TCS CodeVita previous years questions Solved – Matrix Rotation

TCS CodeVita previous years questions – 2014



Problem : Matrix Rotations

You are given a square matrix of dimension N. Let this matrix be called A. Your task is to rotate A in clockwise direction byS degrees, where S is angle of rotation. On the matrix, there will be 3 types of operations viz.

1. Rotation
   

   Rotate the matrix A by angle S, presented as input in form of A S

2. Querying
   

   Query the element at row K and column L, presented as input in form of Q K L

3. Updation
   

   Update the element at row X and column Y with value Z,  presented as input in form of U X Y Z

Print the output of individual operations as depicted in Output Specification

Input Format:
Input will consist of three parts, viz.

1. Size of the matrix (N)

2. The matrix itself (A = N * N)

3. Various operations on the matrix, one operation on each line. (Beginning either with A, Q or U)
-1 will represent end of input.

Note:

• Angle of rotation will always be multiples of 90 degrees only.
• All Update operations happen only on the initial matrix. After update all the previous rotations have to be applied on the updated matrix

Output Format:
For each Query operation print the element present at K-L location of the matrix in its current state.

Constraints:

1<=N<=1000

1<=Aij<=1000

0<=S<=160000

1<=K, L<=N

1<=Q<=100000

Sample Input and Output

SNo.	Input	Output
1	2 1 2 3 4 A 90 Q 1 1 Q 1 2 A 90 Q 1 1 U 1 1 6 Q 2 2 -1	3 1 4 6

Explanation:

Initial Matrix

1 2

3 4

After 90 degree rotation, the matrix will become

3 1

4 2

Now the element at A₁₁ is 3 and A₁₂ is 1.

Again the angle of rotation is 90 degree, now after the rotation the matrix will become

4 3

2 1

Now the element at A₁₁ is 4.

As the next operation is Update, update initial matrix i.e.

6 2

3 4

After updating, apply all the previous rotations (i.e. 180 = two 90 degree rotations).

The matrix will now become

4 3

2 6

Now A₂₂ is 6.

TCS CodeVita previous years questions – Matrix Rotation solution
Sample Solution C proram code :

#include<stdio.h>
int i,j,n,arr1[1000][1000],arr2[1000][1000];
void swaparray()
{
 for(i=0;i<n;i++)
 for(j=0;j<n;j++)
 arr1[i][j]=arr2[i][j];
}
int main()
{
 char opt[4];
 scanf("%d",&n);
 for(i=0;i<n;i++)
 for(j=0;j<n;j++)
 scanf("%d",&arr1[i][j]);
 scanf("%s",opt);
 while(strcmp(opt,"-1"))
 {
 if(opt[0]=='A')
 {
 if(!strcmp(opt,"A90"))
 {
 for(i=0;i<n;i++)
 for(j=0;j<n;j++)
 arr2[j][n-i-1]=arr1[i][j];
 swaparray();
 }
 else if(!strcmp(opt,"A180"))
 {
 for(i=0;i<n;i++)
 for(j=0;j<n;j++)
 arr2[n-i-1][n-j-1]=arr1[i][j];
 swaparray();
 }
 else if(!strcmp(opt,"A270"))
 {
 for(i=0;i<n;i++)
 for(j=0;j<n;j++)
 arr2[n-j-1][i]=arr1[i][j];
 swaparray();
 }
 else if(!strcmp(opt,"A0") || !strcmp(opt,"A360")); //not required
 }
 else if(opt[0]=='Q')
 {
 if(opt[1]-48<n && opt[2]-48<n)
 printf("%d\n",arr1[opt[1]-48][opt[2]-48]);
 }
 else if(opt[0]='U')
 {
 if(opt[1]-48<n && opt[2]-48<n)
 arr1[opt[1]-48][opt[2]-48]=opt[3]-48;
 }
 scanf("%s",opt);
 }
 return 0;
}

//validate the constraints too

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