NCERT Solutions for Class 12 Computer Science (C++) – Arrays

NCERT Solutions for Class 12 Computer Science (C++) – Arrays

Computer ncert solution of class 12 Computer Science (C++) – Arrays, we have covered all the chapter questions. This solution will prove to be beneficial for the board examination as well as for students who are preparing for graduate(B.Tech) in information technology and preparing for Post Graduate(M.C.A).
All the concepts of programming in this whole solution have been made quite accurate and authentic which will help you to make notes and also increase your interest in computer programming. This solution is based on the syllabus of  NCERT Solutions for Class 12 Computer Science (C++) – Arrays, which will prove useful in the board exams and competitive exams as well.

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NCERT Solutions for Class 12 Computer Science (C++) having 14th Chapter whose Chapter wise Solution given below.


TOPIC-1

Data Structures and One-dimensional Arrays

Short Answer Type Questions-I [2 Marks Each]

Question 1.
Observe the following C++ code and find out, which out of the given options (i) to (iv) are the expected correct output. Also assign the maximum and minimum value that can be assigned to the variable ‘Go’.

void main()
 { int X [4] = {100, 75, 10, 125};
 int Go = random(2) + 2;
 cout<<X [i]<<"$$";
 for (int i = Go; i<4; i ++)
 }

(i) 100$$75
(ii) 75$$10$$125$$
(iii) 75$$10$$
(iv) 10$$125$

Answer:
(iv) is correct option
Max. Value = 3
Min. Value = 2

Short Answer Type Questions-II[3 marks each]

Question 1.
Write the difintion of a function grace_score (int score [], int size) in C ++,which should check all the elements of the array and give an increase of 5 to those scores which are less than 40.
Example: if an array of seven integers is as follows:
45,35,85,80,33,27,90
After executing the function, the array content should be changed as follows :
45,40,85,80,38,32,90

Answer:

void grace_score(int score[].int size)
 {
 for(int i=0;i<size;i++)
 {

 score [i]=score[i]+5;
 if (score [i]<40)
 cout< }

 }

Question 2.

Find the output of the following C+ + program:
#include<iostream.h> void repch(char s[]) { for(int i=0;s[i]!=’′;i++) { if(((i%2)!=0) &&(s[i]!=s[i + 1])) { S[i]=’@’; cout<<“Hello”; } else if (s[i]==s [i + 1]) { s[i+1]=” ! ” ; i++ ; } } } void main() { char str[]=”SUCCESS”; cout<<“Original String”<<str; repch(str); cout<<“Changed String”<<str; }

Answer:
Original String SUCCESS
Changed String S@C!ES!



Question 3.
Find the output of the following:

 #include<iostream.h>
 void switchover (int A[],int N, int split)
 {
 for(int K=0; K<N; K++)
 if(K<split)

 void display(int A [],int N)
 A [K] + = K;
 else

 {
 A[K] *=K; }

 cout<<A[K]<<endl;
 for (in K=0; K<N; K++)
 (K%2==0)? Cout<<A[K]<<"%" :
 }

 switchover(H, 6, 3);
 void main()
 { int H [ ] = {30, 40, 50, 20, 10, 5};
 display(H, 6);

 }

Answer:
30%41
52% 60
40% 25

Question 4.
An integer array A [30] [40] is stored along the column in the memory. If the element A[20][25] is stored at 50000, find out the location of A[25] [30].

Answer:

A[i] [j] = B+W x [No. of rows x (I-Lr) + (J-Lc)]
 A[20] [25] = B + 2x [30x (20-0) + (25-0)]
 A[7] [10] = 48750+ 2x[30x(7-0)+(10-0)]
 50000 = B+2x[30x(20-0)+(25-0)]
 B = 48750

 = 49190

Question 5.
An array P[30][20] is stored along the column in the memory with each element requiring 2 bytes of storage. If the base address of the array P is 26500, find out the location of P[20][10].

Answer:
Total number of rows = 30
Total size = 2 bytes
Base Address = 26500
LOC(P[I][J]) = Base Address+((I-LBR)+(J- LBC)*R)W
Assuming Lower Bound of Row(LBR)=0
Lower Bound of Column(LBC)=0
Total number of Rows(R)=30
Size of each element(W)=2
LOC(P[20][10]) = 26500 + ((20-0)+(10-0)*30)*2
LOC(P[20][10]) = 26500 + 640
LOC(P[20][10]) = 27140

Question 6.
Write a function to sort any array of n elements using insertion sort. Array should be passed as argument to the function.

