Water Bottles

Water Bottles - Problem

Math Simulation Easy

There are numBottles water bottles that are initially full of water. You can exchange numExchange empty water bottles from the market with one full water bottle.

The operation of drinking a full water bottle turns it into an empty bottle.

Given the two integers numBottles and numExchange, return the maximum number of water bottles you can drink.

Input & Output

Example 1 — Basic Case

$ Input: numBottles = 9, numExchange = 3

› Output: 13

💡 Note: Drink 9 bottles → 9 empties. Exchange 9÷3=3 new bottles → drink them → 3 empties. Exchange 3÷3=1 new bottle → drink it → 1 empty. Can't exchange 1 empty (need 3). Total: 9+3+1=13.

Example 2 — Different Exchange Rate

$ Input: numBottles = 15, numExchange = 4

› Output: 19

💡 Note: Drink 15 → 15 empties. Exchange 15÷4=3 new, 3 empties left → total 6 empties. Exchange 6÷4=1 new, 2 empties left → total 3 empties. Can't exchange (need 4). Total: 15+3+1=19.

Example 3 — Minimum Case

$ Input: numBottles = 5, numExchange = 5

› Output: 6

💡 Note: Drink 5 bottles → 5 empties. Exchange 5÷5=1 new bottle → drink it → 1 empty. Can't exchange 1 empty (need 5). Total: 5+1=6.

Constraints

1 ≤ numBottles ≤ 100
2 ≤ numExchange ≤ 100

Visualization

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Asked in

a Amazon 15 M Microsoft 8

Brief HTML summary: The key insight is that each exchange reduces empty bottles by (numExchange-1). Best approach is the mathematical formula: total = numBottles + (numBottles-1)÷(numExchange-1). Time: O(1), Space: O(1)

Common Approaches

✓ Mathematical Formula

⏱️ Time: O(1) Space: O(1)

Use the insight that total additional bottles we can get is (numBottles - 1) / (numExchange - 1), since each exchange cycle reduces empties by (numExchange - 1).

Simulation Approach

⏱️ Time: O(log n) Space: O(1)

Drink all initial bottles, then repeatedly exchange empty bottles for new ones until we don't have enough empties to make an exchange.

Mathematical Formula — Algorithm Steps

Calculate maximum possible exchanges using formula
Add to initial bottles for total
Handle edge cases

Visualization

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Step-by-Step Walkthrough

Initial Setup

Start with numBottles=9, numExchange=3

Apply Formula

Additional = (9-1)/(3-1) = 8/2 = 4

Total

9 + 4 = 13 bottles

Code -

solution.c — C

#include <stdio.h>

int solution(int numBottles, int numExchange) {
    // Each exchange reduces empty bottles by (numExchange - 1)
    // Maximum additional bottles = (numBottles - 1) / (numExchange - 1)
    int additionalBottles = (numBottles - 1) / (numExchange - 1);
    return numBottles + additionalBottles;
}

int main() {
    int numBottles, numExchange;
    scanf("%d %d", &numBottles, &numExchange);
    
    int result = solution(numBottles, numExchange);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Single formula calculation without loops

✓ Linear Growth

Space Complexity

O(1)

Only using constant variables

✓ Linear Space

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Mathematical Formula — Algorithm Steps

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Code -

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