Given

nnon-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`

.

## Analysis

To find the trapped water at position i, we need to find the maximum value of the left elements of i and right elements of i. Assuming they are maxLeft[i] and maxRight[i]. The trapped water is min(maxLeft[i], maxRight[i]) – A[i] (if this value is larger than 0).

To find maxLeft and maxRight, we need to scan the array from left to right and from right to left. You can check the details in the code below.

## Code

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public class Solution { public int trap(int[] A) { if (A.length == 0) return 0; int[] maxLeft = new int[A.length]; int[] maxRight = new int[A.length]; int max = A[0]; for (int i = 1; i < A.length; i++) { maxLeft[i] = max; if (A[i] > max) max = A[i]; } max = A[A.length - 1]; for (int i = A.length - 2; i >= 0; i--) { maxRight[i] = max; if (A[i] > max) max = A[i]; } int ret = 0; for (int i = 1; i < A.length - 1; i++) { int trap = Math.min(maxLeft[i], maxRight[i]) - A[i]; if (trap > 0) ret += trap; } return ret; } } |

## Complexity

We only scan the array 3 times. So the complexity is $O(n)$.