**Coincidence Search - Google Top Interview Questions**

### 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]:
candidates.discard(num)
for num in nums[mid + 1 :]:
if num < nums[mid]:
candidates.discard(num)
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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