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Ant on the Boundary

Number: 3311

Difficulty: Easy

Paid? No

Companies: Accenture

Problem Description

An ant is placed on a boundary (position 0) and moves either left or right according to the values given in an array. Positive numbers move the ant to the right and negative numbers move it to the left by the absolute value of the number. The ant is considered to have “returned” to the boundary only if, after completing a move, its ending position is exactly 0. Note that if the ant crosses the boundary during its move (for example, from a positive to a negative side without stopping at 0), it does not count as a return.

Key Insights

  • Simulate the ant's movement by keeping track of its current position.
  • After each move, check if the ant’s position is 0. If yes, increment a return counter.
  • The moves are processed sequentially and no intermediate positions (other than the endpoint) are considered.
  • The problem can be solved in linear time relative to the length of the input array.

Space and Time Complexity

Time Complexity: O(n), where n is the length of the input array. Space Complexity: O(1), as only a few variables are needed to track state.

Solution

The solution involves iterating through the input array and updating a variable representing the ant's current position. We increment the position for positive moves (right) and decrement it for negative moves (left). After updating the position for each move, we check if the ant has landed back on the boundary (position 0). If it has, we increment a counter. Finally, we return the counter.

Key points:

  • Use a simple loop to traverse the array.
  • Maintain a variable (e.g., pos) for the ant's current position.
  • Use another variable (e.g., count) to track the number of returns to the boundary.
  • Ensure to only check the ant’s position after completing each move.

Code Solutions

# Function to count the number of times the ant returns to the boundary
def count_boundary_returns(nums):
    # Initialize current position and return counter
    current_position = 0
    return_count = 0
    
    # Iterate through the moves in the list
    for move in nums:
        # Update current position based on move direction
        current_position += move
        # Check if the ant is at the boundary after the move
        if current_position == 0:
            return_count += 1
    return return_count

# Example usage:
nums = [2, 3, -5]
print(count_boundary_returns(nums))  # Output: 1
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