easy

Question

easy

HG · Accepted Answer

public static int getIndexOfEqualSum(int []a)
    {
        int index = -1; //not found
        
        int sum = 0;
        for (int i = 0; i < a.length; i++) {
            sum += a[i];
        }
        
        int tempSum = 0;
        for (int i = 0; i < a.length; i++) {
            int val = a[i];
            sum -= val;
            if (sum == tempSum) {
                return i;
            }
            tempSum += val;
        }
        
        return index;
    }

Gilles · Answer

@ lurii Mykhailov.

try your algorithm with this:
int[] array = new int[] {3,-2,-1,0};

It seems it does not work... Don't forget we may have negative values.

Anonimo · Answer

int [] newArr = new int[array.length];
int sum;
newArr[0] = array[0];
for(int i = 1 ; i =0  ; j--){
	sum += array[j];

	if(sum == newArray[j-2]){
		return j-1;
	}
}

My solution: The new array creatio can be avoided by calculating the sum again in the 2nd iteration · Answer

public int middleIndex(int[] num) {
        if(num.length==1) return 0;
        int[] sum = new int[num.length]; 
        sum[0] = num[0];
        for(int i=1;i

Iurii Mykhailov · Answer

Here is my implementation, should be a bit faster if the required element is somewhere in the middle:

public static int getIndexOfEqualSumMy(int[] a) {
        int leftSum = a[0];
        int j = a.length - 1;
        int rightSum = a[j];
        for (int i = 0; i != j;) {
            if (leftSum < rightSum) {
                i++;
                leftSum += a[i];
            } else {
                j--;
                rightSum += a[j];
            }
        }
        if (leftSum != rightSum) {
            return -1;
        }
        return j;
    }

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