Trapping Rain Water Hard 0 attempts
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Trapping Rain Water

Hard Dynamic ProgrammingMemoization LeetCode

Given n non-negative integers representing an elevation map, compute how much water it can trap after raining.

Example: height = [0,1,0,2,1,0,1,3,2,1,2,1] → Output: 6

Sample Input
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Sample Output
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Constraints
  • 1 <= n <= 2 * 10^4
  • 0 <= height[i] <= 10^5
Test Cases
Case 1
Args: [[0,1,0,2,1,0,1,3,2,1,2,1]] Expected: 6
Case 2
Args: [[4,2,0,3,2,5]] Expected: 9
Topics
Dynamic ProgrammingMemoization

Two Pointers

function trap(height) {
  let left = 0, right = height.length - 1;
  let leftMax = 0, rightMax = 0, water = 0;
  while (left < right) {
    if (height[left] < height[right]) {
      leftMax = Math.max(leftMax, height[left]);
      water += leftMax - height[left];
      left++;
    } else {
      rightMax = Math.max(rightMax, height[right]);
      water += rightMax - height[right];
      right--;
    }
  }
  return water;
}

Time: O(n) | Space: O(1)

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