programs on arrays.pdf

Solve the given problems on arrays
1D ARRAY:
1. Program to reverse the elements of an array
Input:
arr = {1, 2, 3, 4, 5}
Output:
Reversed array : {5, 4, 3, 2, 1}
Algorithm to reverse an array
 Input the number of elements of an array.
 Input the array elements.
 Traverse the array from the last.
 Print all the elements.
Solution using C
#include<stdio.h>
int main()
{
int n,i, arr[20];
scanf(“%d”,&n);
for(i = 0; i < n; i++)
scanf(“%d”,&arr[i]);
printf(“Reversed array is:”);
for(i = n-1; i >= 0; i–)
printf(“%dt”,arr[i]);
return 0;
}
Solution using python3
n = int(input())
sum = 0
arr = []
for i in range(0,n):
temp = int(input())
arr.append(temp)
print("Reversed array is :",end=" ")
arr.reverse()
print(arr[i],end=" ")
Output:
3
3
2
1
Reversed array is : 1 2 3
2. Program to find the array type (even, odd or mixed array)
Sample Input:
5
2
4
1
3
5
Sample output:
Mixed

Algorithm to find the array type (even, odd or mixed array)
 Input the number of elements of the array.
 If all the elements of the array are odd, display "Odd".
 If all the elements of the array are even, display "Even".
 Else, display "Mixed".
Solution using C
#include<stdio.h>
int main()
{
int n,i, arr[20];
scanf(“%d”,&n);
int odd = 0, even = 0;
for(i = 0; i < n; i++)
for(i = 0; i < n; i++)
{
if(arr[i] % 2 == 1)
odd++;
if(arr[i] % 2 == 0)
even++;
}
if(odd == n)
printf(“Odd”);
else if(even == n)
printf(“Even”);
else
printf(“Mixed”);
return 0;
}
# Python program find the array type (even, odd or
mixed array)
n = int(input())
arr = []
temp = int(input())
arr.append(temp)
odd = 0
even = 0
if(arr[i] % 2 == 1):
odd = odd + 1
else:
even = even + 1
if(odd == n):
print(“Odd”)
elif(even == n):
print(“Even”)
else:
print(“Mixed”);
Output:
5
1 2 3 4 5
Mixed
3. Program to count the number of even and odd elements in an array
Sample Input:
3 (number of elements of the array)
1 (array elements)
2
3
Sample Output:
Odd: 2
Even: 1

Algorithm to count the number of even and odd elements in an array
 Initialize count_odd = count_even = 0.
 Traverse the array and increment count_odd if the array element is odd, else increment
count_even.
 Print count_odd and count_even.
Solution using C
#include
int main()
{
int n,i, arr[20];
scanf(“%d”,&n);
for(i = 0; i < n; i++)
int count_odd =0, count_even = 0;
for( i = 0; i < n; i++)
{
if(arr[i] % 2 == 1)
count_odd++;
else
count_even++;
}
printf(“Odd: %d”,count_odd);
printf(“nEven: %d”,count_even);
return 0;
}
n = int(input())
arr = []
temp = int(input())
arr.append(temp)
count_odd = 0
count_even = 0
if(arr[i] % 2 == 1):
count_odd = count_odd + 1
else:
count_even = count_even + 1
print(“Odd:”, count_odd)
print(“Even:”, count_even)
Output:
5
1 2 3 4 5
Odd : 3
Even : 2
4. Program to check if two arrays are equal or not
Sample Input:
3
3
1
2
3
1
2
3
Sample output:
Same

