How to optimally divide an array into two subarrays so that sum of elements in both are same, otherwise give an error?

Question

How to optimally divide an array into two subarrays so that sum of elements in both subarrays is same, otherwise give an error?

Example 1

Given the array

Answer 1

    public boolean splitBetween(int[] x){
    int sum=0;
    int sum1=0;
    if (x.length==1){
        System.out.println("Not a valid value");
    }

    for (int i=0;i<x.length;i++){
        sum=sum+x[i];
        System.out.println(sum);
        for (int j=i+1;j<x.length;j++){
            sum1=sum1+x[j];
            System.out.println("SUm1:"+sum1);

        }

        if(sum==sum1){
            System.out.println("split possible");
            System.out.println("Sum: " +sum +" Sum1:" + sum1);
            return true;
        }else{
            System.out.println("Split not possible");
        }

        sum1=0;
    }
    return false;   
}

Answer 2

I was asked this question in an interview, and I gave below simple solution, as I had NOT seen this problem in any websiteS earlier.

Lets say Array A = {45,10,10,10,10,5} Then, the split will be at index = 1 (0-based index) so that we have two equal sum set {45} and {10,10,10,10,5}

int leftSum = A[0], rightSum = A[A.length - 1];
int currentLeftIndex = 0; currentRightIndex = A.length - 1

/* Move the two index pointers towards mid of the array untill currentRightIndex != currentLeftIndex. Increase leftIndex if sum of left elements is still less than or equal to sum of elements in right of 'rightIndex'.At the end,check if leftSum == rightSum. If true, we got the index as currentLeftIndex+1(or simply currentRightIndex, as currentRightIndex will be equal to currentLeftIndex+1 in this case). */

while (currentLeftIndex < currentRightIndex)
{
if ( currentLeftIndex+1 != currentRightIndex && (leftSum + A[currentLeftIndex + 1)     <=currentRightSum )
{
 currentLeftIndex ++;
 leftSum = leftSum + A[currentLeftIndex];
}


if ( currentRightIndex - 1 != currentLeftIndex && (rightSum + A[currentRightIndex - 1] <= currentLeftSum)
{
 currentRightIndex --;
 rightSum = rightSum + A[currentRightIndex];
}

}

if (CurrentLeftIndex == currentRightIndex - 1 && leftSum == rightSum)
PRINT("got split point at index "+currentRightIndex);

Answer 3

Tried a different solution . other than Wiki solutions (Partition Problem).

static void subSet(int array[]) {
    System.out.println("Input elements  :" + Arrays.toString(array));

    int sum = 0;
    for (int element : array) {
        sum = sum + element;
    }
    if (sum % 2 == 1) {
        System.out.println("Invalid Pair");
        return;
    }

    Arrays.sort(array);
    System.out.println("Sorted elements :" + Arrays.toString(array));

    int subSum = sum / 2;

    int[] subSet = new int[array.length];
    int tmpSum = 0;
    boolean isFastpath = true;
    int lastStopIndex = 0;
    for (int j = array.length - 1; j >= 0; j--) {
        tmpSum = tmpSum + array[j];
        if (tmpSum == subSum) { // if Match found
            if (isFastpath) { // if no skip required and straight forward
                                // method
                System.out.println("Found SubSets 0..." + (j - 1) + " and "
                        + j + "..." + (array.length - 1));
            } else {
                subSet[j] = array[j];
                array[j] = 0;
                System.out.println("Found..");
                System.out.println("Set 1" + Arrays.toString(subSet));
                System.out.println("Set 2" + Arrays.toString(array));
            }
            return;
        } else {
            // Either the tmpSum greater than subSum or less .
            // if less , just look for next item
            if (tmpSum < subSum && ((subSum - tmpSum) >= array[0])) {
                if (lastStopIndex > j && subSet[lastStopIndex] == 0) {
                    subSet[lastStopIndex] = array[lastStopIndex];
                    array[lastStopIndex] = 0;
                }
                lastStopIndex = j;
                continue;
            }
            isFastpath = false;
            if (subSet[lastStopIndex] == 0) {
                subSet[lastStopIndex] = array[lastStopIndex];
                array[lastStopIndex] = 0;
            }
            tmpSum = tmpSum - array[j];
        }
    }

}

I have tested. ( It works well with positive number greater than 0) please let me know if any one face issue.

Answer 4

A BAD greedy heuristic to solve this problem: try sorting the list from least to greatest, and split that list into two by having list1 = the odd elements, and list2 = the even elements.