Problem 4 - Pattern 3 - Concentric Squares. (C Programming)

I have also uploaded a part 2 of this video which is one level up version, but the concept used there is very easy than this. You may also check out that video at

[prashantsom]

Samsung R&D same question 44 lakh package

[satyamshivam]

lo bhai.. solution without using matrix. Working perfectly fine. Same hi logic hai, thora sa bas modify kiya hai.
Python me hai .. you can convert to C

N = int(input())

# Print for first half

for i in range(1, N+1, 1):
val = N #Value to be printed
for j in range(1, 2*N, 1):
if(i>j):
print(val, end='')
val -= 1
elif ((i+j)>=2*N):
print(val, end='')
val += 1
else:
print(val, end='')

print()


# Print for second half

for i in range(1, N, 1):
val = N #Value to be printed
for j in range(1, 2*N, 1):
if((N-i)>j):
print(val, end='')
val -= 1
elif ((N-i+j)>=2*N):
print(val, end='')
val += 1
else:
print(val, end='')

print()

[rehanalampanah]

Is there any correct solution of this question available ? Everyone is giving wrong answers.

[cagnusmarlsen]

solution:

#include <stdio.h>
int main()
{
int n;
scanf("%d", &n);
int len = n*2 - 1;
for(int i=0;i<len;i++)
{
for(int j=0;j<len;j++)
{
int min = i < j ? i : j;
min = min < len-i ? min : len-i-1;
min = min < len-j-1 ? min : len-j-1; printf("%d ", n-min);
}
printf("\n");
}
return 0;
}

[pcenxyz]

Concentric triangle pey bhi ek video banao na pls

[pragatiagrawal]

Sir, On writing this code why I am not getting the correct output. Please help me out.

[pankajkumar-muuj]

Har 1 questions m logic pakad ni paa Raha hun

[maheshchitty]

2 2
2 1
2 2
I have got above output for the input 2
Sir, Pls explain why??

[pankajkumar-muuj]

Sir kuchh Aisa batao na jisse ki ham his tarah math ko Bana lete h usi tarah c ko bhi Bana le

[rohithchittibommala]

Sir I sent a message again please check

[faizanquazi]

not correct for n=7 ....try it on computer

[UIEC_MANESH_RAM-tbhb]

#include<stdio.h>
#include<stdlib.h>
int main()
{
int n, mid, i, j, p;
printf("\n ENTER ODD NUMBER OF EDGES OF SQUARE = ");
scanf("%d", &n);
mid=n/2;
printf("\n ");
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
if(abs(mid-i)>abs(mid-j))
{
p=abs(mid-i);
}
else
{
p=abs(mid-j);
}
printf("%d ", p+1);
}
printf("\n ");
}
return 0;
}
SIR I THINK THIS CODE IS EASIER TO UNDERSTAND