Algorithm to check if two arrays are equal or not
 Input the number of elements of arr1 and arr2.
 Input the elements of arr1 and arr2.
 Sort arr1 and arr2 using any sorting technique
 If all the elements of arr1 and arr2 are not equal, then print "not Same".
 Else, print " Same".
Solution using C
#include<stdio.h>
int sort(int arr[], int n)
{
int i,j;
for (i = 0; i < n-1; i++)
{
for (j = 0; j < n-i-1; j++)
{
if (arr[j] > arr[j+1])
{
int temp = arr[j];
arr[j] = arr[j+1];
arr[j+1] = temp;
}
}
}
}
int arrays_equal(int arr1[], int arr2[], int n, int
m)
{
int i;
sort(arr1,n);
sort(arr2,m);
for(i = 0; i < n; i++)
{
if(arr1[i] != arr2[i])
{
return 0;
}
}
return 1;
}
int main()
{
int ,arr[20],n,m,i;
scanf(“%d”,&n);
scanf(“%d”,&m);
for(i = 0; i < n; i++)
scanf(“%d”,&arr1[i]);
for(i = 0; i < m; i++)
if(arrays_equal(arr1, arr2, n, m) == 0)
def arrays_equal(arr1, arr2, n, m):
arr1.sort()
arr2.sort()
if(arr1[i] != arr2[i]):
return 0
return 1
n = int(input())
m = int(input())
arr1 = []
arr2 = []
temp = int(input())
arr1.append(temp)
for i in range(0,m):
temp = int(input())
arr2.append(temp)
if(arrays_equal(arr1, arr2, n, m) == 0):
print(“Not Same”)
else:
print(“Same”)

printf(“Not same”);
else
printf(“Same”);
return 0;
}
4. Program to find the sum of perfect square elements in an array
Input: arr = {1, 2, 3, 4, 5, 9}
The perfect squares are 1, 4, 9.
Sum = 1 + 4 + 9 = 14
Output: 14
Algorithm to find the sum of perfect square elements in an array
 Initialize sum = 0.
 Check if the array element is a perfect square.
 If it is a perfect square, sum = sum + num.
 Return sum
Solution using C
#include<stdio.h>
#include<math.h>
int isPerfectSquare(int number)
{
int iVar;
float fVar;
fVar=sqrt((double)number);
iVar=fVar;
if(iVar==fVar)
return number;
else
return 0;
}
int main()
{
int n,i,arr[20],sum=0;
scanf(“%d”,&n);
for(i = 0; i < n; i++)
for(i = 0; i < n; i++)
sum = sum + isPerfectSquare(arr[i]);
printf(“%d”,sum);
return 0;
}
from math import *
def isPerfectSquare(number):
if(sqrt(number) == floor(sqrt(number))):
return number
else:
return 0
n = int(input())
sum = 0
arr1 = []
temp = int(input())
arr1.append(temp)
sum = sum + isPerfectSquare(arr1[i])
print(sum)

Output:
4
1 4 9 16
30
5. Program to find the minimum scalar product of two vectors (dot product)
Sample Input 1:
3 (Number of elements of the array)
1 3 5 (Array 1 elements)
2 4 1 (Array 2 elements)
Sample Output 1:
15
Calculation:
Minimum scalar product = 1 * 4 + 3 * 2 + 5 * 1
= 4 + 6 + 5
= 15
Algorithm to find the minimum scalar product of two vectors
 Input the number of elements of the arrays.
 Input the array 1 and array 2 elements.
 Initialize sum = 0.
 Sort the array 1 in ascending order.
 Sort the array 2 in descending order.
 Repeat from i = 1 to n
o sum = sum + (arr1[i] * arr2[i])
 Return sum.
Solution using C
#include<stdio.h>
int sort(int arr[], int n)
{
int i, j;
for (i = 0; i < n-1; i++)
for (j = 0; j < n-i-1; j++)
if (arr[j] > arr[j+1])
{
int temp = arr[j];
arr[j] = arr[j+1];
arr[j+1] = temp;
}
}
int sort_des(int arr[], int n)
{
int i, j;
for (i = 0; i < n-1; i++)
for (j = 0; j < n-i-1; j++)
if (arr[j] < arr[j+1])
n = int(input())
arr1 = []
arr2 = []
temp = int(input())
arr1.append(temp)
temp = int(input())
arr2.append(temp)
arr1.sort()
arr2.sort(reverse=True)
sum = 0
for i in range(0, n):
sum = sum + (arr1[i] * arr2[i])
print(“Minimum Scalar Product :”,sum)

