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.​
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.
😆😆😆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

codeforces rating system | Codeforces rating Newbie to Legendary Grandmaster

 Codeforces rating system | Codeforces rating Newbie to Legendary Grandmaster- Codeforces is one of the most popular platforms for competitive programmers and  codeforces rating matters a lot  .  Competitive Programming  teaches you to find the easiest solution in the quickest possible way. CP enhances your problem-solving and debugging skills giving you real-time fun. It's brain-sport. As you start solving harder and harder problems in live-contests your analytical and rational thinking intensifies. To have a good codeforces profile makes a good impression on the interviewer. If you have a good  codeforces profile so it is very easy to get a referral for product base company like amazon, google , facebook etc.So in this blog I have explained everything about codeforces rating system. What are different titles on codeforces- based on rating codeforces divide rating into 10 part. Newbie Pupil Specialist Expert Candidate Codemaster Master International Master Grandmaster Internat

Apple Division CSES Problem Set Solution | CSES Problem Set Solution Apple division with code

 Apple Division CSES Problem Set Solution | CSES Problem Set Solution Apple division with code - Apple Division CSES Problem Solution Easy Explanation. Apple division is problem is taken form cses introductory problem set.Let's Read Problem statement first. Problem Statement- Time limit:  1.00 s   Memory limit:  512 MB There are  n n  apples with known weights. Your task is to divide the apples into two groups so that the difference between the weights of the groups is minimal. Input The first input line has an integer  n n : the number of apples. The next line has  n n  integers  p 1 , p 2 , … , p n p 1 , p 2 , … , p n : the weight of each apple. Output Print one integer: the minimum difference between the weights of the groups. Constraints 1 ≤ n ≤ 20 1 ≤ n ≤ 20 1 ≤ p i ≤ 10 9 1 ≤ p i ≤ 10 9 Example Input: 5 3 2 7 4 1 Output: 1 Explanation: Group 1 has weights 2, 3 and 4 (total weight 9), and group 2 has weights 1 and 7 (total weight 8). Join Telegram channel for code discussi

Concert Tickets Cses Problem set solution | Concert Tickets Cses Problem set solution Using multiset

 Concert Tickets Cses Problem set solution- Porblem statement- Time limit:  1.00 s   Memory limit:  512 MB There are  n n  concert tickets available, each with a certain price. Then,  m m  customers arrive, one after another. Each customer announces the maximum price he or she is willing to pay for a ticket, and after this, they will get a ticket with the nearest possible price such that it does not exceed the maximum price. Input The first input line contains integers  n n  and  m m : the number of tickets and the number of customers. The next line contains  n n  integers  h 1 , h 2 , … , h n h 1 , h 2 , … , h n : the price of each ticket. The last line contains  m m  integers  t 1 , t 2 , … , t m t 1 , t 2 , … , t m : the maximum price for each customer. Output Print, for each customer, the price that they will pay for their ticket. After this, the ticket cannot be purchased again. If a customer cannot get any ticket, print  − 1 − 1 . Constraints 1 ≤ n , m ≤ 2 ⋅ 10 5 1 ≤ n , m ≤ 2 ⋅