndex($arr, $N);
     
    // This code is contributed by AnkitRai01
?>

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Output: 

2

 

Approach without overflow 
The above solution might cause overflow. To prevent overflow problem, Take log of all the values of the array. Now, question is boiled down to divide array in two halves with absolute difference of sum is minimum possible. Now, array contains log values of elements at each index. Maintain a prefix sum array B which holds sum of all the values till index i. Check for all the indexes, abs(B[n-1] – 2*B[i]) and find the index with minimum possible absolute value. 

C++

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#include <bits/stdc++.h>
#define ll long long int
using namespace std;
 
// Function to find index
void solve(int Array[], int N)
{
    // Array to store log values of elements
    double Arraynew[N];
    for (int i = 0; i < N; i++) {
        Arraynew[i] = log(Array[i]);
    }
 
    // Prefix Array to Maintain Sum of log values till index i
    double prefixsum[N];
    prefixsum[0] = Arraynew[0];
 
    for (int i = 1; i < N; i++) {
        prefixsum[i] = prefixsum[i - 1] + Arraynew[i];
    }
 
    // Answer Index
    int answer = 0;
    double minabs = abs(prefixsum[N - 1] - 2 * prefixsum[0]);
 
    for (int i = 1; i < N - 1; i++) {
        double ans1 = abs(prefixsum[N - 1] - 2 * prefixsum[i]);
 
        // Find minimum absolute value
        if (ans1 < minabs) {
            minabs = ans1;
            answer = i;
        }
    }
 
    cout << "Index is: " << answer << endl;
}
 
// Driver Code
int main()
{
    int Array[5] = { 1, 4, 12, 2, 6 };
    int N = 5;
    solve(Array, N);
}

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Java

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public class Main
{
    // Function to find index
    public static void solve(int Array[], int N)
    {
        // Array to store log values of elements
        double Arraynew[] = new double[N];
        for (int i = 0; i < N; i++) {
            Arraynew[i] = Math.log(Array[i]);
        }
      
        // Prefix Array to Maintain Sum of log values till index i
        double prefixsum[] = new double[N];
        prefixsum[0] = Arraynew[0];
      
        for (int i = 1; i < N; i++)
        {
            prefixsum[i] = prefixsum[i - 1] + Arraynew[i];
        }
      
        // Answer Index
        int answer = 0;
        double minabs = Math.abs(prefixsum[N - 1] - 2 *
                                 prefixsum[0]);
      
        for (int i = 1; i < N - 1; i++)
        {
            double ans1 = Math.abs(prefixsum[N - 1] - 2 *
                                   prefixsum[i]);
      
            // Find minimum absolute value
            if (ans1 < minabs)
            {
                minabs = ans1;
                answer = i;
            }
        }
      
        System.out.println("Index is: " + answer);
    }
   
   
 // Driver code
    public static void main(String[] args)
    {
        int Array[] = { 1, 4, 12, 2, 6 };
        int N = 5;
        solve(Array, N);
    }
}
 
// This code is contributed by divyesh072019

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Python3

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import math
 
# Function to find index
def solve( Array,  N):
 
    # Array to store log values of elements
    Arraynew = [0]*N
    for i in range( N ) :
        Arraynew[i] = math.log(Array[i])
     
  
    # Prefix Array to Maintain Sum of log values till index i
    prefixsum = [0]*N
    prefixsum[0] = Arraynew[0]
  
    for i in range( 1,  N) :
        prefixsum[i] = prefixsum[i - 1] + Arraynew[i]
     
  
    # Answer Index
    answer = 0
    minabs = abs(prefixsum[N - 1] - 2 * prefixsum[0])
  
    for i in range(1, N - 1):
        ans1 = abs(prefixsum[N - 1] - 2 * prefixsum[i])
  
        # Find minimum absolute value
        if (ans1 < minabs):
            minabs = ans1
            answer = i
  
    print("Index is: " ,answer)
  
# Driver Code
if __name__ == "__main__":
    Array = [ 1, 4, 12, 2, 6 ]
    N = 5
    solve(Array, N)
 
# This code is contributed by chitranayal

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C#

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using System;
using System.Collections;
 
class GFG{
 
// Function to find index
public static void solve(int []Array, int N)
{
     
    // Array to store log values of elements
    double []Arraynew = new double[N];
    for(int i = 0; i < N; i++)
    {
        Arraynew[i] = Math.Log(Array[i]);
    }
   
    // Prefix Array to Maintain Sum of
    // log values till index i
    double []prefixsum = new double[N];
    prefixsum[0] = Arraynew[0];
   
    for(int i = 1; i < N; i++)
    {
        prefixsum[i] = prefixsum[i - 1] +
                        Arraynew[i];
    }
   
    // Answer Index
    int answer = 0;
    double minabs = Math.Abs(prefixsum[N - 1] - 2 *
                             prefixsum[0]);
   
    for(int i = 1; i < N - 1; i++)
    {
        double ans1 = Math.Abs(prefixsum[N - 1] - 2 *
                               prefixsum[i]);
   
        // Find minimum absolute value
        if (ans1 < minabs)
        {
            minabs = ans1;
            answer = i;
        }
    }
   
    Console.WriteLine("Index is: " + answer);
}
 
// Driver code
public static void Main(string []args)
{
    int []Array = { 1, 4, 12, 2, 6 };
    int N = 5;
     
    solve(Array, N);
}
}
 
// This code is contributed by pratham76

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Output: 

Index is: 2

 

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