Phone Interview Questions

There is stream of float values (-inf, inf) which is coming as input and an integer D.

We need to find a set of 3 values which satisfy condition - |a - b| <= D, |b - c| <= D, |a - c| <= D, assuming a,b,c are 3 float values. Print these 3 values and remove them and continue ....

Constraints - All values in stream will be unique. D -> [0, inf)

Eg: Input stream - [1,10,7,-2,8,....], d = 5 Output - (when 8 comes, then print "7 8 10" and remove them and continue)

class Solution {
	private int D;
	void init(int d) {
		this.D = d;
	void func(float item) {
		// implement
	}
}

What data structure should be used here, and what approach can be applied? ()

Given one struct of time interval with using that i have to return in bool that are they overlapping 2 intervals or not. struct TimeInterval{ int id ; int startTime; int endTime; }; bool overlapping( TimeInterva t1, TimeInterva t2){} ()
Given stream of time interval tell that min how many cars will be used to book for all timeintervals;
Given : {{1,3}, {2,5},{6,8},{7,10},{9,10}} output : car1 : {1,3},{6,8},{9,10} car2 : {2,5},{7,10} ()

Imagine we have google drive in Some other country and your code is running in another country.

struct FilesFolders {
	vector<string> files;
	vector<string> folders;
}

FilesFolders FindAllFilesAndFolders(path) {
	// network call in google drive which fetches all files and folders inside this path.
	return FilesFolders;
}

// Implement Search method which return any path which has sub_string in it.
string Search(string path, string sub_string) {

}

Expectation: -> Implement this Search method and give time complexity according to network latency for google drive.
Solution -> A recurrence and focus on Time Complexity.

Follow up ->

New method:
void get_async(string path, callback_func…) {
	// creates a new thread
	// Calls callback_func after 100ms or whenever operation done.
	callback_func(files, folders);
	return;
}

-> Now implement Search using new method get_async() along with time_complexities.

()

inputs:

int n - [0..n-1] digits in the array, 
int[]arr1 - [0, 1, 2] len 3 values will be valid and will use the digits in the range of 0 to n-1
int[]arr2 - [2, 3, 1] len 3 values will be valid and will use the digits in the range of 0 to n-1
arr1 and arr2 can contain duplicate values and also all values can be equal. arr1:[0, 1, 2] arr2:[0, 1, 2]
int t - will be used to calculate upper and lower bound for each digit in the array. `0<=t<=n`.

Function signature

int solve(int n, int t, int[]arr1, int[]arr2) {
}

Return how many distinct combinations (length of 3) we can create. Distinct - order matters [0, 1, 2] [2, 0, 1] is distinct. [0, 1, 2] [0, 1, 2] not distinct E.x

n = 3
arr1[0, 1, 2]
arr2[2, 0, 1]
t = 1

You are given n fingerprints from 0 , 1 ,2 , 3.....(n-1). One has to generate a password of length n or greater than n, such that every fingerprint should be utilised atleast once. Write a general function, where you are given number of fingerprints and password length and print all the possible passwords.

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Last updated 10 months ago

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struct FilesFolders { vector<string> files; vector<string> folders; } FilesFolders FindAllFilesAndFolders(path) { // network call in google drive which fetches all files and folders inside this path. return FilesFolders; } // Implement Search method which return any path which has sub_string in it. string Search(string path, string sub_string) { } Expectation: -> Implement this Search method and give time complexity according to network latency for google drive. Solution -> A recurrence and focus on Time Complexity. Follow up -> New method: void get_async(string path, callback_func…) { // creates a new thread // Calls callback_func after 100ms or whenever operation done. callback_func(files, folders); return; } -> Now implement Search using new method get_async() along with time_complexities.

int n - [0..n-1] digits in the array, int[]arr1 - [0, 1, 2] len 3 values will be valid and will use the digits in the range of 0 to n-1 int[]arr2 - [2, 3, 1] len 3 values will be valid and will use the digits in the range of 0 to n-1 arr1 and arr2 can contain duplicate values and also all values can be equal. arr1:[0, 1, 2] arr2:[0, 1, 2] int t - will be used to calculate upper and lower bound for each digit in the array. `0<=t<=n`.