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Trapping Rain Water

Compute how much rain water an elevation map holds.

Problema

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.

Vincoli

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

Esempi

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

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

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

Output atteso:
long expected = 6;

Spiegazione

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

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

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

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

Output atteso:
long expected = 0;

Spiegazione

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

Riferimenti

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