TrappingRainWater.java

package by.andd3dfx.common;

/**
 * <pre>
 * <a href="https://leetcode.com/problems/trapping-rain-water/description/">Task description</a>
 *
 * 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.
 *
 * Example 1:
 * <img src="https://assets.leetcode.com/uploads/2018/10/22/rainwatertrap.png" />
 *
 * Input: height = [0,1,0,2,1,0,1,3,2,1,2,1]
 * Output: 6
 * Explanation: The above elevation map (black section) is represented
 * by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped.
 *
 * Example 2:
 * Input: height = [4,2,0,3,2,5]
 * Output: 9
 * </pre>
 *
 * @see <a href="https://youtu.be/zGrmT8V8PVE">Video solution</a>
 */
public class TrappingRainWater {

    public static int trap(int[] height) {
        if (height == null || height.length == 0) {
            return 0;
        }

        int n = height.length;
        int[] leftMax = new int[n];
        int[] rightMax = new int[n];
        int water = 0;

        // Calculate leftMax
        leftMax[0] = height[0];
        for (int i = 1; i < n; i++) {
            leftMax[i] = Math.max(height[i], leftMax[i - 1]);
        }

        // Calculate rightMax
        rightMax[n - 1] = height[n - 1];
        for (int i = n - 2; i >= 0; i--) {
            rightMax[i] = Math.max(height[i], rightMax[i + 1]);
        }

        // Calculate trapped water
        for (int i = 0; i < n; i++) {
            water += Math.min(leftMax[i], rightMax[i]) - height[i];
        }

        return water;
    }
}