Rain Catcher - Amazon Top Interview Questions


Problem Statement :


You are given a list of non-negative integers nums where each element represents the height of a hill. Suppose it will rain and all the spaces between two sides get filled up.

Return the amount of rain that would be caught between the hills.

Constraints

n ≤ 100,000 where n is the length of nums

Example 1

Input

nums = [2, 5, 2, 0, 5, 8, 8]

Output

8

Explanation

nums[2] can catch 3 rain drops, and nums[3] can catch 5 for a total of 8.

Example 2

Input

nums = [2, 1, 2]

Output

1

Explanation

We can hold 1 unit of water in middle.

Example 3

Input

nums = [3, 0, 1, 3, 0, 5]

Output

8

Explanation

We can hold 3 units in the first index, 2 in the second, and 3 in the fourth index (we cannot hold 5 since it would run off to the left), so we can catch 8 units of water.


Solution :



title-img 




                        Solution in C++ :

int solve(vector<int>& nums) {
    int n = nums.size();
    if (n == 0) return 0;
    int ans = 0;
    vector<int> left(n, 0);
    vector<int> right(n, 0);
    left[0] = nums[0];
    right[n - 1] = nums[n - 1];
    for (int i = 1; i < n; i++) left[i] = max(nums[i], left[i - 1]);
    for (int i = n - 2; i >= 0; i--) right[i] = max(nums[i], right[i + 1]);
    for (int i = 0; i < n; i++) ans += min(left[i], right[i]) - nums[i];
    return ans;
}
                    


                        Solution in Java :

class Solution {
    public int solve(int[] nums) {
        int n = nums.length;
        if (n < 3) {
            return 0;
        }
        int[] left = new int[n];
        left[0] = nums[0];
        for (int i = 1; i < n; i++) {
            left[i] = Math.max(nums[i], left[i - 1]);
        }
        int water = 0, right = nums[n - 1];
        for (int i = n - 2; i > 0; i--) {
            int side = Math.min(left[i - 1], right);
            if (side > nums[i]) {
                water += side - nums[i];
            }
            right = Math.max(nums[i], right);
        }
        return water;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, nums):
        left, right = [], []
        for i in nums:
            left.append(max(left[-1] if left else -1, i))
        for i in reversed(nums):
            right.append(max(right[-1] if right else -1, i))
        return sum([min(left[i], right[-(i + 1)]) - nums[i] for i in range(len(nums))])


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