mediumMixed
Maximum Component Combinations Solution
Problem Statement
You are given two integer arrays, `resourceA_amounts` and `resourceB_amounts`, representing the quantities of Component A and Component B available in various batches. To synthesize one unit of a final product, `R_A` units of Component A and `R_B` units of Component B are required. Your task is to determine the maximum number of final product units that can be synthesized using the available components. For this problem, `R_A = 27` and `R_B = 13`.
Examples
Example 1:
Input:resourceA_amounts = [100, 50], resourceB_amounts = [20, 30]
Output:undefined
Explanation: Total Component A available is 150 (100 + 50). Total Component B available is 50 (20 + 30). Based on Component A, 150 / 27 = 5 units can be made. Based on Component B, 50 / 13 = 3 units can be made. The maximum is min(5, 3) = 3.
Example 2:
Input:resourceA_amounts = [20], resourceB_amounts = [100]
Output:undefined
Explanation: Total Component A available is 20. Total Component B available is 100. Based on Component A, 20 / 27 = 0 units can be made. Based on Component B, 100 / 13 = 7 units can be made. The maximum is min(0, 7) = 0.
Constraints
- 0 <= resourceA_amounts.length <= 10^5
- 0 <= resourceB_amounts.length <= 10^5
- 0 <= resourceA_amounts[i], resourceB_amounts[i] <= 10^9
- R_A = 27
- R_B = 13
- The sum of all elements in `resourceA_amounts` and `resourceB_amounts` will fit within a 64-bit integer type.
Time: O(N + M) Space: O(1)
The optimal approach leverages the fact that to maximize the minimum yield, we should independently maximize the available quantity of each component. First, find the maximum quantity in `resourceA_amounts` (let's say `max_A`) and the maximum quantity in `resourceB_amounts` (`max_B`). Then, the maximum final product units will be `min(max_A / R_A, max_B / R_B)`.
Run, Test & Submit Code
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Solve on Interactive WorkspaceTested Solutions
import sys
R_A = 27
R_B = 13
def solve():
# Read resourceA_amounts from the first line of stdin
resourceA_amounts_str = sys.stdin.readline().strip()
# Read resourceB_amounts from the second line of stdin
resourceB_amounts_str = sys.stdin.readline().strip()
# Convert space-separated strings to a list of integers
resourceA_amounts = list(map(int, resourceA_amounts_str.split()))
resourceB_amounts = list(map(int, resourceB_amounts_str.split()))
# Calculate the total available Component A
total_A = sum(resourceA_amounts)
# Calculate the total available Component B
total_B = sum(resourceB_amounts)
# Determine the maximum products based on Component A availability
# Using // for integer division which floors the result for positive numbers
max_products_A = total_A // R_A
# Determine the maximum products based on Component B availability
max_products_B = total_B // R_B
# The final product is limited by the component that runs out first
result = min(max_products_A, max_products_B)
print(result)
solve()