Water Bottles - Practice Coding Problems

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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Understanding the Visualization

Input

numBottles=9 full bottles, numExchange=3 empties needed

Process

Drink bottles and exchange empties for new ones

Output

Maximum 13 bottles can be drunk total

Key Takeaway

🎯 Key Insight: Mathematical formula eliminates the need for simulation loops

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

Approach	Time	Space	Notes
✓ Mathematical Formula	O(1)	O(1)	Calculate total exchanges using mathematical formula
Simulation Approach	O(log n)	O(1)	Simulate the drinking and exchanging process step by step

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) {
    if (numExchange == 1) {
        return numBottles;  // Can exchange infinitely
    }
    
    // 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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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Mathematical Formula — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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