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The **bitwise AND** of an array `nums`

is the bitwise AND of all integers in `nums`

.

- For example, for
`nums = [1, 5, 3]`

, the bitwise AND is equal to`1 & 5 & 3 = 1`

. - Also, for
`nums = [7]`

, the bitwise AND is`7`

.

You are given an array of positive integers `candidates`

. Evaluate the **bitwise AND** of every **combination** of numbers of `candidates`

. Each number in `candidates`

may only be used **once** in each combination.

Return *the size of the largest combination of *

`candidates`

`0`

.**Example 1:**

```
Input: candidates = [16,17,71,62,12,24,14]
Output: 4
Explanation: The combination [16,17,62,24] has a bitwise AND of 16 & 17 & 62 & 24 = 16 > 0.
The size of the combination is 4.
It can be shown that no combination with a size greater than 4 has a bitwise AND greater than 0.
Note that more than one combination may have the largest size.
For example, the combination [62,12,24,14] has a bitwise AND of 62 & 12 & 24 & 14 = 8 > 0.
```

**Example 2:**

```
Input: candidates = [8,8]
Output: 2
Explanation: The largest combination [8,8] has a bitwise AND of 8 & 8 = 8 > 0.
The size of the combination is 2, so we return 2.
```

**Constraints:**

`1 <= candidates.length <= 10`

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

^{7}

```
public int largestCombination(int[] A) {
int res = 0, cur = 0;
for (int i = 1; i <= 10000000; i <<= 1) {
cur = 0;
for (int a : A)
if ((a & i) > 0)
cur++;
res = Math.max(res, cur);
}
return res;
}
```

```
int largestCombination(vector<int>& A) {
int res = 0, cur = 0;
for (int i = 1; i <= 10000000; i <<= 1) {
cur = 0;
for (int& a : A)
if (a & i)
cur++;
res = max(res, cur);
}
return res;
}
```

```
def largestCombination(self, A):
return max(sum(1 << i & a > 0 for a in A) for i in range(30))
```

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