### Problem Statement :

```You are given a list of unique integers nums. Return the number of integers that could still be successfully found using a standard binary search.

Binary search in pseudocode:

lo = 0
hi = nums.size - 1

while lo <= hi
mid = floor((lo + hi) / 2)
if nums[mid] == target
return mid
elif nums[mid] < target
lo = mid + 1
else
hi = mid - 1

Constraints

0 ≤ n ≤ 100,000 where n is the length of nums.

Example 1

Input

nums = [1, 5, 3, 2, 9]

Output

3

Explanation

Since if we used binary search to look for 3, we would still find it in the first iteration. We would also
happen to find 1 and 9 after two iterations.```

### Solution :

```                        ```Solution in C++ :

int go(vector<int>& nums, int l, int r, int a, int b) {
if (l > r) return 0;
auto m = (l + r) / 2;
auto mid = nums[m];
return int(a <= mid && mid <= b) + go(nums, l, m - 1, a, min(mid - 1, b)) +
go(nums, m + 1, r, max(a, mid + 1), b);
}

int solve(vector<int>& nums) {
return go(nums, 0, (int)nums.size() - 1, INT_MIN, INT_MAX);
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int solve(int[] nums) {
if (nums.length < 2) {
return nums.length;
}

int ans = 0;
for (int i = 0; i < nums.length; i++) {
int j = Arrays.binarySearch(nums, nums[i]);
if (i == j) {
ans++;
}
}

return ans;
}
}```
```

```                        ```Solution in Python :

def remove_unfindable(nums, candidates):
if nums:
mid = (len(nums) - 1) // 2
for num in nums[:mid]:
if num > nums[mid]:
for num in nums[mid + 1 :]:
if num < nums[mid]:
if any(num in candidates for num in nums[:mid]):
remove_unfindable(nums[:mid], candidates)
if any(num in candidates for num in nums[mid + 1 :]):
remove_unfindable(nums[mid + 1 :], candidates)

class Solution:
def solve(self, nums):
candidates = set(nums)
remove_unfindable(nums, candidates)
return len(candidates)```
```

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