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Trapping Rain Water

Compute how much rain water an elevation map holds.

Problem

Bars of width 1 have the given heights. After rain, water collects above each bar up to the lower of the tallest bars on its left and right. Compute the total trapped water. An O(n) two-pointer solution is expected.

Implement trapped_water(heights).

Input

heights: Bar heights from left to right.

Output

Return the total units of water.

Constraints

0 ≤ length ≤ 200,000; 0 ≤ height ≤ 100,000

Examples

Example 1
Input:
heights = [0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1]

Expected output:
expected = 6
Input:
std::vector<long long> heights = {0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1};

Expected output:
long long expected = 6;
Input:
long[] heights = new long[] {0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1};

Expected output:
long expected = 6;

Explanation

Water fills the dips: 1 + 1 + 2 + 1 + 1 = 6 units.

Example 2
Input:
heights = [3, 2, 1]

Expected output:
expected = 0
Input:
std::vector<long long> heights = {3, 2, 1};

Expected output:
long long expected = 0;
Input:
long[] heights = new long[] {3, 2, 1};

Expected output:
long expected = 0;

Explanation

Heights only go down, so water runs off: 0.

References

This challenge is adapted for CodeUtility. Follow the statement and constraints above.

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