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Given the number of rows and number of columns of the box, the task is to print the Box Number pattern in C, C++, and python.
Examples:
Outer Elements 1 and inner Elements 0:
Example1:
Input:
given number of rows of the box =11 given number of columns of the box =19
Output:
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Outer Elements 0 and inner Elements 1:
Example2:
Input:
given number of rows of the box =11 given number of columns of the box =19
Output:
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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Program to Print Box Number Pattern in C, C++, and Python
Below are the ways to Print Box Number Pattern in C, C++, and python.
- Box Pattern with 1 and 0 characters (Outer Elements 1 and Inner elements 0)
- Box Pattern with 1 and 0 characters (Outer Elements 0 and Inner elements 1)
Method #1: Box Pattern with 1 and 0 characters (Outer Elements 1 and Inner elements 0)
Approach:
- Give the number of rows and number of columns of the box as static or user input and store them in two separate variables.
- Loop from 1 to the number of rows of the box +1 using a For loop.
- Loop from 1 to the number of columns of the box +1 using another Nested For loop.
- Check if the parent iterator value is equal to 1 or the number of rows of the box using the If statement.
- Check if the inner loop iterator value is equal to 1 or the number of columns of the box using the If statement.
- Merge these two conditions in a single If statement using or operator.
- If the condition is true then print 1 with a space character.
- Else print 0 with a space character.
- Print the newline character after the end of the inner For loop.
- The Exit of the Program.
1) Python Implementation
Below is the implementation:
# Give the number of rows and number of columns of the box as static or user input
# and store them in two separate variables.
boxrows = 11
boxcolumns = 19
# Loop from 1 to the number of rows of the box +1 using a For loop.
for m in range(1, boxrows+1):
# Loop from 1 to the number of columns of the box +1 using another Nested For loop.
for n in range(1, boxcolumns+1):
'''Check if the parent iterator value is equal to 1 or the number of rows of the box using the If statement.
Check if the inner loop iterator value is equal to 1 or the number of columns of the box using the If statement.
Merge these two conditions in a single If statement using or operator.
'''
if(m == 1 or m == boxrows or n == 1 or n == boxcolumns):
# If the condition is true then print 1 with a space character.
print("1", end=" ")
else:
# Else print 0 with a space character.
print("0", end=" ")
# Print the newline character after the end of the inner For loop.
print()
Output:
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2) C++ Implementation
Below is the implementation:
#include <iostream>
using namespace std;
int main()
{
// Give the number of rows and number of columns of the
// box as static or user input and store them in two
// separate variables.
int boxrows = 11;
int boxcolumns = 19;
// Loop from 1 to the number of rows of the box +1 using
// a For loop.
for (int m = 1; m <= boxrows; m++) {
// Loop from 1 to the number of columns of the box
// +1 using another Nested For loop.
for (int n = 1; n <= boxcolumns; n++) {
/*Check if the parent iterator value is equal to
1 or the number of rows of the box using the
If statement. Check if the inner loop
iterator value is equal to 1 or the number of
columns of the box using the If statement.
Merge these two conditions in a single If
statement using or operator. */
if (m == 1 || m == boxrows || n == 1
|| n == boxcolumns) {
// If the condition is true then print 1
// with a space character.
cout << "1 ";
}
else {
// Else print 0 with a space character.
cout << "0 ";
}
}
// Print the newline character after the end of the
// inner For loop.
cout << endl;
}
return 0;
}
Output:
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
3) C Implementation
Below is the implementation:
#include <stdio.h>
int main()
{
// Give the number of rows and number of columns of the
// box as static or user input and store them in two
// separate variables.
int boxrows = 11;
int boxcolumns = 19;
// Loop from 1 to the number of rows of the box +1 using
// a For loop.
for (int m = 1; m <= boxrows; m++) {
// Loop from 1 to the number of columns of the box
// +1 using another Nested For loop.
for (int n = 1; n <= boxcolumns; n++) {
/*Check if the parent iterator value is equal to
1 or the number of rows of the box using the
If statement. Check if the inner loop
iterator value is equal to 1 or the number of
columns of the box using the If statement.