Answer:

void insertsort (int a[],int n)
 {
 int p,ptr;
 //Assuming a[0]=int_min i.e. smallest integer
 for(p = 1; p<=n;p++)
 {
 temp=a[p];

 Forward
 ptr=p-1;
 while (temp<a[ptr] )

 a [ptr+1]=a[ptr] ; //Move Element
 {
 ptr--;

 }
 }
 a[ptr+1]=temp; //Insert Element in Proper Place

Question 7.
Write a function NewMAT(int A[][],int r, int c) in C++, which accepts a 2d array of integer and its size as parameters divide all those array elements by 6 which are not in the range 60 to 600(both values inclusive) in the 2d Array.

Answer:

void NewMAT(int A[] [] , int r,int c)
 {
 for (int i = 0; i<r; i+ + )
 A[i][j]/=6
 for (j=0; j=60)&&(A[1] [j)<=600))

OR

A[i] [j] =A[i] [j] /6;
 }



Question 8.
Write the definition of a function Alter (intA[], int N) in C++, which should change all the multiples of 5 in the array to 5 and rest of the elements as 0. For example, if an array of 10 integers is as follows :

A[0] A[l] A[2] A[3] A[4] A[5] A[6] A[7] A[8] A[9]
55 43 20 16 39 90 83 40 48 25

After executing the function, the array content should be changed as follows:

A[0] A[l] A[2] A[3] A[4] A[5] A[6] A[7] A[8] A[9]
5 0 5 0 0 5 0 5 0 5

Answer:

void Alter (int A [] , int N)
 {
 for (int i=0; i<N; i++)
 else
 if (A[i] %5==0)
 A [i] = 5;

 }
 A [i] = 0;

OR
Any other correct equivalent function definition

Question 9.
A two dimensional array P[20] [50] is stored in the memory along the row with each of its element occupying 4 bytes, find the address of the element P[10] [30], if the element P[5] [5] is stored at the memory location 15000.

Answer:
Loc (P [I] [J] along the row =BaseAddress+W [(I-LBR)*C+ (J-LBC)]
(where C is the number of columns, LBR = LBC =0)
LOC (P [5] [5]) = BaseAddress + W* [I*C + J]
15000 = BaseAddress + 4 * [5 * 50 + 5]
= BaseAddress + 4 * [250 + 5]
= BaseAddress + 4*255 = BaseAddress + 1020
BaseAddress = 15000 -1020 = 13980
LOC (P [10] [30]) = 13980 + 4* [10 * 50 + 30]
= 13980 + 4 * 530
= 13980 + 2120
= 16100
OR LOC (P [10] [30])
= Loc (P[5] [5]) + W[ (I-LBR) * C+ (J-LBC)]
= 15000 + 4 [(10 – 5) * 50 + (30 – 5)]
= 15000 + 4 [5 * 50 + 25]
= 15000 + 4 *275
= 15000 + 1100
= 16100
OR
(where C is the number of columns and LBR = LBC = 1)
LOC (P [5] [5])
15000 = BaseAddress + W [(I-1) *C + (J-1)]
= BaseAddress + 4 [4*50 + 4]
= BaseAddress + 4 [200 + 4]
= BaseAddress + 4 * 204
= BaseAddress + 816
BaseAddress = 15000 – 816 = 14184
LOC (P [10] [30])
= 14184 + 4 [(10 -1) * 50 + (30 -1)]
= 14184 + 4 [9*50 + 29]
= 14184 + 4*479
= 14184 + 1916
= 16100

Question 10.

Write the definition of function Change (int P[], int N) in C+ +, which should
change all the multiples of 10 in the array to 10 and rest of the element as 1.
For example, if an arrary of 10 integers is as follows :

P[0]

P[l]

P[2]

P[3]

P[4]

P[5]

P[6]

P[7]

P[8]

P[9]

100

43

20

56

32

91

80

40

45

21

After executing the
function, the array content should be changed as follows :

P[0]

P[l]

P[2]

P[3]

P[4]

P[5]

P[6]

P[7]

P[8]

P[9]

10

1

10

1

1

1

10

10

1

1

Answer:

void Change (int P[ ]
, int N)

 {

 for (int i = 0; i<N; i++)

 if (P [i] % 10 == 0)

 P [i] = 10;

 else.

