Problem Description
Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it can trap after raining.
Key Insights
- Water can be trapped between two heights, and the amount of water above a specific bar is determined by the height of the shortest of the two tallest bars on either side.
- The problem can be solved using various approaches, including two-pointer technique, dynamic programming, or a stack.
- The maximum water level above each bar depends on the tallest bars to the left and right, which can be efficiently calculated.
Space and Time Complexity
Time Complexity: O(n)
Space Complexity: O(1) for the two-pointer approach; O(n) for dynamic programming.
Solution
The problem can be effectively solved using two pointers, where we maintain pointers at both ends of the elevation map and move them inward based on the relative heights. We keep track of the maximum height seen from both ends and compute the trapped water as we iterate through the bars. This approach allows us to determine the water trapped in a single pass, achieving an optimal time complexity.