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 : 0
For index 0: abs((2) – (2 * 1)) = 0
For index 1: abs((2 * 2) – (1)) = 3Input : arr[] = { 3, 2, 5, 7, 2, 9 }
Output : 2
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++
// 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 minimumint 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 codeint main(){ int arr[] = { 3, 2, 5, 7, 2, 9 }; int N = sizeof(arr) / sizeof(arr[0]); cout << findIndex(arr, N); return 0;} |
Java
// 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 |
Python3
# 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 |
C#
// 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 |
PHP
<?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?> |
Javascript
<script> // Javascript 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) { // To store the required index let res = 0; let min_diff = Number.MAX_VALUE; // Prefix product array let prod = new Array(n); prod[0] = a[0]; // Compute the product array for (let i = 1; i < n; i++) prod[i] = prod[i - 1] * a[i]; // Iterate the product array to find the index for (let i = 0; i < n - 1; i++) { let curr_diff = Math.abs(parseInt(prod[n - 1] / prod[i], 10) - prod[i]); if (curr_diff < min_diff) { min_diff = curr_diff; res = i; } } return res; } let arr = [ 3, 2, 5, 7, 2, 9 ]; let N = arr.length; document.write(findIndex(arr, N)); // This code is contributed by suresh07.</script> |
Output
2
Time Complexity: O(N), where n is the size of the given array.
Auxiliary Space: O(N), where n is the size of the given array.
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++
#include <bits/stdc++.h>#define ll long long intusing namespace std;// Function to find indexvoid 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 Codeint main(){ int Array[5] = { 1, 4, 12, 2, 6 }; int N = 5; solve(Array, N);} |
Java
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 |
Python3
import math# Function to find indexdef 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 Codeif __name__ == "__main__": Array = [ 1, 4, 12, 2, 6 ] N = 5 solve(Array, N)# This code is contributed by chitranayal |
C#
using System;using System.Collections;class GFG{// Function to find indexpublic 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 codepublic static void Main(string []args){ int []Array = { 1, 4, 12, 2, 6 }; int N = 5; solve(Array, N);}}// This code is contributed by pratham76 |
Javascript
<script> // Function to find index function solve(array, N) { // Array to store log values of elements let Arraynew = new Array(N); for(let i = 0; i < N; i++) { Arraynew[i] = Math.log(array[i]); } // Prefix Array to Maintain Sum of // log values till index i let prefixsum = new Array(N); prefixsum[0] = Arraynew[0]; for(let i = 1; i < N; i++) { prefixsum[i] = prefixsum[i - 1] + Arraynew[i]; } // Answer Index let answer = 0; let minabs = Math.abs(prefixsum[N - 1] - 2 * prefixsum[0]); for(let i = 1; i < N - 1; i++) { let ans1 = Math.abs(prefixsum[N - 1] - 2 * prefixsum[i]); // Find minimum absolute value if (ans1 < minabs) { minabs = ans1; answer = i; } } document.write("Index is: " + answer + "</br>"); } let array = [ 1, 4, 12, 2, 6 ]; let N = 5; solve(array, N); </script> |
Output
Index is: 2
Time Complexity: O(N), where n is the size of the given array.
Auxiliary Space: O(N), where n is the size of the given array.
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