Constellation - CodeVita 9 | Zonal Round

Constellation 

 Three characters { #, *, . } represents a constellation of stars and galaxies in space. Each galaxy is demarcated by # characters. There can be one or many stars in a given galaxy. Stars can only be in shape of vowels { A, E, I, O, U } . A collection of * in the shape of the vowels is a star. A star is contained in a 3x3 block. Stars cannot be overlapping. The dot(.) character denotes empty space. 

 Given 3xN matrix comprising of { #, *, . } character, find the galaxy and stars within them. 

 Note: Please pay attention to how vowel A is denoted in a 3x3 block in the examples section below. 

 Constraints 

 3 <= N <= 10^5 

 Input

 Input consists of single integer N denoting number of columns. 

 Output 

 Output contains vowels (stars) in order of their occurrence within the given galaxy. Galaxy itself is represented by # character. 

Example 1:

Input:

18

* . * # * * * # * * * # * * * . * .

* . * # * . * # . * . # * * * * * *

* * * # * * * # * * * # * * * * . *

Output: 

U#O#I#EA

Example 2:

Input:

12

* . * # . * * * # . * .

* . * # . . * . # * * *

* * * # . * * * # * . *

Output:

U#I#A

Program:

#include <stdio.h>

int main()

{

  int n,x1,y1;

  scanf("%d",&n);

  char x[3][n];

  for(int i=0;i<3;i++)

{

    for(int j=0;j<n;j++)

{

      scanf("%s",&x[i][j]);

    }

  }

  for(int i=0;i<n;i++)

  {

    if(x[0][i]=='#' && x[1][i]=='#' && x[2][i]=='#')

{

      printf("#");

    }

else if(x[0][i]=='.' && x[1][i]=='.' && x[2][i]=='.')

{}

else

{

      char a,b,c,a1,b1,c1,a2,b2,c2;

      x1 = i;

      a = x[0][x1];

      b = x[0][x1+1];

      c = x[0][x1+2];

      a1 = x[1][x1];

      b1 = x[1][x1+1];

      c1 = x[1][x1+2];

      a2 = x[2][x1];

      b2 = x[2][x1+1];

      c2 = x[2][x1+2];

      if(a=='.' && b=='*' && c=='.' && a1=='*' && b1=='*' && c1=='*' && a2=='*' && b2=='.' && c2=='*')

  { 

      printf("A");

        i = i + 2;

      }

      if(a=='*' && b=='*' && c=='*' && a1=='*' && b1=='*' && c1=='*' && a2=='*' && b2=='*' && c2=='*')

  { 

        printf("E");

        i = i + 2;

      }

      if(a=='*' && b=='*' && c=='*' && a1=='.' && b1=='*' && c1=='.' && a2=='*' && b2=='*' && c2=='*')

  { 

        printf("I");

        i = i + 2;

      }

      if(a=='*' && b=='*' && c=='*' && a1=='*' && b1=='.' && c1=='*' && a2=='*' && b2=='*' && c2=='*')

  { 

        printf("O");

        i = i + 2;

      }

      if(a=='*' && b=='.' && c=='*' && a1=='*' && b1=='.' && c1=='*' && a2=='*' && b2=='*' && c2=='*')

  { 

          printf("U");

        i = i + 2;

      }

    }

  }

}

You can also run it on an online IDE:    

https://ide.geeksforgeeks.org/DPOevh3dug

Your feedback are always welcome! If you have any doubt you can contact me or leave a comment!  Happy Coding !! Cheers!!!

Related Links:

https://codepiggy.blogspot.com/2020/06/square-free-numbers-codevita-9-round-1.html

Parallelograms | Code Vita 2018 | round 1

Parallelograms

Problem Description

A number (N) of lines (extending to infinity) in both directions are drawn on a plane. The lines are specified by the angle (positive or negative) made with the x axis (in degrees). It may be assumed that the different lines are not coincident (they do not overlap each other). The objective is to determine the number of parallelograms in the plane formed by these lines.

