Question

Given an array of numbers, find the index of the smallest array element (the pivot), for which the sums of all elements to the left and to the right are equal. The array may not be reordered. 

Example arr=[1,2,3,4,6] 

• the sum of the first three elements, 1+2+3=6. The value of the last element is 6. 

• Using zero based indexing, arr[3]=4 is the pivot between the two subarrays. 

• The index of the pivot is 3. 

Function Description 

Complete the function balancedSum in the editor below. 

balanced Sum has the following 

parameter(s): 

       int arr(n): an array of integers 

Returns: 

       int: an integer representing the index of the pivot

Constraints 

•  3≤n≤10⁵ 

• 1≤ arr[i]≤ 2 * 10⁴, where 0≤i<n

• It is guaranteed that a solution always exists.

Answer 1

Approach used:
>>Maintain two pointers, leftptr, rightptr
>>Traverse array from both sides at once and calculate left_sum and right_sum to store array sum from right and left.

>>If left_sum < right_sum increment leftptr
>>If left_sum > right_sum decrement rightptr
>>When left_sum = right_sum and leftptr and rightptr have only one element in between which is pivot,
>>Return the pivot
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I have included my code and screenshots in this answer. In case, there is any indentation issue due to editor, then please refer to code screenshots to avoid confusion.

----------------solution.java---------

import java.util.*;
import java.lang.String;

public class solution
{
   public static void main(String[] args)
   {
       int arr1 [] ={1,2,3,4,6};
       int pivot_index = balancedSum(arr1);
       System.out.print("\nThe pivot index for arr1 is: " + pivot_index+"\n");

       int arr2 [] ={10, 5, 5, 1, 9, 3, 1, 2 , 15};
       pivot_index = balancedSum(arr2);
       System.out.print("\nThe pivot index for arr2 is: " + pivot_index+"\n");
   }

   public static int balancedSum(int [] array) //returns pivot index for array
   {
       int leftptr = 0; //declare leftptr and rightptr
       int rightptr = array.length-1;   
       int left_sum = array[leftptr]; //initialize left sum to first element
       int right_sum = array[rightptr]; //initialize left sum to last element
       while(rightptr - leftptr != 2) //do until there is one element between both pointers
       {
           if(left_sum <= right_sum){ //if left sum is less, increment leftptr and increase sum
               leftptr++;
               left_sum = left_sum + array[leftptr];
           }
           else{ //if right sum is less, decrement rightptr and decrease sum
               rightptr--;
               right_sum = right_sum + array[rightptr];
           }
       }
       if(left_sum == right_sum) //if only one element exists in between and both sums are equal, then return pivot
           return (leftptr+1);
       else
           return -1; //else return -1
   }
}

----------------Screenshot solution.java---------

-/java/balanced sum/solution.java - Sublime Text (UNREGISTERED) t ) 11:41 AM * 1 solution.java X import java.util.*; import java.lang.String; public class solution 6 public static void main(String[] args) { int arri [] ={1,2,3,4,6}; int pivot index balancedSum(arrl); System.out.print("\nThe pivot index for arrl is: ". int arr2 [] ={10, 5, 5, 1, 9, 3, 1, 2 , 15}; pivot_index balancedSum(arr2); System.out.print("\nThe pivot index for arr2 is: ". pivot_index+"\n"); pivot_index+"\n"); ها 15 16 117 18 19 20 21 { 24 public static int balancedSum(int [] array) //returns pivot index for array int leftptr 0; //declare leftptr and rightptr int rightptr array.length-1; int left_sum array[leftptr); //initialize left sum to first element int right_sum array[rightptr); //initialize left sum to last element while(rightptr leftptr != 2) //do until there is one element between both pointers { if(left_sum <= right_sum) { //if left sum is less, increment leftptr and increase sum leftptr++; left_sum = left_sum + array[leftptr); else{ //if right sum is less, decrement rightptr and decrease sum rightptr--; right_sum = right_sum + array[rightptr); } } } 33 34 35 36 37 38 39 140 if(left_sum == right_sum) //if only one element exists in between and both sums are equal, then return pivot return (leftptr+1); else return 1; //else return } } Line 37, Column 43; Saved -/java/balanced sum/solution.java (UTF-8) Tab Size: 4 Java

----------------Output---------

11:39 AM W vs@ubuntu:/java/balanced sum vs@ubuntu:-/java/balanced sum$ javac solution.java vs@ubuntu:-/java/balanced sums java solution The pivot index for arrl is: 3 The pivot index for arr2 is: 4 vs@ubuntu:-/java/balanced sums |

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