 P [i] = 1;

 }

OR

Any other correct equivalent function definition

Question 11.

Write code for a function oddEven (int s[], int N) in C+ + , to add 5 in all
the odd values and 10 in all the even values of the array S.
Example : If the original content of the array S is

50

11

19

24

28

The modified content
will be :

60

16

24

34

38

Answer:

void oddEven (int s []
, int N)

 {

 for (int i=0; i<N; i++)

 {

 if(s[i]%2!=0)

 s[i]=s[i]+5;

 else

 s[i]=s [i]+10;

 }

 }

Question 12.

Write the definition for a function void Transfer (int A[6], int B[6]) in C++,
which takes integer arrays, each containing 6 elements as parameters. The
function should exchange all odd places (1st, 3rd and 5th) of the two arrays,
for example
If the array A contains

15

10

12

21

52

76

And if the array B
contains

23

41

67

83

13

53

Then the function
should make the contents of the array A as

15

41

12

83

52

53

And the contents of
array B as

23

10

67

21

13

76

Answer:

void Transfer (int A
[6] , int B[6] )

 {

 int temp;

 for (int i=;i<6; i+ = 2)

 {

 temp=A [i]

 A [i] =B [i]

 B [i] +temp;

 }

 }

Question 13.

Write a user-defined function DisTen (int A[] [4], int N, int M) in C++ to find
and display all the numbers, which are divisible by 10. For example if the
content of array is :

12

20

13

2

10

30

The output should be

20 10 30

Answer:

void DisTen (int A[]
[4] , int N, int M)

 {

 int i, j;

 for (i=0; i<M; i++)

 {

 for (j = 0 ; j <N; j++)

 {

 if (A[i][j]%10==0)

 cout << A [i] [ j] ;

 }

 }

 }

Question 14.

Write code for a function void ChangeOver (int P[], int N) in C++, which
re-positions all the elements of the array by shifting each of them to the next
position and by shifting the last element to the first position.
For example: If the content of the array is :

0

1

2

3

4

12

15

17

13

21

The Chenged Content
will be:

0

1

2

3

4

21

12

15

17

13

Answer:

void ChangeOver (int
P[] , int N)

 {

 int i, j, temp;

 for (i=0; i<N; i = i+1)

 {

 j=j+1;

 temp = P[i];

 P[i] = P [j ] ;

 P [j] = temp;

 }

 }

Question 15.

Write code for a function void Convert (int T[], int Num) in C++, which
re-positions all the elements of the array by shifting each of them one to one
position before and by shifting the first element to the last position.
For example : If the content of the array is :

0

1

2

3

4

22

25

70

32

12

The Chenged Content
will be:

0

1

2

3

4

25

70

32

12

22

Answer:

void Convert (int T[],
int Num)

 {

 int i, j, temp;

 for (i=Num; i > 10; i-=l)

 {

 j = i – 1;

 temp = P [ i ] ;

 P [i] = P[jl ;

 P[j] = temp;

 }

 }

Question 16.

Write a user-defined function DispNTen (int L[] [4], int R, int C) in C++ to
find the display all the numbers, which are not divisible by 10. For example if
the content of array is :

20

17

30

12

19

10

The output should be

17 12 19

Answer:

void DispNTen (int L[]
[4], int R,

 int C)

 {

 int i, j ;

 for (i = 0 ; i<C; i + +)

 {for (j = 0; j<R; j+ + )

 {

 if (A[i] [j]%10= =0)

 cout <<A[i] [j]<<” “;

 }

 }

 }

Question 17.

Write a function Transfer (int A[], int B[], int Size) in C++ to copy the
elements of array A into array B in such a way that all the negative elements
of A appear in the beginning of B, followed by all the positive elements,
followed by all the zeroes maintaining their respective orders in array A. For
example :
If the contents of array A are :
7, -23,3,0, -8, -3,4,0
The contents of array B should be
-23, -8, -3,7,3,4,0

Answer:

void Transfer (int
A[], int B[], int size)

 {

 for (int i = 0; i<size; i++)

 {

 if(A[i]<0) B [i] =A [i] ; if (A [i] > 0)

 B[i+1]=A[i];

 else

 B[i/2]=A[i];

 }

 }

Question 18.