If the lines are given with an angle of 10, 70, 30 and 30, the figure looks like this

L1 is at 10 degrees to the x axis, L2 is at 70 degrees to the x axis, L3 and L4 are at 30 degrees to the x axis. It can be seen that there are no parallelograms formed by these lines

Constraints

N<=50

-89 <= angle for any line <=90

Input Format

The first line of the input consists of a single integer, N.

The next line consists of a series of comma separated integers (positive, zero or negative), each corresponding to a separate line, and giving the angle that it makes with the x axis (in degrees).

Output

The output is a single integer giving the number of parallelograms in the plane formed by the lines

Explanation

Example 1

Input

5

20,20,-20,-20,50

Output

1

Explanation

There are 5 lines (N=5), with angles at 20,20,-20,-20and 50 degrees with the x axis. The figure looks like this

There is one

Hence the output is 1.

Example 2

Input

6

20,20,-20,-20,50,50

Output

3

Explanation

There are 6 lines (N=6) with angles 20,20, -20,-20,50 and 50 degrees with the x axis.. The figure looks like this

There are three parallelograms formed by (L1, L2, L3, L4), (L1, L2, L5, L6) and (L3, L4, L5, L6). Hence the output is 3.

Program:

#include<stdio.h>
int pairfound(int angle[],int i,int n)
{
  int j;
  for(j=i+1;j<n;j++)
  {
    if(angle[i]==angle[j])
      return 1;
  }
  return 0;
}
int main()
{
  int n,i,count=0,pllgm,ang;
  scanf("%d",&n);
  int angle[n];
  for(i=0;i<n;i++)
  {
    scanf("%d,",&ang);
    if(ang>=-89&&ang<=90)
        angle[i]=ang;
  }
  for(i=0;i<n;i++)
  {
    if(pairfound(angle,i,n)==1)
    {
      count++;
    }
  }
  pllgm=(count-1)*count/2;
  printf("%d",pllgm);
  return 0;
}


Output

6

20,20,-20,-20,50,50

3

You can also try it in online IDE: https://ide.geeksforgeeks.org/vVEgz35Wso

Your comments and feedback are welcomed! You can contact me if you have any doubts! Cheers!



Related Links: Reverse Gear

Super Ascii - Code Vita 2014 | round 2

Super Ascii

Problem Decription:

In the Byteland country a string "S" is said to super ascii string if and only if count of each character in the string is equal to its ascii value.

In the Byteland country ascii code of 'a' is 1, 'b' is 2 ...'z' is 26.

Your task is to find out whether the given string is a super ascii string or not.

Input Format:

First line contains number of test cases T, followed by T lines, each containing a string "S".

Output Format:

For each test case print "Yes" if the String "S" is super ascii, else print "No"

Constraints:

1<=T<=100

1<=|S|<=400, S will contains only lower case alphabets ('a'-'z').

Sample Input and Output

SNo.	Input	Output
1	2 bba scca	Yes No

Explanation:

In case 1, viz. String "bba" -
The count of character 'b' is 2. Ascii value of 'b' is also 2.
The count of character 'a' is 1. Ascii value of 'a' is also 1.
Hence string "bba" is super ascii.

Program:

#include <stdio.h>
int main() {
    char s[30];
    int i,num[30]={0},isascii,n;
    scanf("%d",&n);
    while(n--)
    {
    scanf("%s",s);
    i=0;
    isascii=1;
    while(s[i]!='\0')
    {
        if((s[i]>='a')&&(s[i]<='z'))
        num[s[i]-'a']++;
        s[i]='\0';
        i++;
    }
    for(i=0;i<26;i++)
    {
        if((num[i]>0)&&(num[i]!=(i+1)))
        isascii=0;
        num[i]=0;
    }
    if(isascii)
    printf("yes\n");
    else
    printf("no");
    }
    return 0;
}

Output:

2

bba

scca

yes

no

You can also run it on the online IDE: https://ide.geeksforgeeks.org/q64aeIcLxM

Your feedback and comments are welcomed! If you have an doubt can contact me or comment below! Cheers!

