Ant on the Boundary

Ant on the Boundary - Problem

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 & Output

Example 1 — Basic Movement

$ Input: nums = [2,-3,3,-4]

› Output: 0

💡 Note: Ant moves: start(0) → +2(2) → -3(-1) → +3(2) → -4(-2). Never returns to position 0.

Example 2 — Return to Boundary

$ Input: nums = [3,-3,4,-4]

› Output: 2

💡 Note: Ant moves: start(0) → +3(3) → -3(0) → +4(4) → -4(0). Returns to boundary twice.

Example 3 — No Returns

$ Input: nums = [1,2,3]

› Output: 0

💡 Note: Ant moves: start(0) → +1(1) → +2(3) → +3(6). Always moving right, never returns to 0.

Constraints

1 ≤ nums.length ≤ 50
-100 ≤ nums[i] ≤ 100
nums[i] ≠ 0

Visualization

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Asked in

G Google 15 a Amazon 12

The key insight is that the ant returns to the boundary whenever the cumulative sum of movements equals zero. Best approach uses prefix sum to track position. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force Simulation

⏱️ Time: O(n) Space: O(1)

Start at position 0 and simulate each move. After each step, check if the ant is back at position 0 (the boundary).

Prefix Sum

⏱️ Time: O(n) Space: O(1)

Maintain a running sum (prefix sum) of movements. Each time the sum equals zero, the ant has returned to the boundary.

Brute Force Simulation — Algorithm Steps

Initialize position to 0 and counter to 0
For each number in array, update position
If position equals 0, increment counter

Visualization

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Step-by-Step Walkthrough

Start

Ant begins at position 0 (boundary)

Move

Process each number, update position

Count

Increment counter when position equals 0

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums, int numsSize) {
    int position = 0;
    int count = 0;
    
    for (int i = 0; i < numsSize; i++) {
        position += nums[i];
        if (position == 0) {
            count++;
        }
    }
    
    return count;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Simple parsing for [1,2,3] format
    int nums[1000];
    int size = 0;
    char* token = strtok(line + 1, ",]");
    
    while (token != NULL) {
        nums[size++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    printf("%d\n", solution(nums, size));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the array

✓ Linear Growth

Space Complexity

O(1)

Only using two variables for position and count

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Simulation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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