Trapping Rain Water


Problem Statement :


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.

 

Example 1:

Input: height = [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6
Explanation: The above elevation map (black section) is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped.
Example 2:

Input: height = [4,2,0,3,2,5]
Output: 9

Constraints:

n == height.length
1 <= n <= 2 * 104
0 <= height[i] <= 105


Solution :



title-img 


                            Solution in C :

#define min(a,b) ((a) < (b) ? (a) : (b))

int trap(int* height, int heightSize){
    int i, j;
    int* leftHighVal = (int*)malloc(heightSize * sizeof(int));
    int* rightHighVal = (int*)malloc(heightSize * sizeof(int));
    
    leftHighVal[0] = height[0];
    rightHighVal[heightSize-1] = height[heightSize-1];
    
    for(i = 1; i < heightSize; i++ ){
        if(height[i] > leftHighVal[i-1])
            leftHighVal[i] = height[i];    
        else
            leftHighVal[i] = leftHighVal[i-1];
    }

    for(i = (heightSize-2); i >= 0; i--){
        if(height[i] > rightHighVal[i+1])
            rightHighVal[i] = height[i];    
        else
            rightHighVal[i] = rightHighVal[i+1];
    }
    
    int ans = 0;
    
    for(i = 0; i < heightSize; i++){
        ans = ans + min( leftHighVal[i], rightHighVal[i] ) - height[i];
    }
    
    free(leftHighVal);
    free(rightHighVal);
    return ans;
}
                        


                        Solution in C++ :

class Solution {
public:
    //total water is trapped into the bars
    int trap(vector<int>& h) {
        int l=0,r=h.size()-1,lmax=INT_MIN,rmax=INT_MIN,ans=0;
        while(l<r){
            lmax=max(lmax,h[l]);
            rmax=max(rmax,h[r]);
            ans+=(lmax<rmax)?lmax-h[l++]:rmax-h[r--];
        }
        return ans;
    }
};
                    


                        Solution in Java :

public class Solution {
    public int trap(int[] h) {
        int l = 0, r = h.length - 1, lmax = Integer.MIN_VALUE, rmax = Integer.MIN_VALUE, ans = 0;
        while (l < r) {
        lmax = Math.max(lmax, h[l]);
        rmax = Math.max(rmax, h[r]);
        ans += (lmax < rmax) ? lmax - h[l++] : rmax - h[r--];
        }
        return ans;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
	def trap(self, height: List[int]) -> int:
		lh, rh = height[0], height[-1]
		l, r = 0, len(height)-1
		res = 0

		while l < r:
			if height[l] < height[r]:
				res += max(lh-height[l] ,0)
				lh = max(lh,height[l])
				l += 1
			else:
				res += max(rh-height[r] ,0)
				rh = max(rh,height[r])
				r -= 1

		return res


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