Related Link: Date Time 2018

Codu And Sum Love 2018

Codu And Sum Love

Problem Description

```

Scanner sc = new Scanner(System.in);

long sum = 0;

int N = sc.nextInt();

for (int i = 0; i < N; i++) {

final long x = sc.nextLong(); // read input

String str = Long.toString((long) Math.pow(1 << 1, x));

str = str.length() > 2 ? str.substring(str.length() - 2) : str;

sum += Integer.parseInt(str);

}

System.out.println(sum%100);

```

Given N number of x’'s, perform logic equivalent of the above Java code and print the output

Constraints

 1<=N<=10^7 0<=x<=10^18

Input Format

 First line contains an integer N

Second line will contain N numbers delimited by space

Output

 Number that is the output of the given code by taking inputs as specified above

Explanation

Example 1

Input

 4 8 6 7 4

 Output

 64

Example 2

Input

3

1 2 3

Output

14

Program

#include <stdio.h>
long power(long k)
{
    long i,j=1;
    for(i=1;i<=k;i++)
        j=j*2;
        return j;
}
long length(long k)
{
    long ct=0,n1;
    while(k)
    {
        ct++;
        k=k/10;
    }
    return ct;
}
long reduce(long p1,long r1)
{
    long ct=0,k,cnt,rev=0,a[1000],h=0,i,sum1=0;
    while(p1)
    {
      k=p1%10;
      a[h]=k;
      h++;
      p1=p1/10;
     }
      for(i=1;i>=0;i--)
      sum1=sum1*10+a[i];
    return sum1;
}
int main() {
 long x,n,sum=0,r[70],p,c,f=0,i;
 scanf("%ld",&n);
 for(i=0;i<n;i++)
 {
    scanf("%ld",&x);
    p=power(x);
    c=length(p);
    if(c>2)
    {
    r[f++]=reduce(p,c);
    }
    else
    r[f++]=p;
 }
 for(i=0;i<n;i++)
 
 sum=sum+r[i];
   printf("%ld",(sum%100));
 
 
 return 0;
}

Output:

4 8 6 7 4

64

You can also try running it on online IDE: https://ide.geeksforgeeks.org/G3gXokhTbW

Your doubts and feedback are welcomed! you can comment it below Cheers!

Related Link: Super Ascii

Bride Hunting -Code Vita 2018 | round 1 | TCS Code Vita Solution 2018 | Bride Hunting TCS Code Vita solution

Bride Hunting

Problem Description

Sam is an eligible bachelor. He decides to settle down in life and start a family. He goes bride hunting.

He wants to marry a girl who has at least one of the 8 qualities mentioned below:-

1) The girl should be rich.

2) The girl should be an Engineer/Doctor.

3) The girl should be beautiful.

4) The girl should be of height 5.3".

5) The girl should be working in an MNC.

6) The girl should be an extrovert.

7) The girl should not have spectacles.

8) The girl should be kind and honest.

He is in search of a bride who has some or all of the 8 qualities mentioned above. On bride hunting, he may find more than one contenders to be his wife.

In that case, he wants to choose a girl whose house is closest to his house. Find a bride for Sam who has maximum qualities. If in case, there are more than one contenders who are at equal distance from Sam’'s house; then

print "“Polygamy not allowed”".

In case there is no suitable girl who fits the criteria then print “"No suitable girl found"”

Given a Matrix N*M, Sam's house is at (1, 1). It is denoted by 1. In the same matrix, the location of a marriageable Girl is also denoted by 1. Hence 1 at location (1, 1) should not be considered as the location of a marriageable Girl’s location.

The qualities of that girl, as per Sam’'s criteria, have to be decoded from the number of non-zero neighbors (max 8-way) she has. Similar to the condition above, 1 at location (1, 1) should not be considered as the quality of a Girl. See Example section to get a better understanding.

Find Sam, a suitable Bride and print the row and column of the bride, and find out the number of qualities that the Bride possesses.

NOTE: - Distance is calculated in number of hops in any direction i.e. (Left, Right, Up, Down and Diagonal)

Constraints

2 <= N,M <= 10^2

Input Format

First Line contains the row (N) and column (M) of the houses.

