antOnBoundary.java

/**3028. Ant on the Boundary
 * An ant is on a boundary. It sometimes goes left and sometimes right.

You are given an array of non-zero integers nums. The ant starts reading nums from the first element of it to its end. At each step, it moves according to the value of the current element:
If nums[i] < 0, it moves left by -nums[i] units.
If nums[i] > 0, it moves right by nums[i] units.
Return the number of times the ant returns to the boundary.

Notes:
There is an infinite space on both sides of the boundary.
We check whether the ant is on the boundary only after it has moved |nums[i]| units. In other words, if the ant crosses the boundary during its movement, it does not count.
 * 
 * 
 * Input: nums = [2,3,-5]
Output: 1
Explanation: After the first step, the ant is 2 steps to the right of the boundary.
After the second step, the ant is 5 steps to the right of the boundary.
After the third step, the ant is on the boundary.
So the answer is 1.
 * 
 * 
 * Input: nums = [3,2,-3,-4]
Output: 0
Explanation: After the first step, the ant is 3 steps to the right of the boundary.
After the second step, the ant is 5 steps to the right of the boundary.
After the third step, the ant is 2 steps to the right of the boundary.
After the fourth step, the ant is 2 steps to the left of the boundary.
The ant never returned to the boundary, so the answer is 0.
 * 
 */

public class antOnBoundary {
    public static int solution(int[] arr){
        int count = 0, sum = 0;
        for(int element : arr){
            sum += element;

            if(sum == 0) count++;
        }
        return count;
    }
    public static void main(String[] args) {
        int[] arr = new int[]{3,2,-3,-4};
        System.out.println("Total Count : " + solution(arr));
    }
}