package com.fishercoder.solutions;

/**
 * 42. Trapping Rain Water
 * Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

 For example,
 Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

 The above elevation map 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.
 */

public class _42 {

    public static class Solution1 {
        /**O(n) time and O(1) space, awesome!*/
        /**
         * reference: https://discuss.leetcode.com/topic/22976/my-accepted-java-solution
         */
        public int trap(int[] height) {
            if (height == null || height.length <= 2) {
                return 0;
            }

            int max = height[0];
            int maxIndex = 0;
            for (int i = 0; i < height.length; i++) {
                if (height[i] > max) {
                    max = height[i];
                    maxIndex = i;
                }
            }

            int water = 0;

            int leftMax = height[0];
            for (int i = 0; i < maxIndex; i++) {
                if (height[i] > leftMax) {
                    leftMax = height[i];
                } else {
                    water += leftMax - height[i];
                }
            }

            int rightMax = height[height.length - 1];
            for (int i = height.length - 1; i > maxIndex; i--) {
                if (height[i] > rightMax) {
                    rightMax = height[i];
                } else {
                    water += rightMax - height[i];
                }
            }

            return water;
        }
    }
}
package com.fishercoder.solutions;

/**
 * 42. Trapping Rain Water
 * Given n non-negative integers representing an elevation map where the width of each bar is 1,
 * compute how much water it is able to trap after raining.

 For example,
 Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

 The above elevation map 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.
 */

public class _42 {

    public static class Solution1 {
        /**O(n) time and O(1) space, awesome!
         *
         * 1. first scan to find the max height index
         * 2. then scan from left up to max index and find all the water units up to the max height
         * 3. then scan from right down to max index and find all the water units down to the max height
         * 4. return the sum of those above two
         *
         * reference: https://discuss.leetcode.com/topic/22976/my-accepted-java-solution
         */
        public int trap(int[] height) {
            if (height == null || height.length <= 2) {
                return 0;
            }

            int max = height[0];
            int maxIndex = 0;
            for (int i = 0; i < height.length; i++) {
                if (height[i] > max) {
                    max = height[i];
                    maxIndex = i;
                }
            }

            int water = 0;

            int leftMax = height[0];
            for (int i = 0; i < maxIndex; i++) {
                if (height[i] > leftMax) {
                    leftMax = height[i];
                } else {
                    water += leftMax - height[i];
                }
            }

            int rightMax = height[height.length - 1];
            for (int i = height.length - 1; i > maxIndex; i--) {
                if (height[i] > rightMax) {
                    rightMax = height[i];
                } else {
                    water += rightMax - height[i];
                }
            }

            return water;
        }
    }
}
refactor 42