Next N lines contain the data about girls and their qualities.

Output

It will contain the row and column of the bride, and the number of qualities that Bride possess separated by a colon (i.e. :).

Explanation

Example 1

Input:

2 9

1 0 1 1 0 1 1 1 1

0 0 0 1 0 1 0 0 1

Output:

1:7:3

Explanation:

The girl and qualities are present at (1,3),(1,4),(1,6),(1,7),(1,8),(1,9),(2,4),(2,6),(2,9).

The girl present at (1,3) has 2 qualities (i.e. (1,4)and (2,4)).

The girl present at (1,4) has 2 qualities.

The Bride present at (1,6) has 2 qualities.

The Bride present at (1,7) has 3 qualities.

The Bride present at (1,8) has 3 qualities.

The Bride present at (1,9) has 2 qualities.

The Bride present at (2,4) has 2 qualities.

The Bride present at (2,6) has 2 qualities.

The Bride present at (2,9) has 2 qualities.

As we see, there are two contenders who have maximum qualities, one is at (1,7) and another at (1,8).

The girl who is closest to Sam's house is at (1,7). Hence, she is the bride.

Hence, the output will be 1:7:3.

Example 2

Input:

6 6

1 0 0 0 0 0

0 0 0 0 0 0

0 0 1 1 1 0

0 0 1 1 1 0

0 0 1 1 1 0

0 0 0 0 0 0

Output:

4:4:8

Explanation:

The bride and qualities are present at (3,3),(3,4),(3,5),(4,3),(4,4),(4,5),(5,3),(5,4),(5,5)

The Bride present at (3,3) has 3 qualities (i.e. (3,4),(4,3) and (4,4)).

The Bride present at (3,4) has 5 qualities.

The Bride present at (3,5) has 3 qualities.

The Bride present at (4,3) has 5 qualities.

The Bride present at (4,4) has 8 qualities.

The Bride present at (4,5) has 5 qualities.

The Bride present at (5,3) has 3 qualities.

The Bride present at (5,4) has 5 qualities.

The Bride present at (5,5) has 3 qualities.

As we see, the girl present in (4,4) has maximum number of Qualities. Hence, she is the bride.

Hence, the output will be 4:4:8.

Solution in C:

 
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
// Function to check if given indices are valid in the matrix
int isValid(int row, int col, int N, int M) {
return (row >= 0 && row < N && col >= 0 && col < M);
}
// Function to count the number of qualities a girl possesses
int countQualities(int matrix[][100], int row, int col, int N, int M) {
int count = 0;
for (int i = row - 1; i <= row + 1; i++) {
for (int j = col - 1; j <= col + 1; j++) {
if (isValid(i, j, N, M) && !(i == row && j == col) && matrix[i][j] == 1) {
count++;
}
}
}
return count;
}
// Function to find the suitable bride for Sam
void findBride(int matrix[][100], int N, int M) {
int minDistance = N + M; // Initialize to maximum possible distance
int maxQualities = 0;
int brideRow = -1, brideCol = -1;
// Loop through the matrix to find the bride with maximum qualities
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
if (matrix[i][j] == 1) {
int qualities = countQualities(matrix, i, j, N, M);
if (qualities > maxQualities) {
maxQualities = qualities;
brideRow = i;
brideCol = j;
} else if (qualities == maxQualities) {
// If multiple brides have the same number of qualities,
// find the one with minimum distance to Sam's house
int distance = abs(0 - i) + abs(0 - j);
if (distance < minDistance) {
minDistance = distance;
brideRow = i;
brideCol = j;
} else if (distance == minDistance) {
printf("Polygamy not allowed\n");
return;
}
}
}
}
}
// If no suitable bride found
if (brideRow == -1 || brideCol == -1) {
printf("No suitable girl found\n");
} else {
printf("%d:%d:%d\n", brideRow + 1, brideCol + 1, maxQualities);
}
}
int main() {
int N, M;
scanf("%d %d", &N, &M);
int matrix[100][100];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
scanf("%d", &matrix[i][j]);
}
}
findBride(matrix, N, M);
return 0;
}
 

