**Randomized Binary Search - Google Top Interview Questions**

### Problem Statement :

Consider a modified binary search, where instead of picking the middle element as the pivot, we pick it randomly between the low and the high indices. Given a list of integers nums that's not necessarily sorted, return the number of elements that will always be found using this modified binary search. Constraints n ≤ 100,000 where n is the length of nums Example 1 Input nums = [1, 10, 5, 20] Output 2 Explanation We can always find the elements 1 and 20. For the element 5, if we had picked index 1 as the first pivot, we wouldn't be able to find it. Similarly for 10, if we had picked index 2 as the first pivot, it wouldn't be found. Example 2 Input nums = [0, 0] Output 0 Explanation For the first element, if we search for 0 we may have index 1 as the pivot and miss it. Similarly, if we search for the second element, we may get index 0 as the pivot. In both cases, we miss the elements.

### Solution :

` ````
Solution in C++ :
int maxp[100005];
int mins[100005];
int solve(vector<int>& v) {
if (v.size() == 0) {
return 0;
}
int n = v.size();
maxp[0] = v[0];
for (int i = 1; i < n; i++) {
maxp[i] = max(maxp[i - 1], v[i]);
}
mins[n - 1] = v[n - 1];
for (int i = n - 2; i >= 0; i--) {
mins[i] = min(mins[i + 1], v[i]);
}
int ret = 0;
for (int i = 0; i < n; i++) {
bool prefixgood = i == 0 || v[i] > maxp[i - 1];
bool suffixgood = i == n - 1 || v[i] < mins[i + 1];
ret += prefixgood && suffixgood;
}
return ret;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] nums) {
int max = Integer.MIN_VALUE, min = Integer.MAX_VALUE, ct = 0;
boolean[] beg = new boolean[nums.length];
for (int i = 0; i < nums.length; i++) {
if (nums[i] > max)
beg[i] = true;
max = Math.max(max, nums[i]);
}
for (int i = nums.length - 1; i > -1; i--) {
if (nums[i] < min && beg[i] == true)
ct++;
min = Math.min(min, nums[i]);
}
return ct;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, nums):
# all elements on the left should be smaller than cur element
# all elements on the right should be greater than cur element
leftMax = [i for i in nums]
rightMin = [i for i in nums]
# the number of occurance has to be unique!
freq = collections.Counter(nums)
for i in range(1, len(nums)):
leftMax[i] = max(leftMax[i], leftMax[i - 1])
for i in range(len(nums) - 2, -1, -1):
rightMin[i] = min(rightMin[i], rightMin[i + 1])
res = 0
# now at each position check!!!
# if leftMax is = to rightMin then we are always going to be found!
# but make sure you are not having any duplicate
for i in range(0, len(nums)):
if freq[nums[i]] == 1 and leftMax[i] == rightMin[i]:
res += 1
return res
```

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