Problem Description

You have n oranges and want to eat all of them in the minimum number of days. Each day, you can choose exactly one of the following actions:

1. Eat one orange - Reduce the number of oranges by 1
2. Eat half of the oranges - If the current number of oranges n is divisible by 2, you can eat n/2 oranges (leaving n/2 oranges remaining)
3. Eat two-thirds of the oranges - If the current number of oranges n is divisible by 3, you can eat 2*(n/3) oranges (leaving n/3 oranges remaining)

Given an integer n representing the initial number of oranges, return the minimum number of days needed to eat all n oranges.

For example, if you have 10 oranges:

• Day 1: Eat 1 orange, leaving 9 oranges
• Day 2: Since 9 is divisible by 3, eat 6 oranges (2×3), leaving 3 oranges
• Day 3: Since 3 is divisible by 3, eat 2 oranges, leaving 1 orange
• Day 4: Eat the last orange

This would take 4 days total.

The challenge is to find the optimal sequence of actions that minimizes the total number of days needed to consume all oranges.

Intuition

The key insight is that eating oranges one by one is inefficient when we have large numbers. Instead, we should leverage the division operations (dividing by 2 or 3) as much as possible since they allow us to reduce the number of oranges dramatically in a single day.

Consider if we have 100 oranges. Eating them one by one would take 100 days. But if we first eat 1 orange to get to 99, then eat 2*(99/3) = 66 oranges to get to 33, we've eliminated 67 oranges in just 2 days!

This suggests a greedy strategy: we want to reach numbers divisible by 2 or 3 as quickly as possible, then use the division operations. For any number n:

• To make it divisible by 2, we need n % 2 days of eating one orange
• To make it divisible by 3, we need n % 3 days of eating one orange

Once we reach a number divisible by 2 or 3, we can perform the division operation in 1 day, reducing our problem to a smaller instance: n/2 or n/3 oranges.

This naturally leads to a recursive approach where for each n, we consider:

1. Eat n % 2 oranges one by one to make it divisible by 2, then divide by 2 (total: n % 2 + 1 days plus the days needed for n/2)
2. Eat n % 3 oranges one by one to make it divisible by 3, then divide by 3 (total: n % 3 + 1 days plus the days needed for n/3)

We choose whichever path gives us the minimum total days. The base cases are when n < 2, where we simply need n days to eat the remaining oranges.

Since the same values of n might be reached through different paths, memoization helps avoid recalculating the same subproblems, making the solution efficient.

Learn more about Memoization and Dynamic Programming patterns.

Solution Approach

The solution implements a memoization search (also known as top-down dynamic programming) to efficiently solve this problem.

Implementation Details:

1. Main Function Structure: We define a recursive function dfs(n) that returns the minimum number of days needed to eat n oranges.

2. Base Case: When n < 2:

   • If n = 0, we need 0 days (no oranges left)
   • If n = 1, we need 1 day (eat the last orange)
   • So we simply return n

3. Recursive Case: For n >= 2, we calculate:

   1 + min(n % 2 + dfs(n // 2), n % 3 + dfs(n // 3))

   Breaking this down:

   • Option 1: n % 2 + dfs(n // 2)
     • First eat n % 2 oranges one by one to make n divisible by 2
     • Then perform the division by 2 operation (1 additional day)
     • Recursively solve for n // 2 oranges
   • Option 2: n % 3 + dfs(n // 3)
     • First eat n % 3 oranges one by one to make n divisible by 3
     • Then perform the division by 3 operation (1 additional day)
     • Recursively solve for n // 3 oranges
   • The 1 + accounts for the day spent performing the division operation
   • We take the minimum of these two options

4. Memoization: The @cache decorator (from Python's functools) automatically memoizes the function results. This means:

   • Each unique value of n is computed only once
   • Subsequent calls with the same n return the cached result immediately
   • This prevents exponential time complexity from redundant calculations

Why This Works:

The algorithm exploits the fact that division operations (by 2 or 3) are much more efficient than eating oranges one by one. By always choosing the path that minimizes the total days (including the days needed to reach a divisible number), we guarantee finding the optimal solution.

The memoization ensures that even though the recursion tree might revisit the same values of n through different paths, we only compute each value once, giving us an efficient solution with time complexity roughly O(log²n) for the number of unique states visited.

Example Walkthrough

Let's trace through the algorithm with n = 10 oranges to see how the memoized recursion finds the optimal solution.

Starting with dfs(10):

Since 10 ≥ 2, we calculate:

• Option 1 (divide by 2): 10 % 2 + dfs(10 // 2) = 0 + dfs(5)
• Option 2 (divide by 3): 10 % 3 + dfs(10 // 3) = 1 + dfs(3)

We need to evaluate both dfs(5) and dfs(3).