Alternate Solution (Brute Force):

#include<stdio.h>
int main()
{
int n,m,i,g[50][50],j,p,q,max=0,cnt=0,k=1,c=0,u=1,x[30],y[30],t1,min=0, sc[50],e,f,ct=0,a[50],count=0,t2=0,t=0;
scanf("%d %d",&n,&m);
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
scanf("%d",&g[i][j]);
} }
g[1][1]=0;
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{ cnt=0;
if(g[i][j]==1)
{
t++;
for(p=i-1;p<=i+1;p++)
{
for(q=j-1;q<=j+1;q++)
{
if(g[p][q]==1)
{ cnt++;
} } }
cnt=cnt-1;
a[k]=cnt;
k++;
} } }
for(k=1;k<=t;k++)
{ if(a[k]>max) max=a[k];
}
if(max==0)
{ printf("No suitable girl found"); goto x; }
for(k=1;k<=t;k++)
{ if(a[k]==max)
c++; }
for(k=1;k<=t;k++)
{ t2=0;
if(a[k]==max)
{ for(i=1;i<=n;i++)
{ for(j=1;j<=m;j++)
{ if(g[i][j]==1) t2++;
if(t2==k)
{ x[u]=i; y[u]=j; u++;
} } } } }
t1=u-1;
if(c==1)
printf("%d:%d:%d",x[1],y[1],max);
else
{ for(u=1;u<=t1;u++)
{ e=x[u]-1;
f=y[u]-1;
if(e>=f)
{ sc[u]=e;
}
else sc[u]=f;
}
min=sc[1];
for(u=1;u<=t1;u++)
{ if(sc[u]<min) min=sc[u];
}
for(u=1;u<=t1;u++)
{ if(sc[u]==min) count++;
}
if(count>1)
printf("Polygamy not allowed");
if(count==1)
{ for(u=1;u<=t1;u++)
{ if(sc[u]==min)
printf("%d:%d:%d",x[u],y[u],max);
} } }
x: return 0;
}

OUTPUT:

6 6
1 0 0 0 0 0
0 0 0 0 0 0
0 0 1 1 1 0
0 0 1 1 1 0
0 0 1 1 1 0
0 0 0 0 0 0

4:4:8

Hey guys you can run this program in the online ide in this link: 

https://ide.geeksforgeeks.org/online-c-compiler/e9bd0a6a-12a8-459f-82dc-ebc592e1d16e

Your feedback are welcomed and can comment or post your doubts in the comment Cheers!

Related Links: 

https://codepiggy.blogspot.com/2024/03/super-market-problem-tcs-code-vita-2023.html

Zombie World

Zombie World

Zoya has developed a new game called Zombie World. The objective of the game is to kill all zombies in given amount of time. More formally,

-         N represents the total number of zombies in the current level

-         T represents the maximum time allowed for the current level

-         P represents the initial energy level a player starts with

-         Ei defines the energy of the i-th zombie

-         D defines the minimum energy the player needs, to advance to the next level

When a player energy is greater than or equal to the i-th zombie's energy, the player wins. Upon winning, the player will be awarded with an additional energy equal to the difference between current zombie energy and the player energy.

One unit of time will be taken to complete the fight with a single zombie.

Rules of the game:-

-         At any given time, a player can fight with only one zombie

-         Player is allowed to choose any one zombie to fight with.

Your task is to determine whether the player will advance to the next level or not, if he plays optimally.

Input Format:

The first line contains the number of test cases (K)

Each test case consists of three parts:

1. The total number of zombies (N) and the maximum time allowed (T)
2. Array of size N, which represents the energy of zombies (E)
3. The initial energy level a player (P) and the minimum energy required to advance (D)

Output Format:
Print "Yes" if a player can advance to the next level else print "No".

