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You are given a **0-indexed** integer array `nums`

. In one operation you can replace any element of the array with **any two** elements that **sum** to it.

- For example, consider
`nums = [5,6,7]`

. In one operation, we can replace`nums[1]`

with`2`

and`4`

and convert`nums`

to`[5,2,4,7]`

.

Return *the minimum number of operations to make an array that is sorted in non-decreasing order*.

**Example 1:**

```
Input: nums = [3,9,3]
Output: 2
Explanation: Here are the steps to sort the array in non-decreasing order:
- From [3,9,3], replace the 9 with 3 and 6 so the array becomes [3,3,6,3]
- From [3,3,6,3], replace the 6 with 3 and 3 so the array becomes [3,3,3,3,3]
There are 2 steps to sort the array in non-decreasing order. Therefore, we return 2.
```

**Example 2:**

```
Input: nums = [1,2,3,4,5]
Output: 0
Explanation: The array is already in non-decreasing order. Therefore, we return 0.
```

**Constraints:**

`1 <= nums.length <= 10`

^{5}`1 <= nums[i] <= 10`

^{9}

```
public long minimumReplacement(int[] A) {
long x = 1000000000, res = 0, k;
for (int i = A.length - 1; i >= 0; --i) {
k = (A[i] + x - 1) / x;
x = A[i] / k;
res += k - 1;
}
return res;
}
```

```
long long minimumReplacement(vector<int>& A) {
long n = A.size(), x = 1e9, res = 0, k;
for (int i = n - 1; i >= 0; --i) {
k = (A[i] + x - 1) / x;
x = A[i] / k;
res += k - 1;
}
return res;
}
```

```
def minimumReplacement(self, A):
x = A[-1]
res = 0
for a in reversed(A):
k = (a + x - 1) // x
x = a // k
res += k - 1
return res
```

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