{
int temp = arr[j];
arr[j] = arr[j+1];
arr[j+1] = temp;
}
}
int main()
{
int n ,i, arr1[n], arr2[n];;
scanf(“%d”,&n);
for(i = 0; i < n ; i++)
for(i = 0; i < n ; i++)
sort(arr1, n);
sort_des(arr2, n);
int sum = 0;
for(i = 0; i < n ; i++)
sum = sum + (arr1[i] * arr2[i]);
printf(“%d”,sum);
return 0;
}
6. Program to find the smallest and largest elements in an array
For example, consider the array.
arr = {1, 2, 3, 4, 5}
Smallest element : 1
Largest element : 5
Algorithm to find the smallest and largest numbers in an array
 Initialize small = large = arr[0]
 Repeat from i = 1 to n
 if(arr[i] > large)
 large = arr[i]
 if(arr[i] < small)
 small = arr[i]
 Print small and large.

Solution using C
#include<stdio.h>
int main()
{
int a[50],i,n,large,small;
scanf(“%d”,&n);
for(i=0;i<n;++i)
scanf(“%d”,&a[i]);
large=small=a[0];
for(i=1;i<n;++i)
{
if(a[i]>large)
large=a[i];
if(a[i]<small)
small=a[i];
}
printf(“nThe smallest element is
%dn”,small);
printf(“nThe largest element is
%dn”,large);
return 0;
}
arr = []
n = int(input())
numbers = int(input())
arr.append(numbers)
print("Maximum element : ", max(arr),
"nMinimum element :" , min(arr))
Output:
5
1 2 3 4 5
The smallest element is 1
The largest element is 5
7. Program to print all the distinct elements in an array
SAMPLE INPUT:
 9 = size of an array
 2 3 4 5 6 1 2 3 4 = array elements
SAMPLE OUTPUT:
 2 3 4 5 61
Algorithm to print distinct numbers in an array
 Declare and input the array elements.
 Traverse the array from the beginning.
 Check if the current element is found in its previous elements the array again.
 If it is found, then do not print that element.
 Else, print that element and continue.

Solution using C
#include <stdio.h>
void distict_elements(int a[], int n);
int main()
{
int size_array, i, arr[20];
scanf(“%d”, &size_array);
for(i=0; i<size_array; i++)
scanf(“%d”, &arr[i]);
distict_elements(arr, size_array);
return 0;
}
void distict_elements(int a[], int n)
{
int i, j;
for (i=0; i<n; i++)
{
for (j=0; j<i; j++)
{
if (a[i] == a[j])
break;
}
if (i == j)
printf(“%dn “, a[i]);
}
}
def distinct(arr, num):
arr1 = []
for i in range(0, num):
d = 0
for j in range(0, i):
if (arr[i] == arr[j]):
d = 1
break
if (d == 0):
arr1.append(arr[i])
print(“Distinct elements:”)
print (arr1)
n = int(input())
arr = []
for i in range(n):
arr.append(int(input()))
print(“Input array: “)
print (arr)
distinct(arr, n)
Output:
5
1 1 2 3 4
1
2
3
4
8. Program to check if the given arrays are disjoint
For example :
arr1 = {1,2,3,4,5}
arr2 = {6,7,8,9}
arr1 and arr2 elements are unique and hence they are said to be disjoint.
arr3 = {1,2,3,4,5}
arr4 = {4,5,6,7}
arr3 and arr4 are not disjoint as they have elements 4 and 5 in common.
Input:

