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Pour Water Between Buckets to Make Water Levels Equal

Number: 2273

Difficulty: Medium

Paid? Yes

Companies: Deutsche Bank

Problem Description

Given n buckets containing different amounts of water and a loss percentage incurred during any water transfer, the goal is to equalize the water level in all buckets. For any pour, if k gallons are transferred from one bucket to another, loss percent of k is spilled. You can pour fractional gallons. Determine the maximum possible water level that can be achieved uniformly in all buckets after optimally transferring water.

Key Insights

  • Use binary search to guess the candidate water level x.
  • For buckets with more water than x, calculate the available excess water after accounting for loss: (bucket - x) * (1 - loss/100).
  • For buckets with less water than x, compute the required water: (x - bucket).
  • The candidate x is achievable if the sum of transferable water from buckets with excess is at least equal to the sum of required water for buckets in deficit.
  • Continue refining the guess until the error tolerance (10^-5) is met.

Space and Time Complexity

Time Complexity: O(n log(maxBucketsValue/precision)) where n is the number of buckets. Space Complexity: O(1) extra space aside from input.

Solution

We use a binary search approach over the final water level x. The search range is [min(buckets), max(buckets)], though in some cases it might be more efficient to consider [0, max(buckets)] since pouring can only reduce water loss but never add extra water. In each iteration:

  1. For a middle value x, calculate how much water extra is available from buckets having more than x after accounting for the loss percentage.
  2. Also sum up how much water is needed by the buckets below the x level.
  3. If the available water is at least as much as the required water, then x might be achievable, and we move the search to higher values. Otherwise, lower x.
  4. The process is repeated until x is determined within the acceptable error margin.

The primary data structure used is an array traversal, and the algorithmic technique is binary search with greedy checking.

Code Solutions

# Python solution with detailed line-by-line comments

def maxWaterLevel(buckets, loss):
    # Loss rate expressed as a fraction (e.g., 80% -> 0.8)
    loss_rate = loss / 100.0
    # Set precision tolerance for binary search termination
    tolerance = 1e-5
    # Define binary search boundaries:
    # Lower bound can be 0. Upper bound is the maximum water available in any bucket.
    low, high = 0, max(buckets)
    
    # Binary search loop
    while high - low > tolerance:
        mid = (low + high) / 2.0  # Candidate water level
        extra = 0.0  # Total water that can be donated from buckets with water above mid
        needed = 0.0  # Total water required for buckets below mid
        
        # Iterate over each bucket to determine extra water or needed water
        for water in buckets:
            if water > mid:
                # Calculate transferable water after loss
                extra += (water - mid) * (1 - loss_rate)
            else:
                # Calculate water needed to reach candidate level mid
                needed += mid - water
        
        # Check if the extra water is sufficient for all deficits
        if extra >= needed:
            low = mid  # Candidate level is achievable; try for a higher level
        else:
            high = mid  # Candidate level not achievable; try lowering the water level
    
    # Return the maximum achievable water level within tolerance
    return low

# Example usage:
buckets1 = [1, 2, 7]
loss1 = 80
print("{:.5f}".format(maxWaterLevel(buckets1, loss1)))  # Expected output: 2.00000

buckets2 = [2, 4, 6]
loss2 = 50
print("{:.5f}".format(maxWaterLevel(buckets2, loss2)))  # Expected output: 3.50000

buckets3 = [3, 3, 3, 3]
loss3 = 40
print("{:.5f}".format(maxWaterLevel(buckets3, loss3)))  # Expected output: 3.00000
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