花花酱 LeetCode 1553. Minimum Number of Days to Eat N Oranges

There are n oranges in the kitchen and you decided to eat some of these oranges every day as follows:

• Eat one orange.
• If the number of remaining oranges (n) is divisible by 2 then you can eat  n/2 oranges.
• If the number of remaining oranges (n) is divisible by 3 then you can eat  2*(n/3) oranges.

You can only choose one of the actions per day.

Return the minimum number of days to eat n oranges.

Example 1:

Input: n = 10
Output: 4
Explanation: You have 10 oranges.
Day 1: Eat 1 orange,  10 - 1 = 9.  
Day 2: Eat 6 oranges, 9 - 2*(9/3) = 9 - 6 = 3. (Since 9 is divisible by 3)
Day 3: Eat 2 oranges, 3 - 2*(3/3) = 3 - 2 = 1. 
Day 4: Eat the last orange  1 - 1  = 0.
You need at least 4 days to eat the 10 oranges.

Example 2:

Input: n = 6
Output: 3
Explanation: You have 6 oranges.
Day 1: Eat 3 oranges, 6 - 6/2 = 6 - 3 = 3. (Since 6 is divisible by 2).
Day 2: Eat 2 oranges, 3 - 2*(3/3) = 3 - 2 = 1. (Since 3 is divisible by 3)
Day 3: Eat the last orange  1 - 1  = 0.
You need at least 3 days to eat the 6 oranges.

Example 3:

Input: n = 1
Output: 1

Example 4:

Input: n = 56
Output: 6

Constraints:

• 1 <= n <= 2*10^9

Solution: Greedy + DP

Eat oranges one by one to make it a multiply of 2 or 3 such that we can eat 50% or 66.66…% of the oranges in one step.
dp(n) := min steps to finish n oranges.
base case n <= 1, dp(n) = n
transition: dp(n) = 1 + min(n%2 + dp(n/2), n % 3 + dp(n / 3))
e.g. n = 11,
we eat 11%2 = 1 in one step, left = 10 and then eat 10 / 2 = 5 in another step. 5 left for the subproblem.
we eat 11%3 = 2 in two steps, left = 9 and then eat 9 * 2 / 3 = 6 in another step, 3 left for the subproblem.
dp(11) = 1 + min(1 + dp(5), 2 + dp(3))

T(n) = 2*T(n/2) + O(1) = O(n)
Time complexity: O(n) // w/o memoization, close to O(logn) in practice.
Space complexity: O(logn)

Solution 2: BFS

if x % 2 == 0, push x/2 onto the queue
if x % 3 == 0, push x/3 onto the queue
always push x – 1 onto the queue