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.

LeetCode Problem - 3028

class Solution {
    // This method returns the count of points where the sum of elements before the point is zero.
    public int returnToBoundaryCount(int[] nums) {
        // Initialize the sum and count variables to zero.
        int sum = 0;
        int count = 0;

        // If the array has only one element, return 0.
        if (nums.length == 1) {
            return 0;
        }

        // Iterate through the array.
        for (int i = 0; i < nums.length; i++) {
            // If the current element is negative, add it to the sum.
            if (nums[i] < 0) {
                sum += nums[i];
                // If the sum becomes zero, increment the count.
                if (sum == 0) {
                    count++;
                }
            }
            // If the current element is positive, add it to the sum.
            else if (nums[i] > 0) {
                sum += nums[i];
                // If the sum becomes zero, increment the count.
                if (sum == 0) {
                    count++;
                }
            }
        }
        // Return the count of points where the sum of elements before the point is zero.
        return count;
    }
}