- Prepare
All Platforms Programming Aptitude Syllabus Interview Preparation Interview Exp. Off Campus - Prime Video
- Prime Mock

- Interview Experience
- Prime VideoNew
- Prime Mock
- Discussion Forum
- Nano Degree
- Prime Video
- Prime Mock

# Chakravyuha Problem (TCS Codevita)

# Chakravyuha Problem – TCS Codevita

Chakravyuha Problem is one of the problem that was asked in previous year TCS Codevita Exam. A Chakravyuha has a very well-defined co-ordinate system. Each point on the co-ordinate system is manned by a certain unit of the army. The Commander-In-Chief is always located at the center of the army to better co-ordinate his forces. The only way to crack the Chakravyuha is to defeat the units in sequential order.

## Problem Description

During the battle of Mahabharat, when Arjuna was far away in the battlefield, Guru Drona made a Chakravyuha formation of the Kaurava army to capture Yudhisthir Maharaj. Abhimanyu, young son of Arjuna was the only one amongst the remaining Pandava army who knew how to crack the Chakravyuha. He took it upon himself to take the battle to the enemies.

Abhimanyu knew how to get power points when cracking the Chakravyuha. So great was his prowess that rest of the Pandava army could not keep pace with his advances. Worried at the rest of the army falling behind, Yudhisthir Maharaj needs your help to track of Abhimanyu’s advances. Write a program that tracks how many power points Abhimanyu has collected and also uncover his trail.

A Chakravyuha has a very well-defined co-ordinate system. Each point on the co-ordinate system is manned by a certain unit of the army. The Commander-In-Chief is always located at the center of the army to better co-ordinate his forces. The only way to crack the Chakravyuha is to defeat the units in sequential order.

A Sequential order of units differs structurally based on the radius of the Chakra. The radius can be thought of as length or breadth of the matrix depicted above. The structure i.e. placement of units in sequential order is as shown below

The entry point of the Chakravyuha is always at the (0,0) co-ordinate of the matrix above. This is where the 1st army unit guards. From (0,0) i.e. 1st unit Abhimanyu has to march towards the center at (2,2) where the 25th i.e. the last of the enemy army unit guards. Remember that he has to proceed by destroying the units in sequential fashion. After destroying the first unit, Abhimanyu gets a power point. Thereafter, he gets one after destroying army units which are multiples of 11. You should also be a in a position to tell Yudhisthir Maharaj the location at which Abhimanyu collected his power points.

**Input Format:**

- First line of input will be length as well as breadth of the army units, say N.

**Output Format:**

- Print NxN matrix depicting the placement of army units with unit numbers delimited by (\t) Tab character
- Print total power points collected
- Print coordinates of power points collected in sequential fashion (one per line )

**Constraints:**

- 0 < N <= 100

**Sample Input:**

**5**

**Sample Output:**

1 2 3 4 5

16 17 18 19 6

15 24 25 20 7

14 23 22 21 8

13 12 11 10 9

Total Power points: 3

(0,0)

(4,2)

(3,2)

## Solution for Charkravyuha Problem

#include<bits/stdc++.h>

#define speed ios_base::sync_with_stdio(false); cin.tie(NULL);

#define booster cout.tie(NULL)

using namespace std;

int dp[100][100];

int main()

{

speed;

int n,c=1,co,road,ans=1;

string str="(0,0)\n";

cin>>n;

int nn=n*n;

if(n&1)

{

co=-1;

int i=n/2,j=n/2,i1,j1;

while(nn)

{

if(c&1)

{j1=j+co;road=abs(co);

for(j;j!=j1;j+=co/abs(co))

dp[i][j]=nn--;

co*=-1;

}

else {i1=i+co;road=abs(co);

for(i;i!=i1;i+=co/abs(co))

dp[i][j]=nn--;

co+=co/abs(co);

}

c++;

}

}

else

{

co=1;

int i=n/2,j=n/2 -1,i1,j1;

while(nn)

{

if(c&1)

{j1=j+co;road=abs(co);

for(j;j!=j1;j+=co/abs(co))

dp[i][j]=nn--;

co*=-1;

}

else {i1=i+co;road=abs(co);

for(i;i!=i1;i+=co/abs(co))

dp[i][j]=nn--;

co+=co/abs(co);

}

c++;

}

}

for(int ii=0;ii<n;ii++)

{for(int jj=0;jj<n;jj++)

{

cout<<dp[ii][jj]<<" ";

if(dp[ii][jj]%11==0)

{

ans++;

str+= "(" + to_string(ii) +","+ to_string(jj)+")\n";

}

}

cout<<endl;

}

cout<<"Total Power points : "<<ans<<endl<<str;

return 0;

}

import java.util.ArrayList;

import java.util.Scanner;

public class ChaalravyuhProblem

{

public static void main (String[]args)

{

Scanner scr = new Scanner (System.in);

int n = scr.nextInt ();

int a[][] = new int[n][n];

int lc = 0;

int lr = 0;

int hc = a[0].length - 1;

int hr = a.length - 1;

int pp = 1;

int nel = (hc + 1) * (hr + 1);

int ct = 1;

ArrayList < Integer > al = new ArrayList <> ();

al.add (0);

al.add (0);

while (lc < hc && lr < hr)

{

for (int i = lc; i <= hc && ct <= nel; i++)

{

a[lr][i] = ct;

if (a[lr][i] % 11 == 0)

{

pp++;

al.add (lr);

al.add (i);

}

ct++;

}

lr++;

for (int i = lr; i <= hr && ct <= nel; i++)

{

a[i][hc] = ct;

if (a[i][hc] % 11 == 0)

{

pp++;

al.add (i);

al.add (hc);

}

ct++;

}

hc--;

for (int i = hc; i >= lc && ct <= nel; i--)

{

a[hr][i] = ct;

if (a[hr][i] % 11 == 0)

{

pp++;

al.add (hr);

al.add (i);

}

ct++;

}

hr--;

for (int i = hr; i >= lr && ct <= nel; i--)

{

a[i][lc] = ct;

if (a[i][lc] % 11 == 0)

{

pp++;

al.add (i);

al.add (lc);

}

ct++;

}

lc++;

}

for (int i = 0; i < n; i++)

{

for (int j = 0; j < n; j++)

{

System.out.print (a[i][j] + " ");

}

System.out.println ();

}

System.out.println ("Total Power points :" + pp);

for (int i = 0; i < al.size (); i += 2)

{

System.out.println ("(" + al.get (i) + "," + al.get (i + 1) + ")");

}

}

}

Login/Signup to comment