Problem Description

An ant starts at a boundary (position 0) and moves left or right based on values in an array nums.

Given an array of non-zero integers nums, the ant reads each element sequentially and moves:

• Left by |nums[i]| units if nums[i] < 0
• Right by nums[i] units if nums[i] > 0

The task is to count how many times the ant returns to the boundary (position 0) after completing each move.

Key Points:

• The boundary is at position 0, with infinite space on both sides
• The ant only counts as returning to the boundary if it lands exactly on position 0 after a move
• If the ant crosses the boundary during a move but doesn't stop there, it doesn't count

Solution Explanation:

The solution uses prefix sums to track the ant's position after each move. Starting at position 0:

• Each element in nums represents a displacement from the current position
• The cumulative sum at any point represents the ant's position relative to the boundary
• When the cumulative sum equals 0, the ant has returned to the boundary

The code accumulate(nums) generates all prefix sums, and sum(s == 0 for s in accumulate(nums)) counts how many times these prefix sums equal zero, which corresponds to the number of returns to the boundary.

For example, if nums = [2, 3, -5]:

• After move 1: position = 2
• After move 2: position = 2 + 3 = 5
• After move 3: position = 5 + (-5) = 0 (returns to boundary)
• Result: 1 return to boundary

Intuition

The key insight is recognizing that the ant's position at any moment is simply the sum of all movements it has made so far. Since the ant starts at position 0 (the boundary), we can track its position using a running total.

Think of it like a number line where:

• Starting point (boundary) = 0
• Moving right adds positive values to our position
• Moving left adds negative values to our position

After processing each element in nums, the ant's current position equals the sum of all elements processed so far. Whenever this sum equals 0, the ant has returned to its starting point (the boundary).

This naturally leads us to think about prefix sums - a technique where we maintain the cumulative sum of elements as we traverse the array. Each prefix sum represents the ant's position after that many moves.

For example, with nums = [3, -2, -1]:

• After 1st move: position = 0 + 3 = 3
• After 2nd move: position = 3 + (-2) = 1
• After 3rd move: position = 1 + (-1) = 0 ← back at boundary!

The problem essentially becomes: "Count how many prefix sums equal zero." This is why the solution uses accumulate(nums) to generate all prefix sums and counts the zeros among them. The elegance of this approach is that it transforms a movement simulation problem into a simple counting problem on prefix sums.

Learn more about Prefix Sum patterns.

Solution Approach

The solution uses the prefix sum pattern to efficiently track the ant's position throughout its journey.

Implementation Details:

1. Using accumulate() function:

   • Python's accumulate() from the itertools module generates running totals
   • For nums = [2, 3, -5], accumulate(nums) produces: [2, 5, 0]
   • Each value represents the ant's position after that many moves

2. Counting zeros in prefix sums:

   • The expression s == 0 for s in accumulate(nums) creates a generator that yields True when a prefix sum equals 0, False otherwise
   • sum() counts the True values (since True = 1 in Python)
   • This gives us the total number of times the ant returns to the boundary

Why this works:

• Starting position = 0
• Position after move i = nums[0] + nums[1] + ... + nums[i]
• When this sum = 0, the ant is back at the boundary

Time and Space Complexity:

• Time: O(n) where n is the length of nums - we traverse the array once
• Space: O(1) extra space - accumulate() is a generator that yields values one at a time

The beauty of this solution lies in its simplicity - instead of simulating each movement and checking the position, we recognize that the problem is fundamentally about finding zeros in the sequence of partial sums. The one-liner solution elegantly combines iteration, accumulation, and counting into a single expression.

Example Walkthrough

Let's walk through the solution with nums = [2, 3, -5, 4, -2]:

Step-by-step movement tracking:

Starting position: 0 (at the boundary)

1. Move 1: Read nums[0] = 2

   • Move right by 2 units
   • New position: 0 + 2 = 2
   • Not at boundary (position ≠ 0)

2. Move 2: Read nums[1] = 3

   • Move right by 3 units
   • New position: 2 + 3 = 5
   • Not at boundary (position ≠ 0)

3. Move 3: Read nums[2] = -5

   • Move left by 5 units
   • New position: 5 + (-5) = 0
   • At boundary! (Count = 1)

4. Move 4: Read nums[3] = 4

   • Move right by 4 units
   • New position: 0 + 4 = 4
   • Not at boundary (position ≠ 0)

5. Move 5: Read nums[4] = -2

   • Move left by 2 units
   • New position: 4 + (-2) = 2
   • Not at boundary (position ≠ 0)

Using the prefix sum approach:

When we apply accumulate(nums) to [2, 3, -5, 4, -2], we get:

• After move 1: prefix sum = 2
• After move 2: prefix sum = 2 + 3 = 5
• After move 3: prefix sum = 5 + (-5) = 0 ✓
• After move 4: prefix sum = 0 + 4 = 4
• After move 5: prefix sum = 4 + (-2) = 2

The prefix sums are: [2, 5, 0, 4, 2]

Counting the zeros in this sequence: There is 1 zero, which means the ant returns to the boundary 1 time.

The solution sum(s == 0 for s in accumulate(nums)) does exactly this - it generates each prefix sum and counts how many equal zero, giving us the answer: 1.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of nums. The accumulate function iterates through all elements once to compute the running sum, and the sum function with generator expression iterates through the accumulated values once, resulting in a total of O(n) time.

The space complexity is O(1). While accumulate returns an iterator that generates values on-the-fly, the generator expression (s == 0 for s in accumulate(nums)) is consumed directly by sum without storing all intermediate results in memory. Only a constant amount of extra space is used for the running sum and counter variables.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Misunderstanding the Boundary Position

A common mistake is thinking the ant returns to the boundary whenever it crosses position 0, rather than when it lands exactly on position 0.

Incorrect thinking: If the ant moves from position -3 to position 2 (with a move of +5), counting this as a boundary return because it "crossed" position 0.

Correct approach: Only count when the cumulative sum equals exactly 0. The ant must stop at the boundary, not just pass through it.

2. Off-by-One Error with Initial Position

Some might incorrectly count the starting position as a "return" to the boundary.

Incorrect implementation:

position = 0
count = 1  # Wrong: counting initial position
for num in nums:
    position += num
    if position == 0:
        count += 1

Correct implementation:

position = 0
count = 0  # Start with 0, only count actual returns
for num in nums:
    position += num
    if position == 0:
        count += 1

3. Memory Issues with Large Arrays

While accumulate() is memory-efficient as a generator, converting it to a list unnecessarily consumes memory.

Inefficient approach:

# Creates entire list in memory
prefix_sums = list(accumulate(nums))
return sum(s == 0 for s in prefix_sums)

Efficient approach:

# Generator-based, memory-efficient
return sum(s == 0 for s in accumulate(nums))

4. Forgetting to Import accumulate

The accumulate function must be imported from itertools. Missing this import will cause a NameError.

Solution: Always include from itertools import accumulate at the beginning of your solution, or use the manual approach:

def returnToBoundaryCount(self, nums: List[int]) -> int:
    position = 0
    count = 0
    for num in nums:
        position += num
        if position == 0:
            count += 1
    return count

5. Integer Overflow Concerns

In languages with fixed integer sizes, large cumulative sums might cause overflow. While Python handles arbitrary precision integers automatically, be aware of this when implementing in other languages.

For other languages: Consider using appropriate data types (like long long in C++) when dealing with potentially large cumulative sums.