Merge these two conditions in a single If
statement using or operator. */
if (m == 1 || m == boxrows || n == 1
|| n == boxcolumns) {
// If the condition is true then print 1
// with a space character.
printf("1 ");
}
else {
// Else print 0 with a space character.
printf("0 ");
}
}
// Print the newline character after the end of the
// inner For loop.
printf("\n");
}
return 0;
}
Output:
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Method #2: Box Pattern with 1 and 0 characters (Outer Elements 0 and Inner elements 1)
Approach:
- Give the number of rows and number of columns of the box as static or user input and store them in two separate variables.
- Loop from 1 to the number of rows of the box +1 using a For loop.
- Loop from 1 to the number of columns of the box +1 using another Nested For loop.
- Check if the parent iterator value is equal to 1 or the number of rows of the box using the If statement.
- Check if the inner loop iterator value is equal to 1 or the number of columns of the box using the If statement.
- Merge these two conditions in a single If statement using or operator.
- If the condition is true then print 0 with a space character.
- Else print 1 with a space character.
- Print the newline character after the end of the inner For loop.
- The Exit of the Program.
1) Python Implementation
Below is the implementation:
# Give the number of rows and number of columns of the box as static or user input
# and store them in two separate variables.
boxrows = 11
boxcolumns = 19
# Loop from 1 to the number of rows of the box +1 using a For loop.
for m in range(1, boxrows+1):
# Loop from 1 to the number of columns of the box +1 using another Nested For loop.
for n in range(1, boxcolumns+1):
'''Check if the parent iterator value is equal to 1 or the number of rows of the box using the If statement.
Check if the inner loop iterator value is equal to 1 or the number of columns of the box using the If statement.
Merge these two conditions in a single If statement using or operator.
'''
if(m == 1 or m == boxrows or n == 1 or n == boxcolumns):
# If the condition is true then print 0 with a space character.
print("0", end=" ")
else:
# Else print 1 with a space character.
print("1", end=" ")
# Print the newline character after the end of the inner For loop.
print()
Output:
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2) C++ Implementation
Below is the implementation:
#include <iostream>
using namespace std;
int main()
{
// Give the number of rows and number of columns of the
// box as static or user input and store them in two
// separate variables.
int boxrows = 11;
int boxcolumns = 19;
// Loop from 1 to the number of rows of the box +1 using
// a For loop.
for (int m = 1; m <= boxrows; m++) {
// Loop from 1 to the number of columns of the box
// +1 using another Nested For loop.
for (int n = 1; n <= boxcolumns; n++) {
/*Check if the parent iterator value is equal to
1 or the number of rows of the box using the
If statement. Check if the inner loop
iterator value is equal to 1 or the number of
columns of the box using the If statement.
Merge these two conditions in a single If
statement using or operator. */
if (m == 1 || m == boxrows || n == 1
|| n == boxcolumns) {
// If the condition is true then print 0
// with a space character.
cout << "0 ";
}
else {
// Else print 1 with a space character.
cout << "1 ";
}
}
// Print the newline character after the end of the
// inner For loop.
cout << endl;
}
return 0;
}
Output:
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
3) C Implementation
Below is the implementation:
#include <stdio.h>
int main()
{
// Give the number of rows and number of columns of the
// box as static or user input and store them in two
// separate variables.
int boxrows = 11;
int boxcolumns = 19;
// Loop from 1 to the number of rows of the box +1 using
// a For loop.
for (int m = 1; m <= boxrows; m++) {
// Loop from 1 to the number of columns of the box
// +1 using another Nested For loop.
for (int n = 1; n <= boxcolumns; n++) {
/*Check if the parent iterator value is equal to
1 or the number of rows of the box using the
If statement. Check if the inner loop
iterator value is equal to 1 or the number of
columns of the box using the If statement.
Merge these two conditions in a single If
statement using or operator. */
if (m == 1 || m == boxrows || n == 1
|| n == boxcolumns) {
// If the condition is true then print 0
// with a space character.
printf("0 ");
}
else {
// Else print 1 with a space character.
printf("1 ");
}
}
// Print the newline character after the end of the
// inner For loop.
printf("\n");
}
return 0;
}
Output:
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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