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.

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. Thanks Marcos for contributing this image!


Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6


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

        int result = 0;
        // leftMax represents the highest bar from left
        int leftMax = Integer.MIN_VALUE;
        // rightMax represents the highest bar from right
        int rightMax = Integer.MIN_VALUE;

        // use two pointers to scan the entire array until they meet with each other
        // Key points: any bars in the middle of leftMax bar and rightMax bar will not influence
        // how much water can current position trap
        for (int left = 0, right = height.length - 1; left <= right;) {
            leftMax = Math.max(leftMax, height[left]);
            rightMax = Math.max(rightMax, height[right]);
            //how much can current position trap depends on the shorter bar 
            if (leftMax < rightMax) {
                //DO NOT FORGET to subtract bar height of current position
                result += leftMax - height[left];
            else {
                result += rightMax - height[right];
        return result;

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s