4
1 2 3 4
3
6 7 8
given arrays are disjoint
Algorithm to check if the given arrays are disjoint
 Use two loops.
 Traverse the array 1 using the outer loop.
 Use the inner loop to check if the elements in array 2 are found in array 1.
 If at least one element of array 2 is found in array 1, return false. Else return true.
Solution using C
#include <stdio.h>
int disjoint_arrays(int arr1[], int arr2[], int n,
int m)
{
int i,j;
for(i = 0;i<n;i++)
{
for(j=0;j<m;j++)
{
if(arr1[i] == arr2[j])
return -1;
}
}
return 1;
}
int main()
{
int m,n,arr1[20],arr2[20],i,j;
printf("Enter the size of array 1n");
scanf("%d",&n);
printf("Input Array 1 elements n");
for(i=0;i<n;i++)
scanf("%d",&arr1[i]);
printf("Enter the size of array 2 n");
scanf("%d",&m);
printf("Input Array 2 elements n");
for(i=0;i<m;i++)
scanf("%d",&arr2[i]);
int res = disjoint_arrays(arr1,arr2,n,m);
if(res == -1)
printf("The arrays are not disjointn");
else
printf("The arrays are disjointn");
return 0;
}
def disjoint_arrays(arr1,arr2):
for i in range(0,len(arr1)):
for j in range(0,len(arr2)):
if(arr1[i] == arr2[j]):
return -1
return 1;
arr1 = [1,2,3,4,5]
arr2 = [6,7,8,9,10]
res = disjoint_arrays(arr1,arr2)
if(res == -1):
print(“The arrays are not disjoint”)
else:
print(“The arrays are disjoint”)

9. Program to find all symmetric pairs in an array. Two pairs (p,q) and (r,s) are said to be
symmetric when q is equal to r and p is equal to s. For example, (5,10) and (10,5) are symmetric
pairs
For example,
Consider a 2D array,
Input:
arr [6] [2] = {{1, 2}, {3, 4}, {5, 6}, {2, 1}, {4, 3},{10,11}}
Output:
{1,2} and {2,1} are symmetric
{3,4} abd {4,3} are symmetri
This problem can be solved in two different ways.
Method 1: Using two loops, one loop to traverse the pairs and the other loop to check if similar pair is
existing.
Method 2: An efficient way to solve this problem is to use hashing. Insert each array pairs into the hash
table with the first element of the pair as the key and the second element as the value. Traverse the hash
table to check if the pairs are found again.
Algorithm to find all symmetric pairs in an array
Input the array from the user.
1. Use two loops.
2. One loop for traversing the array and the other loop to check if symmetric pair is found in the
array.
3. If symmetric pair is found, print the pairs.
Algorithm to find symmetric pairs in an array using hashing
1. Input the array from the user.
2. From all the array pairs, the first element is used as the key and the second element is used as
the value.
3. Traverse all the pairs one by one.
4. For every pair, check if its second element is found in the hash table.
5. If yes, then compare the first element with the value of the matched entry of the hash table.
6. If the value and the first element match, it is displayed as a symmetric pair.
7. Else, insert the first element as the key and second element as value and repeat the same.

10. Program to insert, delete and search an element in an array
Input format:
 Input consists of 3 integers and 1 array.
 Input the size of the array.
 Input the position where the element should be inserted.
 Input the element to be inserted in that position.
Sample Input:
5 (size of the array)
1 (array elements)
2
3
4
5
4 (Position)
10 (Element to be inserted)
Sample Output:
Array after insertion is:
1
2
3
10
4
5
Algorithm to insert, delete and search an element in an array
Insertion
 Input the array elements, the position of the new element to be inserted and the new element.
 Insert the new element at that position and shift the rest of the elements to right by one position.
Deletion
 Input the array elements, the position of the new element to be inserted and the new element.
 Delete the element and shift the rest of the elements to left by one position.
Search
 Input the array elements, the element to be searched.
 Traverse the array and check if the element is present in the array.

