How could we get the index of an array where left and right subarrays sum the same‽

Question

I have completed the following programming exercise: Equal Side of an Array. The statement is the following:

You are going to be given an array of integers. Your job is to take that array and find an index N where the sum of the integers to the left of N is equal to the sum of the integers to the right of N. If there is no index that would make this happen, return -1.

For example:

Let's say you are given the array {1,2,3,4,3,2,1}: Your function will return the index 3, because at the 3rd position of the array, the sum of left side of the index ({1,2,3}) and the sum of the right side of the index ({3,2,1}) both equal 6.

Let's look at another one. You are given the array {1,100,50,-51,1,1}: Your function will return the index 1, because at the 1st position of the array, the sum of left side of the index ({1}) and the sum of the right side of the index ({50,-51,1,1}) both equal 1.

Last one: You are given the array {20,10,-80,10,10,15,35} At index 0 the left side is {} The right side is {10,-80,10,10,15,35} They both are equal to 0 when added. (Empty arrays are equal to 0 in this problem) Index 0 is the place where the left side and right side are equal.

Note: Please remember that in most programming/scripting languages the index of an array starts at 0.

Input: An integer array of length 0 < arr < 1000. The numbers in the array can be any integer positive or negative.

Output: The lowest index N where the side to the left of N is equal to the side to the right of N. If you do not find an index that fits these rules, then you will return -1.

Note: If you are given an array with multiple answers, return the lowest correct index.

I have read the following answer provided by the user JensPiegsa. Here you have the link to it.

import java.util.stream.IntStream;

public class Kata {
  public static int findEvenIndex(int[] arr) {
    return IntStream.range(0, arr.length)
        .filter(n -> IntStream.of(arr).limit(n).sum() == IntStream.of(arr).skip(n + 1).sum())
        .findFirst().orElse(-1);
  }
}

I was wondering if there is a way to instead of looping all over the Instream filtering which left and right subarrays are equal and then returning the first one; just break the execution when we return the first equal.

I would like to know how a functional solution would look like, when we just get the left and right equal subarray, without looping all over it.

With loops I have thought it could be:

public class Kata {
  public static int findEvenIndex(int[] arr) {
    int left = 0, right = 0;

    for(int i = 0; i < arr.length; i++, left = 0, right = 0){
      for(int j = 0; j < i; j++){
        left += arr[j];
      }
      for(int k = arr.length - 1; k > i; k--){
        right += arr[k];
      }
      if(left == right) return i;  
    }
    return -1;
  }
}

How could it be done in just a functional sentence?

I have also read: Is there a subarray that sums to a target? Find the index of the subarray whose sum is minimum Make sums of left and right sides of array equal by removing subarray

Answer 1

Solution

1.if want get Sum between indexes from subarray. so you have try this code i am using java it's working fine.

public class SumSubArray{
    
  public int subArraySum(int arr[], int n, int sum)
    {
        int curr_sum = arr[0], start = 0, i;

        for (i = 1; i <= n; i++) {
            while (curr_sum > sum && start < i - 1) {
                curr_sum = curr_sum - arr[start];
                start++;
            }

            if (curr_sum == sum) {
                int p = i - 1;
                System.out.println("Sum found between indexes from " + start
                    + " to " + p);
                return 1;
            }

            if (i < n)
                curr_sum = curr_sum + arr[i];
        }

        System.out.println("No subarray found");
        return 0;
    }
    

     public static void main(String[] args)
    {
        SumSubArray arraysum = new SumSubArray();
        int arr[] = { 4, 2, 5, 8, 33, 20, 10, 21,25,50 };
        int n = arr.length;
        int sum = 15;
        arraysum.subArraySum(arr, n, sum);
    }
}

Answer 2

if you want a not O(n^2) solution:

    public Integer getfirstIndexEqual(Integer [] nums){
        Integer sum = 0;
        for (int i = 0; i < nums.length; i++){
            sum += nums[i];
        }
        Integer half = 0;
        for (int i=0; i< nums.length; i++){
            if (half.floatValue() == (sum-nums[i]) / 2){
                return i;
            }
            half += nums[i];
        }
        return -1;
    }