A two dimensional array ARR[50] [20] is stored in the memory along the row with
each of its elements occupying 4 bytes. Find the address of the element
ARR[30][10], if the element ARR[10] [5] is stored at the memory location 15000.

Answer:

Loc (ARR[I] [J] along the row
= BaseAddress + W [(I – LBR) *C + (J – LBC)]
(where C is the number of columns, LBR = LBC = 0 LOC (ARR [10] [5])
= BaseAddress + W [I *C + J]
15000 = BaseAddress + 4 [10 * 20 + 5]
= BaseAddress + 4 [200 + 5]
= BaseAddress + 4 x 205
= BaseAddress + 820
BaseAddress = 15000 – 820
= 14180
LOC (ARR [30] [10]) = 14180 + 4 [30 * 20 + 10]
= 14180 + 4 * 610
= 14180 + 2440
= 16620
OR
LOC (ARR [30] [10]) = LOC (ARR [10] [5] + W [(I – LBR) * C + (J – LBC)]
= 15000 + 4 [(30-10) *20 + (10 – 5)]
= 15000 + 4 [20*20 + 5]
= 15000 + 4 *405
= 15000 + 1620
= 16620
OR
Where C is the number of columns and LBR=LBC=1
LOC (ARR[10] [5])
15000 = BaseAddress + W [(I -1) * C + (J-l)]
= BaseAddress + 4 [9*20 + 4]
= BaseAddress + 4[180 + 4]
= BaseAddress + 4 * 184
= BaseAddress + 736
BaseAddress = 15000 – 736
= 14264
LOC (ARR [30] [10])
= 14264 + 4 [(30 -1) * 20 + (10 -1)]
= 14264 + 4[29 *20 + 9]
= 14264 + 4 [580 + 9]
= 14264 + 4*589
= 14264 + 2356
= 16620

Question
19.

Write a function SORTSCORE() in C++ to sort an array of structure IPL in
descending order of score using selection sort.

Note : Assume the following definition of structure IPL:

struct IPL:

 {

 int Score;

 char Teamname[20];

 } ;

Answer:

void
SORTSCORE (IPL a[] , int n)

{ int largest;

IPL temp;

for ( int K = 0; K <n-1; K++)
{ largest = K;
for ( int j = K+1; j< n; j++) { if ( a[j].score > a[largest]. score)
{ largest = j;
}
}
temp = a[K];
a[K] = a[largest];
a [largest] = temp;
}
}

Question
20.

Write a function in C++ TWOTOONE() which accepts two array X[ ], Y[ ] and their
size n as argument. Both the arrays X[ ] and Y[ ] have the same number of
elements. Transfer the content from two arrays X[ ], Y[ ] to array Z[ ]. The even
places (0,2,4…) of array Z[ ] should get the contents from the array X[ ] and
odd places (1,3,5…) of array Z[ ] should get the contents from the array Y[ ].

Example : If the X[ ] array contains 30,60,90 and the Y[ ] array contains
10,20,50. Then Z[ ] should contain 30,10,60,20,90,50.

Answer:

 void TWOTOONE (int x[] , int n)

 {

 int z[n] ;

 for (int I =0;i<n;i++)

 {

 if(i%2 == 0)

 z [i] = x [i] ;

 else

 z [i] = y [i] ;

 }

 }

Question
21.

Write code for a function void EvenOdd (int T[], int C) in C+ + , to add 1 in
all the odd values and 2 in all the even values of the array T.

Example : If the original content of the array T is

T[0]

T[l]

T[2]

T[3]

T[4]

35

12

16

69

26

The
modified content will be:

T[0]

T[l]

T[2]

T[3]

T[4]

36

14

18

70

28

Answer:

 void EvenOdd (intT [] , int(C)

 {

 for (int i=0; i<C; i++)

 {

 if (T[i] %2==0)

 T [i] + = 2;

 else

 T[i] + = 1;

 }

 }

Question
22.

