In this guide, you will get 3028. Ant on the Boundary LeetCode Solution with the best time and space complexity. The solution to Ant on the Boundary problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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 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.
Example 1:
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.
3028. Ant on the Boundary LeetCode Solution in C++
class Solution {
public:
int returnToBoundaryCount(vector<int>& nums) {
partial_sum(nums.begin(), nums.end(), nums.begin());
return ranges::count(nums, 0);
}
}; /* code provided by PROGIEZ */
3028. Ant on the Boundary LeetCode Solution in Java
class Solution {
public int returnToBoundaryCount(int[] nums) {
int ans = 0;
int prefix = 0;
for (final int num : nums) {
prefix += num;
if (prefix == 0)
++ans;
}
return ans;
}
} // code provided by PROGIEZ
3028. Ant on the Boundary LeetCode Solution in Python
class Solution:
def returnToBoundaryCount(self, nums: list[int]) -> int:
return sum(prefix == 0 for prefix in itertools.accumulate(nums)) # code by PROGIEZ
