Trapping Rain Water

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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
Input: height = [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6
Example 2
Input: height = [4,2,0,3,2,5]
Output: 9
Constraints:

Contents

Approach 1: Prefix/Suffix Max Arrays

Water above position i = min(maxLeft[i], maxRight[i]) - height[i]. Pre-compute the running maximum from the left and from the right.

Java
public int trap(int[] height) {
    int n = height.length;
    int[] maxLeft  = new int[n]; // maxLeft[i]  = max height in [0..i]
    int[] maxRight = new int[n]; // maxRight[i] = max height in [i..n-1]

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

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

    int water = 0;
    for (int i = 0; i < n; i++) {
        water += Math.min(maxLeft[i], maxRight[i]) - height[i];
    }
    return water;
}

Approach 2: Two Pointers O(1) Space

We can avoid the extra arrays. Water at position i depends on the minimum of the max from the left and max from the right. We process the side with the smaller max first: that side's water is already determined.

Java
public int trap(int[] height) {
    int left = 0, right = height.length - 1;
    int leftMax = 0, rightMax = 0;
    int water = 0;

    while (left < right) {
        if (height[left] <= height[right]) {
            // left side is the limiting factor
            if (height[left] >= leftMax) {
                leftMax = height[left]; // update max
            } else {
                water += leftMax - height[left]; // trapped water
            }
            left++;
        } else {
            // right side is the limiting factor
            if (height[right] >= rightMax) {
                rightMax = height[right];
            } else {
                water += rightMax - height[right];
            }
            right--;
        }
    }
    return water;
}

Complexity Analysis
ApproachTimeSpace
Prefix/Suffix MaxO(n)O(n)
Two PointersO(n)O(1)