Write a function SWAP2 CHANGE (int P[], int N) in C++ to modify the content of
the array in such a way that the elements, which are multiples of 10 swap with
the value present in the very next position in the array.

For example:

If the content of array P is
91,50,54,22,30,54
The content of array P should become
91,54,50,22,54,30

Answer:

 void SWAP 2CHANGE (int P[], int N)

 {

 for (int i=0; i<N; i++)

 {

 if (P [i]% 10==0)

 {

 int temp=0;

 temp=P[i];

 P[i]=P [i+1] ;

 P[i+1]=temp;

 i++; //Go to next element

 }

 }

 }

Question
23.

Write a function SWAP2BEST (int ARR[], int Size) in C+ + to modify the content
of the array in such a way that the elements, which are multiples of 10 swap
with the value present in the very next position in the array.

For example:

If the content of array ARR is
90,56,45,20,34,54
The content of array ARR should become
56,90,45,34,20,54

Answer:

 void SWAP2BEST (int ARR [] , int Size)

 {

 for (int n=0; n<size; n++)

 {

 if(ARR[n]%10 = =0)

 {

 int temp = 0;

 temp = ARR [n];

 ARR[n] = ARR[n+1];

 ARR [n+1] = temp;

 n++;

 }

 }

 }

Question
24.

Write the definition of a function AddTax(float Amt[], int N) in C + + , which
should modify each element of the array Amt having N elements, as per the
following rules :

Existing
Value of Amt

Amt
to be changed to

If
less than 50,000

Add
25% in the existing value

If
> =50,000 and <1,00,000

Add
20% in the existing value

If>
=1,00,000

Add
15% in the existing value

Answer:

 void AddTax (Float Amt[] , int N)

 {

 for (int i=0; i<N; i++)

 if(Amt[i]<50,000) Amt[i]+=0-25*Amt[i]; else
if (Amt[i]>=50000&&Amt[i] <100000)

    Amt[i]+=0.2*Amt[i]; else
if(Amt[i]>=1,00000)

 Amt[i]+=0.15*Amt[i];

 }

 

TOPIC-2

Two-dimensional Arrays

Short Answer Type
Questions-I[2 marks each]

Question
1.

Write
a function Alternate (int A[] [3], int N, int M) in C + + to display all
alternate elements from two-dimensional array A(starting from A[0][0]).

For example:

If the array is containing :

23 54 76
37 19 28
62 13 19
The output will be
23 76 19 62 19

Answer:

void
Alternative (int A[] [3], intN, int M)

 {

 int T=0 ; I<N; I++)

 for (int J=0 ; J {

 if (T%2 = 0)

 COUt << A[I] [J]<<” ” ;

 T++ ;

 }

 }

 OR

 void Alternative (int A[] [3] , intN, int M)

 {

 int *P=&A [0] [0] ;

 for (int I = 0; I<N*M ; I+=2)

 {

 Cout<<*p<<” ” ;

 { I+=2 ;

 }

 }

OR

Any other equivalent correct answer acceptable

Question
2.

Write a function in C++ to print the sum of all the non-negative elements
present on both the diagonal of a two dimensional array passed as the argument
to the function.

Answer:

void
Diagonal (int A[] [20] , int N)

 {

 int K, Sdiag=0;

 for(int K=0;K<N;K++) if ( A[K] [K] > 0)

 Sdiag+=A [K] [K] ;

 for(int K= 0;K<N;K++) if ( A[N-K-1][K] >
0 )

 Sdiag+=A[N-K-1] [K] ;

 cout<<“Sum of all the non-negative
elements on both the diagonals= “<<Sdiag;

 }

Question
3.

Write a user-defined function swap_row(int ARR[][3],int R,int C) in C++ to swap
the first row values with the last row values : For example:

10

20

30

40

50

60

70

80

90

Then
after function call,the content of the array should be:

70

80

90

40

50

60

10

20

30

Answer:

voidswap_row
(intARR [] [3] , int R,int() )

 {

 for(int i=0 J=0: J<C;J++)

 {

 int temp=ARR[i][j];

 ARR[i][j]=ARR[R-1][j];

 ARR[R-1][j]=temp;

 }

 }

Question
4.

