# 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```
```

## Dynamic Array

Create a list, seqList, of n empty sequences, where each sequence is indexed from 0 to n-1. The elements within each of the n sequences also use 0-indexing. Create an integer, lastAnswer, and initialize it to 0. There are 2 types of queries that can be performed on the list of sequences: 1. Query: 1 x y a. Find the sequence, seq, at index ((x xor lastAnswer)%n) in seqList.

## Left Rotation

A left rotation operation on an array of size n shifts each of the array's elements 1 unit to the left. Given an integer, d, rotate the array that many steps left and return the result. Example: d=2 arr=[1,2,3,4,5] After 2 rotations, arr'=[3,4,5,1,2]. Function Description: Complete the rotateLeft function in the editor below. rotateLeft has the following parameters: 1. int d

## Sparse Arrays

There is a collection of input strings and a collection of query strings. For each query string, determine how many times it occurs in the list of input strings. Return an array of the results. Example: strings=['ab', 'ab', 'abc'] queries=['ab', 'abc', 'bc'] There are instances of 'ab', 1 of 'abc' and 0 of 'bc'. For each query, add an element to the return array, results=[2,1,0]. Fun

## Array Manipulation

Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu

## Print the Elements of a Linked List

This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node's data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print. Function Description: Complete the printLinkedList function in the editor below. printLinkedList has the following parameter(s): 1.SinglyLinkedListNode

## Insert a Node at the Tail of a Linked List

You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink