**Number of Decrements to Reach Zero - Google Top Interview Questions**

### Problem Statement :

You are given an integer n. In one operation you can either Decrement n by one If n is even, decrement by n / 2 If n is divisible by 3, decrement by 2 * (n / 3) Return the minimum number of operations required to decrement n to zero. Constraints n ≤ 10 ** 9 Example 1 Input n = 15 Output 5 Explanation Since n = 15 is divisible by 3 we decrement by 10 = 2 * (15 / 3) to get 5. Then we decrement by one to get 4. Then we decrement by 2 since n is even. Then we decrement n by one twice to get 0

### Solution :

` ````
Solution in C++ :
unordered_map<int, int> dp;
int solve(int n) {
if (n == 0) return 0;
if (n == 1) return 1;
if (n < 0) assert(false);
if (dp.count(n)) return dp[n];
int md3 = n % 3;
int md2 = n % 2;
dp[n] = min(md2 + 1 + solve((n - md2) / 2), md3 + 1 + solve((n - md3) / 3));
return dp[n];
}
```

` ````
Solution in Python :
class Solution:
@functools.lru_cache(None)
def solve(self, n):
if n <= 1:
return n
return min(1 + int(n % 3) + self.solve(n // 3), 1 + int(n % 2) + self.solve(n // 2))
```

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