Write a user-defined function SumLast 3 (int A [] [4], int N, int M) in C + +
to find and display the sum of all the values, which are ending with 3 (i.e.,
units place is 3). Fdr example if the content of array is :

33

13

92

99

3

12

The
output should be 49

Answer:

void
SumLast 3 (int A[] [4], int N, int M)

 {

 int Sum=0;

 for(int i=0; i<N; i++)

 for(int j = 0; j <M; j ++)

 {

 If (((A[i] [j]-3)% 10==0))

 Sum=Sum+A[i][j];

 }

 cout<<sum;

 }

Question
5.

Write a user-defined function AddEnd2(int A[] [4], int N, int M) in C+ + to
find and display the

sum of all the values, which are ending with 2 (i.e., units place is 2).

For example, if the content of arrray is :

22

16

12

19

5

2

The
output should be 36

Answer:

void
AddEnd2(int A[] [4], int N, int M)

 {

 int Sum=0;

 for(int i=0; i<N; i++)

 for(int j = 0; j<m;j++)

 if(((A[i] [j]-2)%10 = = 0))

 Sum=Sum + A[i] [j] ;

 }

 cout<<sum;

 }

Question
6.

Write a function in C++ which accepts a 2D array of integers and its size
arguments and displays the elements which lie on minor diagonal. [Top right to
bottom left diagonal] [Assuming the 2D array to be square matrix with odd
dimension

i.e. 3.3 , 5.5 , 7.7, etc …]

For example:
If the 2D array is
6 7 8
1 3 6
7 9 3
The following should be displayed :
8
3
7

Answer:

void
diag (int x[] [] , int m, int n)

 {

 for (int i=m; i>0 ; i— —)

 {

 for(int j =0; j<n ; j++ 

{

 if(i==j)

 cout<<” “<<n[i] [j] ;

 }

 }

 }

Question
7.

Write a user-defined function int Sum Single(int A[4][4])in C++, which finds
and returns the sum of all numbers present in the first row of the array, for
example, if the array contains

1

13

10

9

29

17

2

21

14

3

12

31

15

16

25

52

Then
the function should return 33.

Answer:

int
SumSingle (int A[4] [4])

 {

 int Sum=0;

 for(int i=0 ; i for (int j=0 ; j  Sum=Sum+ A[i] [j]

 return Sum;

 }

Question
8.

Write a function SKIPEACH(int H[] [3], int C, int R) in C+ + to display all
alternate elements from two-dimensional array H (starting from H [0] [0]).

For example:

If the array is containing:
12 45 67
33 90 76
21 43 59
The output will be
12 67 90 21 59

Answer:

void
SKIPEACH (int H[ ] [3], int C, int R)

 {

 for (int i = 0; i C; i + + )

 { for (int j = 0, j < R; j++)

 {

 if ( (i + j)%2 = =0)

 cout<<H[i] [j]<< ” ” ;

 }

 }

Short Answer Type
Questions-II[3 marks each]

Question
1.

An array P[15][10] is stored along the column in the memory with each element
requiring 4 bytes of storage. If the base address of array P is 1400, find out
the location of P[8][5].

Answer:

Loc (P[8][5]) = B+W((I-Lr)+(J-Lc)M)

= 1400+4((8-0)+(5-0)15)

= 1732.

Question
2.

Given an array A[10][12] whose base address is 10000. Calculate the memory
location of A[2][5] if each element occupies 4 bytes and array is stored
columnwise.

Answer:

w=4

b=10000

m=10
n[i][i] = [b+ w(i-l1) + m(j-l2)
n[2][5] = [10000+ 4(2-0) + 10(5-0)]
n[2][5] = [10000 +8 +50 ]
n[2][5] = 10058

Question
3.

An array T[-1..35][-2..15] is stored in the memory along the row with each
element occupying 4 bytes. Find out the base address and address of element
T[20][5], if an element T[2][2] is stored at the memory location 3000. Find the
total number of elements stored in T and number of bytes allocated to T.

Answer:

Take T[2][2] as the base address [BA],

sr=starting row, sc=starting column;

Along the row
A[i][j] = BA + size[nc*(i-sr) + (j-sc)]
T[20][5] = 3000 + 4 [ 18 * (20-2) + (5-2)]
= 3000 + 4[(18 * 18) + (3)]
= 3000 + 4[324 + 3]
= 3000 + 4(327)
= 3000 + 1308
= 4308
nc = no. of columns sr=2 sc=2
Total number of elements stored in T=37*18=666
Total number of bytes allocated to T=666*4=2664

Question
4.

