Skip to main content

Josephus problem Solution with code and explanation

Josephus problem Solution-

Read Josephus Story -here



Problem Statement

Given the total number of persons n and a number k which indicates that k-1 persons are skipped and kth person is killed in circle in a fixed direction.รข€‹
The task is to choose the safe place in the circle so that when you perform these operations starting from 1
st place in the circle, you are the last one remaining and survive.



Example 1:

Input:
n = 3 k = 2
Output: 3
Explanation: There are 3 persons so 
skipping 1 person i.e 1st person 2nd 
person will be killed. Thus the safe 
position is 3

Josephus problem Solution -
Josephus servived  because he knew maths and recursion that's why he stand at 40th place.
I also know recursion so I can also survive when I will face a similar problem So you should also know How to survive in such situation for that you also have to learn recursion.
๐Ÿ˜†๐Ÿ˜†๐Ÿ˜†XD
 i have to think how can i break 
this problem into subproblem 
People are standing in a circle so counting form very first place 
i have to delete every k-1 th node from current index
For example, if n = 5 and k = 2, then the safe position is 3. Firstly, 
the person at position 2 is killed, then person at position 4 is killed,
 then person at position 1 is killed. Finally, the person at position 5
 is killed. So the person at position 3 survives.
If n = 7 and k = 3, then the safe position is 4. The persons at positions 3, 6, 2, 7, 5, 1 are
 killed in order, and person at position 4 survives.

i have simple approach i will first add all person in a vector and delete i will kill every k-1 th person and start countin again from new position
Now think about the smallest problem 
when I have only one person in my vector this will my solution no matter what is the value of k 
so let's write base case 

if(v.size()==0){
return v[0];
}

Now Think about induction step
If i am standing in a generic position in my vactor what will my intension 
to delete kth node from my posion if it is out of box start count from begin .
so i will find next index which i have to delete and then i will call my solve funcion which will return 
me final ans
induction step 
index=(index+k)%v.size();
v.erase(v.begin()+index);
return solve(v,index,k,ans);


final code is here- must share with your friends  



Comments

Popular posts from this blog

Vaccine Distribution Codechef December Long challenge solution 2020

 Vaccine Distribution  Codechef December Long challenge solution 2020- This Problem vaccine distribution is taken from December long challenge 2020. lets read problem statement. Problem statement- Finally, a COVID vaccine is out on the market and the Chefland government has asked you to form a plan to distribute it to the public as soon as possible. There are a total of  N N  people with ages  a 1 , a 2 , … , a N a 1 , a 2 , … , a N . There is only one hospital where vaccination is done and it is only possible to vaccinate up to  D D  people per day. Anyone whose age is  ≥ 80 ≥ 80  or  ≤ 9 ≤ 9  is considered to be  at risk . On each day, you may not vaccinate both a person who is at risk and a person who is not at risk. Find the smallest number of days needed to vaccinate everyone. Input The first line of the input contains a single integer  T T  denoting the number of test cases. The description of  T T  test cases follows. The first line of each test case contains two space-separ

Codechef February long challenge solution 2021

   Codechef February  long challenge solution 2021- Hey guys Welcome Again, A small update CodeChef February  long challenge solution 2021 starting from 1st week of beb  2021 . Codechef long challenge is one of the popular contests of CodeChef you should participate in codechef Feb  long challenge.This is first contest of codechef in 2021 so good luck. And wish you happy new year. Contest link- https://www.codechef.com/FEB21/?itm_medium=navmenu&itm_campaign=FEB21 I will post the solution after Contest.Stay tuned. About CodeChef  February  long challenge solution 2021: CodeChef Long Challenge is a 10-day monthly coding contest where you can show off your computer programming skills. The significance being - it gives you enough time to think about a problem, try different ways of attacking the problem, read the concepts etc. If you’re usually slow at solving problems and have ample time at hand, this is ideal for you. We also put in a lot of efforts in getting quality problems, w

Merge Two sorted array Without Extra Space

Problem statement-  Given two sorted arrays arr1[] and arr2[] of sizes N and M in non-decreasing order. Merge them in sorted order without using any extra space. Modify arr1 so that it contains the first N elements and modify arr2 so that it contains the last M elements.    Example 1: Input:  N = 4, arr1[] = [1 3 5 7]  M = 5, arr2[] = [0 2 6 8 9] Output:  arr1[] = [0 1 2 3] arr2[] = [5 6 7 8 9] Explanation: After merging the two  non-decreasing arrays, we get,  0 1 2 3 5 6 7 8 9.   Example 2: Input:  N = 2, arr1[] = [10, 12]  M = 3, arr2[] = [5 18 20] Output:  arr1[] = [5 10] arr2[] = [12 18 20] Explanation: After merging two sorted arrays  we get 5 10 12 18 20. Your Task: You don't need to read input or print anything. You only need to complete the function merge() that takes arr1, arr2, N and M as input parameters and modifies them in-place so that they look like the sorted merged array when concatenated. Expected Time Complexity:  O((n+m) log(n+m)) Expected Auxilliary Space: O(1