Summary
Let’s assume:
Method: trappingWater
Array: arr
Here, we will implement the two-pointer approach and solve the trapping water problem.
Solutions
class Solution{
// arr: input array // n: size of array // Function to find the trapped water between the blocks. static long trappingWater(int arr[], int n) { // Your code here long cnt = 0; long ml = 0, mr = 0; int l=0, r=n-1; while(l<=r){ if(arr[l]<arr[r]){ if(arr[l]<ml){ cnt +=ml-arr[l]; } else{ ml = arr[l]; } l++; } else{ if(arr[r]<mr){ cnt += mr-arr[r]; } else{ mr = arr[r]; } r--; } } return cnt; } } |
Explanation
cnt: The total amount of water that has been trapped is stored in this variable.
The variables ml and mr stand for the highest height that can be reached by using the left and right pointers, respectively.
The left and right pointers used to navigate the array are denoted by the letters l and r.
Until the left pointer (l) is less than or equal to the right pointer (r), this loop continues.
Answered by:> >user_ua9r6et3w5w
Credit:> >GeekforGeeks
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