 Display "Yes" if it is present in the array, else display "No".
Insertion:
Solution using C
#include <stdio.h>
int main()
{
int n,arr[20],i,pos;
scanf(“%d”,&n);
for(i = 0; i < n; i++)
scanf(“%d”,&pos);
int ele;
scanf(“%d”,&ele);
if(pos > n)
printf(“Invalid Input”);
else
{
for (i = n – 1; i >= pos ; i–)
arr[i+1] = arr[i];
arr[pos] = ele;
printf(“Array after insertion
is:n”);
for (i = 0; i <= n; i++)
printf(“%dn”, arr[i]);
}
return 0;
}
n = int(input())
sum = 0
arr = [1,2,3,4,5]
pos = int(input())
ele = int(input())
arr.insert(pos,ele)
print(arr[i], end = ” “)
Deletion:
Solution using C
#include <stdio.h>
int main()
{
int array[10], position, c, n;
printf(“Enter the number of elements of
the array : “);
scanf(“%d”, &n);
printf(“nInput the array elements : “);
for (c = 0; c < n; c++)
scanf(“%d”, &array[c]);
printf(“nEnter the position : “);
scanf(“%d”, &position);
if (position >= n+1)
printf(“nDeletion not possible.n”);
else
arr = [1,2,3,4,5]
pos = int(input())
arr.remove(arr[pos])
for i in range(0,n-1):
print(arr[i], end = ” “)

{
for (c = position ; c < n – 1; c++)
array[c] = array[c+1];
printf(“nArray after deletion : “);
for (c = 0; c < n – 1; c++)
printf(“%dn”, array[c]);
}
return 0;
}
Searching:
Solution using C
#include <stdio.h>
#include <stdlib.h>
int main()
{
int array[10], ele, c, n;
printf(“Enter the number of elements of
the array : “);
scanf(“%d”, &n);
printf(“nInput the array elements : “);
for (c = 0; c < n; c++)
scanf(“%d”, &array[c]);
printf(“nEnter element : “);
scanf(“%d”, &ele);
for(c = 0; c < n ; c++)
{
if(array[c] == ele)
{
printf(“nElement
foundn”);
exit(0);
}
}
printf(“element not foundn”);
return 0;
}
n = int(input())
arr = []
temp = int(input())
arr.append(temp)
ele = int(input())
if(arr[i] == ele):
print("Element Found")
exit()
print("element not found")

11. Program to sort the array of elements
Solution using C
#include<stdio.h>
int main()
{
int arr[20],n,i,j,temp;
printf("Enter array size:");
scanf("%d",&n);
printf("Enter %d elements:n",n);
for(i=0;i<n;i++)
scanf("%d",&arr[i]);
printf("Before sorting array elements
are:n");
for(i=0;i<n;i++)
printf("%dt",arr[i]);
for(i=0;i<n-1;i++)
{
for(j=i+1;j<n;j++)
{
if(arr[i]>arr[j])
{
temp=arr[i];
arr[i]=arr[j];
arr[j]=temp;
}
}
}
printf("nAfter sorting array elements
are:n");
for(i=0;i<n;i++)
printf("%dt",arr[i]);
return 0;
}
n = int(input())
arr = []
temp = int(input())
arr.append(temp)
print("before sorting")
print("nafter sorting")
arr.sort();
OUTPUT:
Enter array size:5
Enter 5 elements:
3 4 2 5 1
Before sorting array elements are:
3 4 2 5 1
After sorting array elements are:
1 2 3 4 5

2D ARRAY:
12. Program to find the maximum element in each row of a matrix
Input:
3 3 (Order of the matrix - number of rows and columns)
1 4 9
3 5 1
2 8 5
Output:
9
5
8
Algorithm:
 Input the order of the matrix.
 Input the matrix elements.
 For row = 0 to n-1
 Find the maximum element in the row
 Print the maximum element
Solution using C
#include<stdio.h>
void maxi_row(int mat[][], int m, int n)
{
int i = 0, j,max;
for(i=0;i<n;i++)
{
max=0;
for ( j = 0; j < n; j++)
{
if (mat[i][j] > max)
max = mat[i][j];
}
printf("%dn",max);
}
}
int main()
{
int m, n, i, j, mat[20][20];;
scanf("%d %d",&m,&n);
for(i = 0; i < m; i++)
{
for(j = 0; j < n; j++)
scanf("%d",&mat[i][j]);
}
maxi_row(mat,m,n);
return 0;
}
def maxi_row(arr):
no_of_rows = len(arr)
no_of_column = len(arr[0])
for i in range(no_of_rows):
max = 0
for j in range(no_of_column):
if arr[i][j] > max :
max = arr[i][j]
print(max)
mat = [[1,2,3],[4,5,6],[7,8,9]]
maxi_row(mat)