Constraints:

1<=K<=10

1<=N<=50

1<=Ei<=500

1<=T<=100

1<=D<=2000

1<=P<=500

Sample Input and Output

SNo.	Input	Output
1	1 2 3 4 5 5 7	Yes

PROGRAM:

#include<stdio.h>
int main()
{
 int n,t,e[20],i,pe,me,k;
 scanf("%d",&k);
 while(k)
 {
 scanf("%d",&n);
 scanf("%d",&t);
 for(i=0;i<n;i++)
 scanf("%d",&e[i]);
 scanf("%d",&pe);
 scanf("%d",&me);
 if(t<n)
 goto x;
 else
 {
  for(i=0;i<n;i++)
  {
   if(pe>=e[i])
   {
    pe=pe+(pe-e[i]);
   }
  }
 if(pe<=me)
 printf("yes\n");
 else
x: printf("no\n");
 }
 
    k--;
 }
 return 0;
}

OUTPUT: 

1
2 3
4 5
5 7
Yes

You can directly run it on a IDE: https://ide.geeksforgeeks.org/3un9lTbXuV

You can comment your feedback and doubts if any cheers!

Related Link:Bride Hunting

Minimum Product Array

         Minimum Product Array



TCS codevita 2016 round 1:
        The task is to find the minimum sum of Products of two arrays of the same size, given that k modifications are allowed on the first array. In each modification, one array element of the first array can either be increased or decreased by 2.Note- the product sum is Summation (A[i]*B[i]) for all i from 1 to n where n is the size of both arrays.

Input Format:
         First line of the input contains n and k delimited by white space Second line contains the Array A (modifiable array) with its values delimited by spaces Third line contains the Array B (non-modifiable array) with its values delimited by spaces.



Output Format:

Output the minimum sum of products of the two arrays.

Constraints:

1 ≤ N ≤ 10^50 ≤ |A[i]|, |B[i]| ≤ 10^50 ≤ K ≤ 10^9

Sample       Input                   Output


1.                3 5                            -31

                   1 2 -3

                  -2 3 -5



2.               5 3                              25

                  2 3 4 5 4

                  3 4 2 3 2




Explanation for sample 1:

     Here total numbers are 3 and total modifications allowed are 5. So we modified A[2], which is -3 and increased it by 10 (as 5 modifications are allowed). Now final sum will be
(1 * -2) + (2 * 3) + (7 * -5)
-2 + 6 - 35
-31

-31 is our final answer.




Explanation for sample 2:

     Here total numbers are 5 and total modifications allowed are 3. So we modified A[1], which is 3 and decreased it by 6 (as 3 modifications are allowed).
Now final sum will be
(2 * 3) + (-3 * 4) + (4 * 2) + (5 * 3) + (4 * 2)
6 - 12 + 8 + 15 + 8
25

25 is our final answer. 




PROGRAM:

#include<stdio.h>
int main()
{
 int m[10],i,min,max,k,n,p=1,sum=0,u[10],mins;
 scanf("%d",&n);
 scanf("%d",&k);
 for(i=0;i<n;i++)
 scanf("%d",&u[i]);
 for(i=0;i<n;i++)
 scanf("%d",&m[i]);
 min=m[0];
 max=m[0];
 for(i=0;i<n;i++)
 {
  if(m[i]<min)
  min=m[i];
  if(m[i]>max)
  max=m[i];
 }
 if (min<0&&max<0)
 mins=min<max?min:max;
 else if(min>0&&max>0)
 mins=max>min?max:min;
 else
 {
 if((min-max)<-(2*min))
  mins=min;
  else 
  mins=max;
 }
 for(i=0;i<n;i++)
 {
  if(mins==m[i]&&mins>0)
   {

u[i]=u[i]-(2*k);
   goto x;
   }
   if(mins==m[i]&&mins<0)
    { 
    u[i]=u[i]+(2*k);
    goto x;
    }
  }
x:for(i=0;i<n;i++)
 {
 p=u[i]*m[i];
 sum=sum+p;
 }
 printf("%d",sum);
 return 0;
 
}

OUTPUT:

5 3                              

2 3 4 5 4

3 4 2 3 2

25

You can directly run it on a IDE: https://ide.geeksforgeeks.org/lP9ISaZ9yx
You can comment your feedback and doubts if any cheers!

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