42. Trapping Rain Water 🌟🌟🌟

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.

Brute force

• For every index find the max water that can be trapped.
• This can be done with the formula min(maxLeft(i),maxRight(i))-a[i]
• TC: O(N^2), 2nd loop for finding left and right max.
• SC: O(1)

PrefixSum Optimized

• We can pre-compute left max in an array and right max from back into the another array, so the 2 loop is not necessary.
• TC: O(N)
• SC: O(2*N)–>O(N)

2-pointer

• We can find left max and right max with 2 pointer approach.
• Explained in code.
• TC: O(N)
• SC: O(1)

Code

class Solution {
public:
    int trap(vector<int>& height)
    {
        int n = height.size();
        int left = 0;
        int right = n - 1;
        int res = 0;
        int maxleft = 0, maxright = 0;

        while (left <= right) {
            if (height[left] <= height[right]) { // update max left
                if (height[left] >= maxleft) maxleft = height[left];
                else res += maxleft - height[left];
                left++;
            } else { // update max right
                if (height[right] >= maxright) maxright = height[right];
                else res += maxright - height[right];
                right--;
            }
        }
        return res;
    }
};