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 :



title-img 




                        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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