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
Solution1:
BFS
class Solution {
public int minDays(int n) {
Deque<Integer> d = new ArrayDeque();
Set<Integer> set = new HashSet();
d.offer(n);
set.add(n);
int step = 0;
while (!d.isEmpty()) {
int size = d.size();
for (int i = 0; i < size; i ++) {
int curr = d.poll();
if (curr == 0) {
return step;
}
if (curr % 2 == 0) {
if (set.add(curr / 2)) {
d.offer(curr / 2);
}
}
if (set.add(curr - 1)) {
d.offer(curr - 1);
}
if (curr % 3 == 0) {
if (set.add(curr / 3)) {
d.offer(curr / 3);
}
}
}
step ++;
}
return -1;
}
}
dp
class Solution {
Map<Integer, Integer> map = new HashMap();
public int minDays(int n) {
if (n <= 1) return n;
if (map.get(n) != null) {
return map.get(n);
}
int res = 1 + Math.min(n % 2 + minDays(n / 2), n % 3 + minDays(n / 3));
map.put(n, res);
return res;
}
}