Computing dfs(5):

• Option 1: 5 % 2 + dfs(5 // 2) = 1 + dfs(2)
• Option 2: 5 % 3 + dfs(5 // 3) = 2 + dfs(1)

Computing dfs(3):

• Option 1: 3 % 2 + dfs(3 // 2) = 1 + dfs(1)
• Option 2: 3 % 3 + dfs(3 // 3) = 0 + dfs(1)

Computing dfs(2):

• Option 1: 2 % 2 + dfs(2 // 2) = 0 + dfs(1)
• Option 2: 2 % 3 + dfs(2 // 3) = 2 + dfs(0)

Base cases:

• dfs(1) = 1 (eat the last orange)
• dfs(0) = 0 (no oranges left)

Working backwards:

• dfs(2) = 1 + min(0 + 1, 2 + 0) = 1 + min(1, 2) = 2
• dfs(3) = 1 + min(1 + 1, 0 + 1) = 1 + min(2, 1) = 2
• dfs(5) = 1 + min(1 + 2, 2 + 1) = 1 + min(3, 3) = 4
• dfs(10) = 1 + min(0 + 4, 1 + 2) = 1 + min(4, 3) = 4

The optimal path for 10 oranges takes 4 days:

1. Day 1: Eat 1 orange (10 → 9)
2. Day 2: Divide by 3, eat 6 oranges (9 → 3)
3. Day 3: Divide by 3, eat 2 oranges (3 → 1)
4. Day 4: Eat the last orange (1 → 0)

Notice how memoization helps: if we later need dfs(3) or dfs(5) again in a different branch, we already have the answer cached!

Solution Implementation

Time and Space Complexity

The time complexity is O(log^2 n), and the space complexity is O(log^2 n).

Time Complexity Analysis: The algorithm uses memoization to solve the problem recursively. At each step, we either:

• Reduce n to n // 2 (after handling n % 2 operations)
• Reduce n to n // 3 (after handling n % 3 operations)

The key insight is that the problem space forms a tree where each node n can branch to n // 2 and n // 3. The depth of this tree is O(log n) because we're repeatedly dividing by 2 or 3.

Due to memoization with @cache, each unique value is computed only once. The number of unique subproblems is bounded by the number of distinct values we can reach by repeatedly dividing by 2 and 3. This creates approximately O(log^2 n) unique states, as we can reach a value through different combinations of dividing by 2 and 3.

Therefore, the time complexity is O(log^2 n).

Space Complexity Analysis: The space complexity consists of:

1. The recursion call stack depth: O(log n) in the worst case
2. The memoization cache storing all unique subproblems: O(log^2 n) entries

The dominant factor is the cache storage, making the overall space complexity O(log^2 n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Naive BFS/DFS Without Memoization Leading to TLE

The Pitfall: Many developers initially attempt a straightforward BFS or DFS approach that explores all three options (eat 1, divide by 2, divide by 3) at each step without memoization. This creates an exponential search space.

# WRONG: Naive recursive approach without memoization
def minDays_wrong(n):
    if n == 0:
        return 0
    if n == 1:
        return 1
  
    result = 1 + minDays_wrong(n - 1)  # Always try eating 1
    if n % 2 == 0:
        result = min(result, 1 + minDays_wrong(n // 2))
    if n % 3 == 0:
        result = min(result, 1 + minDays_wrong(n // 3))
  
    return result

This approach will timeout for large values of n because it recalculates the same subproblems multiple times and explores the "eat 1 orange" path too extensively.

The Solution: Add memoization and avoid exploring the "eat 1" option directly in recursion. Instead, only eat individual oranges to reach the nearest multiple of 2 or 3:

from functools import cache

@cache
def minDays(n):
    if n < 2:
        return n
    # Only consider paths through division by 2 or 3
    return 1 + min(n % 2 + minDays(n // 2), 
                   n % 3 + minDays(n // 3))

2. Using Bottom-Up DP with Large Arrays

The Pitfall: Attempting to solve this with traditional bottom-up dynamic programming by creating a DP array of size n+1:

# WRONG: Memory limit exceeded for large n
def minDays_wrong(n):
    if n <= 1:
        return n
  
    dp = [0] * (n + 1)  # This will cause MLE for n = 10^9
    dp[1] = 1
  
    for i in range(2, n + 1):
        dp[i] = dp[i - 1] + 1
        if i % 2 == 0:
            dp[i] = min(dp[i], dp[i // 2] + 1)
        if i % 3 == 0:
            dp[i] = min(dp[i], dp[i // 3] + 1)
  
    return dp[n]

For large values of n (up to 2×10^9 in this problem), this approach will cause memory limit exceeded.

The Solution: Use top-down memoization which only stores the states that are actually visited. The number of unique states is much smaller (approximately O(log²n)) compared to O(n).

3. Incorrect Understanding of the Operations

The Pitfall: Misunderstanding what "eat half" or "eat two-thirds" means. Some might think:

• Eating half means reducing n to n/2 (incorrect)
• Eating two-thirds means reducing n to n/3 (correct, but confusing wording)

The problem states that when n is divisible by 3, you eat 2n/3 oranges, leaving n/3 oranges. This is equivalent to dividing n by 3.

The Solution: Clearly understand that:

• Operation 2: n → n/2 (when n is even)
• Operation 3: n → n/3 (when n is divisible by 3)
• Operation 1: n → n-1 (always available)

4. Not Handling the Modulo Operations Correctly

The Pitfall: Forgetting to account for the days needed to make n divisible by 2 or 3:

# WRONG: Doesn't account for days to reach divisible numbers
@cache
def minDays_wrong(n):
    if n < 2:
        return n
    return 1 + min(minDays_wrong(n // 2), minDays_wrong(n // 3))

This incorrectly assumes we can always divide by 2 or 3, even when n is not divisible by these numbers.

The Solution: Always include the modulo operations to count the days needed to eat oranges one by one until reaching a divisible number:

return 1 + min(n % 2 + minDays(n // 2), 
               n % 3 + minDays(n // 3))

This ensures we count all the days correctly: the days to reach a divisible number plus one day for the division operation itself.