An array A[20] [30] is stored along the row in the memory with each element
requiring 4 bytes of storage. If the base address of array A is 32000, find out
the location of A[15][10]. Also, find the total number of elements present in
this array.

Answer:

Array A[20] [30]

Base Address = 32000

Size of element(W)=4 bytes
A[I][J]=A[15][10] hence I=15 and J=10
Total Row (R)=20
Total Column (C)=30
A [I] IJ]=B + W[C(I-0)+(J-0)]
A[15][10]=32000 +4[30(15-0)+(10-0)]
= 32000+4[30(15)+10]
= 32000+4[450+10]
= 32000+4[460]
= 32000+1840
= 33840

Question
5.

An array T[25][20] is stored along the row in the memory with each element
requiring 2 bytes of storage. If the base address of array T is 42000, find out
the location of T[10][15]. Also, find the total number of elements present in
this array.

Answer:

Array A[25] [20]

Base Address = 42000

Size of element(W)= 2 bytes
A[I][J]=A[10][15] hence I=10 and J=15
Total Row (R) = 25
Total Column (C)= 20
A [I] [J] =B+ W[C(I-0)+(J-0)]
A[10] [15] =42000 +2[20(10-0)+(15-0)]
= 42000+2(20(10+15)]
= 42000+2[200+15]
= 42000+2(215]
= 42000+430]
= 42430

Question
6.

An array S[10][15] is stored in the memory with each elements requiring 2 bytes
of storage. If the base address of array S is 25000, determine the location of
S[5] [10] if the array is S stored along the column.

Answer:

S[10][15]          //Base address; 25000

W=2 bytes

S[5][10] =?
m(row)=0 i=5
n(column)=15 j=10
Array is stored in the memory along the column:
P[i][j] = Base address + u[i+mxj]
s[5][10] = 25000+2[5+10×10]
= 25000+2(5+100] [1]
= 25000+2[105]
= 25000+210
= 25210

Question
7.

An array T[15][10] is stored along the row in the memory with each element
requiring 8 bytes of storage. If the base address of array T is 14000, find out
the location of T[10][7].

Answer:

Base Address = 14000 = B

T [I][J] = B + W [C[I-Ir] + (J-Jc)]

= 14000 + 8 [10(10-0)+(7-0)]
= 14000 + 8 x 107
= 14000 + 856
= 14856

Question
8.

Each element of the array A[8][6] is stored using 4 bytes of memory. If the
element A[2][4] is stored at location 936, find the address of A[5][1]. Assume
that the array is stored in Row Major.

Answer:

nr=8 nc=6 size = 4 bytes

A[2][4] = 936 A[5][1] = ?

Formula for row-major
A[i][j] = BA + size * ((i – sr) * nc + (j – sc))
A[5][1] = A[2][4] + 4 * ((5 – 2) * 6 + (1 – 4))
= 936 + 4* ((3) *6-3)
= 936 + 4 * 15
= 936 + 60
A[5][l] = 996

Question
9.

Write a function REVCOL (in P[] [5], int N, int int M) in C++ to display the
content of a two dimensional array, with each column content in reverse order.

Note : Array may contain any number of rows. For example, if the content of
array is as follows :

15

12

56

45

51

13

91

92

87

63

11

23

61

46

81

The
function should display output as:

11 23 61 46 81

13 91 92 87 63

15 12 56 45 51

Answer:

void
REVCOL (int P[ ] [5], int N, int M)

 {

 for (int I = N – 1; I>=0; I- -)

 {

 for (int J = 0; J<M; J++)

 Cout<<P[l] [J];

 cout<<end1;

 }

 }

 OR

 void REVCOL (int P [] [5], int N, int M)

 {

 for (int I = 0; I<N/2; I++)

 {

 for (int J = 0; J<M; J++)

 {

 int T = P [I] [J] ;

 P [1] [J] = P [N – I – 1] [J] ;

 P[N-I-1] [J] = T;

 }

 }

 for (I = 0; I<N; I++)

 {

 for (int J=0; J<M; J++)

 cout<<P [I] [J] ;

 

cout<<end1;

}

 }

Question
10.