13. Program to find the minimum element in each row of a matrix
Input:
3 3 (Order of the matrix - number of rows and columns)
1 4 9
3 5 1
2 8 5
Output:
1
1
2
Algorithm:
 For row = 0 to n-1
 Find the minimum element in the row
 Print the minimum element
Solution using C
#include<stdio.h>
#include<limits.h>
void min_row(int mat[][20], int m, int n)
{
int i = 0, j,min;
for(i=0;i<n;i++)
{
min = INT_MAX;
for ( j = 0; j < n; j++)
{
if (mat[i][j] < min)
min = mat[i][j];
}
printf("%dn",min);
}
}
int main()
{
int m, n, i, j, mat[20][20];;
scanf("%d %d",&m,&n);
for(i = 0; i < m; i++)
{
for(j = 0; j < n; j++)
scanf("%d",&mat[i][j]);
}
min_row(mat,m,n);
return 0;
}
def maxi_row(arr):
no_of_rows = len(arr)
no_of_column = len(arr[0])
for i in range(no_of_rows):
min = 99999
for j in range(no_of_column):
if arr[i][j] < min :
min = arr[i][j]
print(min)
mat = [[1,2,3],[4,5,6],[7,8,9]]
maxi_row(mat)

14.Program to find transpose of a matrix
Input:
3 3
1 2 3
4 5 6
7 8 9
Output:
1 4 7
2 5 8
3 6 9
Algorithm:
 For i = 0 to row_size
o For j= 0 to col_size
 Transpose[j][i]=matrix[i][j]
 Display the transpose matrix
Solution using C
#include <stdio.h>
int main()
{
int m, n, i, j, matrix[10][10], transpose[10][10];
scanf("%d%d", &m, &n);
for (i = 0; i < m; i++)
for (j = 0; j < n; j++)
scanf("%d", &matrix[i][j]);
for (i = 0; i < m; i++)
for (j = 0; j < n; j++)
transpose[j][i] = matrix[i][j];
printf("Transpose of the matrix:n");
for (i = 0; i < n; i++)
{
for (j = 0; j < m; j++)
printf("%dt", transpose[i][j]);
printf("n");
}
return 0;
}
matrix=[]
row=int(input("Enter number of rows:"))
column=int(input("Enter number of columns:"))
for i in range(row):
matrix.append([])
for j in range(column):
num=int(input("Enter the element:"))
matrix[i].append(num)
print('n')
print("Input matrix:")
for i in range (row):
for j in range (column):
print (matrix[i][j], end=" ")
print('n')
transpose=[]
for j in range(column):
transpose.append([])
t_num=matrix[i][j]
transpose[j].append(t_num)
print('n')
print("Transpose matrix:")
for j in range (column):
print (transpose[i][j], end=" ")
print('n')

15. Program to print the sum of boundary elements of a matrix
For example, consider the matrix given below.
1 2 3
4 5 6
7 8 9
Boundary Matrix
1 2 3
4 6
7 8 9
Sum of boundary elements: 1 + 2 + 3 + 4 + 6 + 7 + 8 + 9 = 40
Algorithm to print the sum of boundary elements of a matrix
 Print all the boundary elements of the matrix.
 Find the sum of the boundary elements.
 Print sum
Solution using C
#include <stdio.h>
#include<stdio.h>
int main()
{
int m, n, sum = 0;
printf(“nEnter the order of the matrix : “);
scanf(“%d %d”,&m,&n);
int i, j;
int mat[m][n];
printf(“nInput the matrix elementsn”);
for(i = 0; i < m; i++)
{
for(j = 0; j < n; j++)
scanf(“%d”,&mat[i][j]);
}
printf(“nBoundary Matrixn”);
for(i = 0; i < m; i++)
{
for(j = 0; j < n; j++)
{
if (i == 0 || j == 0 || i == n – 1 || j == n – 1)
{
printf(“%d “, mat[i][j]);
sum = sum + mat[i][j];
}
else
printf(” “);
}
printf(“n”);
}
printf(“nSum of boundary is %d”, sum);
return 0;
}
mat = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
sum = 0
m = 3
n = 3
print(“nBoundary Matrixn”)
for i in range (0, m):
for j in range (0, n):
if (i == 0 or j == 0 or i == n – 1 or j
== n – 1):
print(mat[i][j], end = ” “)
sum = sum + mat[i][j]
else:
print(end = ” “)
print(“n”)
print(“nSum of boundary is “, sum)

