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Trapping Rain Water

Certification: Advanced Level Accuracy: 0% Submissions: 0 Points: 15

Write a C program to calculate how much water can be trapped after raining given n non-negative integers representing an elevation map where the width of each bar is 1. Given an array of non-negative integers representing an elevation map, 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]
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  • Output: 6
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  • Explanation: 
    • The elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. 
    • Water can be trapped at indices where there are higher bars on both sides. 
    • Total water trapped = 6 units
    •
Example 2
    
  
  • Input: height = [4,2,0,3,2,5]
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  • Output: 9
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  • Explanation: 
    • The elevation map is [4,2,0,3,2,5]. 
    • Water gets trapped between the bars. 
    • Total water trapped = 9 units
    •
Constraints
    
  
  • n == height.length
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  • 1 ≤ n ≤ 2 * 10^4
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  • 0 ≤ height[i] ≤ 3 * 10^4
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  • Time Complexity: O(n)
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  • Space Complexity: O(1)
    •
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Hint
Solution

Solution Hints

    
  
  • Use two pointers approach from both ends of the array
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  • Keep track of maximum height seen so far from left and right
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  • Move the pointer with smaller maximum height
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  • Calculate trapped water at each position using the minimum of left and right maximums
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  • Continue until both pointers meet
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Steps to solve by this approach:

 Step 1: Initialize two pointers at the beginning (left = 0) and end (right = n-1) of array
 Step 2: Initialize leftMax and rightMax to track maximum heights seen from respective sides
 Step 3: Compare heights at left and right pointers to determine which side to process
 Step 4: If current height is greater than or equal to respective max, update the max
 Step 5: If current height is less than respective max, add trapped water (max - current height)
 Step 6: Move the pointer from the side with smaller height towards center
 Step 7: Continue until both pointers meet and return total trapped water

Submitted Code :