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Ant on the Boundary - LeetCode 3028 Solution

Ant on the Boundary - Complete Solution Guide

Ant on the Boundary is LeetCode problem 3028, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 whethe

Detailed Explanation

The problem describes an ant moving along a boundary. The ant's movement is dictated by an array of integers, `nums`. Positive numbers indicate a rightward movement by that amount, and negative numbers indicate a leftward movement. The ant starts at the boundary (position 0). The goal is to determine how many times the ant returns to the boundary during its journey across the number array. The ant only checks its position relative to the boundary after completing each move, not during the move itself.

Solution Approach

The provided solution uses a simple iterative approach. It initializes the ant's position (`pos` or `position`) to 0 and a counter (`count`) to 0. The algorithm iterates through the `nums` array. In each iteration, it updates the ant's position by adding the current number from the array. After each update, it checks if the ant's position is 0. If it is, the `count` is incremented. Finally, the function returns the `count`, representing the number of times the ant returned to the boundary.

Step-by-Step Algorithm

  1. Step 1: Initialize the ant's position (`pos`) to 0 and the return count (`count`) to 0.
  2. Step 2: Iterate through the input array `nums`.
  3. Step 3: For each number `num` in `nums`, add `num` to `pos` (update the ant's position).
  4. Step 4: Check if `pos` is equal to 0. If it is, increment `count`.
  5. Step 5: After iterating through the entire array, return `count`.

Key Insights

  • Insight 1: The problem can be solved by tracking the ant's current position. The key is to realize that returning to the boundary simply means the ant's position becomes 0 again.
  • Insight 2: A simple iterative approach, traversing the array and updating the position is sufficient. No complex data structures are needed.
  • Insight 3: The problem's constraint that numbers are non-zero simplifies the solution. We don't need to handle 0 values specially.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Simulation, Prefix Sum.

Companies

Asked at: Accenture.