Patterns:
1. Program to print solid square star pattern
SAMPLE INPUT:
5
SAMPLE OUTPUT:
n x n matrix filled with „*‟
* * * * *
* * * * *
* * * * *
* * * * *
* * * * *
Algorithm:
 Input the number n
 Repeat from i = 0 to n-1
o Repeat from j = 0 to n-1
 Print *
Solution using C
#include <stdio.h>
int main()
{
int i, j,n;
scanf(“%d”,&n);
for(i = 0; i <n; i++)
{
for(j = 0; j < n; j++)
{
printf(“*”);
}
printf(“n”);
}
return 0;
}
n = int(input())
for j in range(0,n):
print("*", end = " ")
print()
Output:
5
* * * * *
* * * * *
* * * * *
* * * * *
* * * * *

2. Program to print Hallow square star pattern
SAMPLE INPUT:
5
SAMPLE OUTPUT:
* * * * *
* *
* *
* *
* * * * *
Algorithm:
o Repeat from j = 0 to n-1
 If i = 0 or i=n-1 or j = 0 or j=n-1 then print *
 Else print (“ “)
Print(“n”)
Solution using C
#include <stdio.h>
int main()
{
int i, j, n;
scanf(“%d”,&n);
for (i = 0; i <n; i++)
{
for (j = 0; j < n; j++)
{
if (i==0 || i==n-1 || j==0 || j==n-1)
printf(“*”);
else
printf(” “);
}
printf(“n”);
}
return 0;
}
n = int(input())
for i in range(0, n):
for j in range(0, n):
if (i == 0 or i == n-1 or j == 0 or j == n-1):
print("*", end=" ")
else:
print(" ", end=" ")
print()
Output:
5
* * * * *
* *
* *
* *
* * * * *

3. Program to print half pyramid pattern using stars
SAMPLE INPUT:
5
SAMPLE OUTPUT:
*
* *
* * *
* * * *
* * * * *
Algorithm:
o Repeat from j = 0 to i
 Print *
Print(“n”)
Solution using C
#include<stdio.h>
int main()
{
int i, j,n;
scanf("%d",&n);
for(i = 1; i <= n; i++)
{
for(j = 1; j <= i; j++)
{
printf("* ");
}
printf("n");
}
return 0;
}
n = int(input())
for i in range(1,n+1):
for j in range(1,i+1):
print("*", end =" ")
print()
OUTPUT:
5
*
* *
* * *
* * * *
* * * * *
4. Program to print pyramid pattern printing using numbers

SAMPLE INPUT:
5
SAMPLE OUTPUT:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Algorithm:
o Repeat from j = 0 to i
 Print j
Print(“n”)
Solution using C
#include<stdio.h>
int main()
{
int i, j,n;
scanf("%d",&n);
for(i = 1; i <= n; i++)
{
for(j = 1; j <= i; j++)
{
printf("%d ",j);
}
printf("n");
}
return 0;
}
n = int(input())
for i in range(1,n+1):
for j in range(1,i+1):
print(j, end =" ")
print()
OUTPUT:
5
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5

programs on arrays.pdf

Recommended

Recommended

More Related Content

Similar to programs on arrays.pdf

Similar to programs on arrays.pdf (20)

Recently uploaded

Recently uploaded (20)

programs on arrays.pdf