Write a function REVROW (int N, int M) in C + + to display the content of a two
dimensional array, with each row content in reverse order.

For example, if the content of array is as follows:

15

12

56

45

51

13

91

92

87

63

11

23

61

46

81

The
function should display output as follows:

51 45 56 12 15

63 87 92 91 13

81 46 61 23 81

Answer:

void
REVROW (int P[] [5] , int N, int M)

 {

 for (int I =0;I<N;I++) {for (int J=M-1;
J>=0; J- -)

 cout<<P[I] [J];

 cout<<end1;

 }

 }

 OR

 void REVROW (int P [] [5] , int N, int M)

 {

 for (int I=0; I<N; I++)

 for (int J=0; J<M/2; J++)

 {

 int T = P [I] [J] ;

 P [I] [j] = P[I] [M – J – 1] ;

 P [I] [M – J – 1] = T;

 }

 }

 for (I = 0; I<N; I++)

 {

 for (int J = 0; J<M; J++)

 Cout<<P[I] [J] ;

 cout<<end1;

 }

 }

Question
11.

S[50][20] is a two dimensional array, which is stored in the memory along the
row with each of its element occupying 4 bytes, find the address of the element
S[5][15], if the element S[8][10] is stored at the memory location 62,000.

Answer:

LOC
(S [I] [J] )

= Reference Address+W [ (I-Lr) *C+ (J-Lc) ]

LOC (S [5] [15] )

= LOC(5 [8] [10]+4[(15-10)*50+(5-8)]
= 62000 + 4 [5*50-3]
= 62000 + 988
= 62988

Question
12.

Write definition for a function SHOWMID (int P[][5], int R, int C) in C++ to
display the elements of first, third and fifth column from a two dimenstional
array P having R number of rows and C number of columns.

For example, if the content of array is as follows :

515

102

113

801

145

303

141

129

202

121

285

189

100

760

112

The
function should display the following as output:

515 303 285

113 129 100

145 121 112

Answer:

void
SHOWMID (intP[] [5],intR,int C)

 {

 for(int J=0; J<C; J++)

 Cout<<P[R/2][J]<<” “;

 cout<<end1;

 for (int I=0;I<R;I++)

 cout<<P[I] [C/2]<<” ” ;

 }

Long Answer Type
Questions[4 marks each]

Question
1.

T[20][50] is a two dimensional array, which is stored in the memory along the
row with each of its element occupying 4 bytes, find the address of the element
T[15][5], if the element T[10][8] is stored at the memory location 52000.

Answer:

T
[20] [50]

Along the row

A [i] [j] = BA + size [((i – sr)*nc + j- sc)]

T [10] [8] = 52000
T [15] [5] =52000 + 4* ((15 – 10)*50 + (5-8))
= 52000 + 4* (250 + -3)
= 52000 + 4 * 247
= 52000 + 988
= 52988
Address = 52988

Question
2.

Write definition for a function SHOWMID (int P[][5]), int R, int C) in C++ to
display the elements of middle row and middle column from a two dimensional
array P having R number of rows and C number of columns.

For example, If the content of array is as follows:

115

112

116

101

125

103

101

121

102

101

185

109

109

160

172

The
function should display the following as output:

103 101 121 102 101

116 121 109

Answer:

void
SHOWMID (int P[] [5] , int R, int C)

 {

 for(int J=0;J<C;J++)

 cout<<P[R/2] [J] <<” ” ;

 cout<<end1;

 for(int I=0;I<R;I++)

 cout<<P[I][C/2]<<” “;

 }

Question
3.

R[20][50] is a two dimensional array, which is stored in the memory along the
row with each of its element occupying 8 bytes, find the address of the element
R[5][15], if the element R[8][10] is stored at the memory location 45,000.

Answer:

R[20]
[50]

Along the row

A[i][j] = BA + size [(i – Sr)nc + (j-Sc)]

R[8][10] = 45000
R[5][15] = 45000+8* ((5 – 8)* 50 + (15 -10))
= 45000 + 8* (- 3*50 + 5)
= 4500

NCERT Solutions for Class 12 Computer Science

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