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# Minimum Numbers of Function Calls to Make Target Array

![](/files/-MFeRVLsuGBvhkkf0Y4w)

Your task is to form an integer array `nums` from an initial array of zeros `arr` that is the same size as `nums`.

Return the minimum number of function calls to make `nums` from `arr`.

The answer is guaranteed to fit in a 32-bit signed integer.

**Example 1:**

```
Input: nums = [1,5]
Output: 5
Explanation: Increment by 1 (second element): [0, 0] to get [0, 1] (1 operation).
Double all the elements: [0, 1] -> [0, 2] -> [0, 4] (2 operations).
Increment by 1 (both elements)  [0, 4] -> [1, 4] -> [1, 5] (2 operations).
Total of operations: 1 + 2 + 2 = 5.
```

**Example 2:**

```
Input: nums = [2,2]
Output: 3
Explanation: Increment by 1 (both elements) [0, 0] -> [0, 1] -> [1, 1] (2 operations).
Double all the elements: [1, 1] -> [2, 2] (1 operation).
Total of operations: 2 + 1 = 3.
```

**Example 3:**

```
Input: nums = [4,2,5]
Output: 6
Explanation: (initial)[0,0,0] -> [1,0,0] -> [1,0,1] -> [2,0,2] -> [2,1,2] -> [4,2,4] -> [4,2,5](nums).
```

**Example 4:**

```
Input: nums = [3,2,2,4]
Output: 7
```

**Example 5:**

```
Input: nums = [2,4,8,16]
Output: 8
```

**Constraints:**

* `1 <= nums.length <= 10^5`
* `0 <= nums[i] <= 10^9`

```java
class Solution {
    public int minOperations(int[] nums) {
        int count = 0;
        // N*Log(Range)
        while (true) {
            boolean allZeros = true;
            for (int i = 0; i < nums.length; i++) {
                count += nums[i] % 2;
                nums[i] /= 2;
                if (nums[i] != 0)
                    allZeros = false;
            }
            if (allZeros)
                break;
            else
                // Count for diving the array by 2
                count++;
        }
        return count;
    }
}
```