Problem-Solving in C (Part-1): Nested Loop, Function, and Pointer

Pattern

Problem Statement

You will be given a positive integer N, you need to print a pattern shown below using this N.

For example: If N=5, the pattern will look like below.

See the sample test cases for more clarification.

Input Format

• Input will contain only N.

Constraints

1. 1 <= N <= 100

Output Format

• Output the pattern.

Sample Input 0

4

Sample Output 0

   #
  ---
 #####
-------
 #####
  ---
   #

Sample Input 1

1

Sample Output 1

#

Sample Input 2

7

Sample Output 2

      #
     ---
    #####
   -------
  #########
 -----------
#############
 -----------
  #########
   -------
    #####
     ---
      #

Try on your own.

Photo by Gema Saputera on Unsplash

Answer:

#include<stdio.h>
#include<limits.h>
#include<string.h>
int main(){
    int n, s, k;
    scanf("%d", &n);
    s=n-1;
    k=1;
    for(int i=1;i<=(2*n)-1;i++){
        for(int j=1;j<=s;j++){
            printf(" ");
        }
        if (i%2==1){
        for(int j=1;j<=k;j++){
            printf("#");
        }
        }
        else if (i%2==0){
            for(int j=1;j<=k;j++){
            printf("-");
        }
        }
        if(i<=n-1){
            s--;
            k=k+2;
        }
        else{
            s++;
            k-=2;
        }
        if(i<n ||i>=n){
        printf("\n");
        }
    }
    return 0;
}

Count Before One

Problem Statement

You will be given an array A and the size of that array N. Take input in the main function. You need to write a function named count_before_one() which receives that array of integers and the size of that array and return type will be a 32-bit integer. The function counts the number of elements in that array until you find 1 and returns that count. Print that count in the main function.

Input Format

• The first line will contain N, the size of the array.
• The next line will contain the array A.

Constraints

1. 1 <= N <= 1000
2. 0 <= A[i] <= 1000; Here 0 <= i < N

Output Format

• Output the number of elements before the first 1.

Sample Input 0

5
5 4 3 1 2

Sample Output 0

3

Sample Input 1

5
1 2 3 4 5

Sample Output 1

0

Try on your own first.

Photo by Ilya Pavlov on Unsplash

Answer:

#include<stdio.h>
#include<limits.h>
#include<string.h>
int count_before_one(int ar[], int x){ 
    int element=0;
    for(int i=0;i<x;i++){
        if(ar[i]==1){
            return element;
        }
        else{
            element++;
        }
    }
    return element;
}
int main(){
    int n;
    scanf("%d", &n);
    int s[n];
    for(int i=0;i<n;i++){
        scanf("%d", &s[i]);
    }
    int z= count_before_one(s,n);
    printf("%d", z);
    return 0;
}

Pattern 2

Problem Statement

You will be given a positive integer N, you need to print a pattern shown below using this N.

For example: If N=5, the pattern will look like below.

See the sample test cases for more clarification.

Input Format

• Input will contain only N.

Constraints

1. 1 <= N <= 9

Output Format

• Output the pattern.

Sample Input 0

1

Sample Output 0

1

Sample Input 1

4

Sample Output 1

   1
  21
 321
4321

Sample Input 2

5

Sample Output 2

    1
   21
  321
 4321
54321

Try on your own.

Photo by Fotis Fotopoulos on Unsplash

Answer:

#include<stdio.h>
#include<limits.h>
#include<string.h>
int main(){
    int n;
    scanf("%d", &n);
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n-i;j++){
            printf(" ");
        }
        for(int k=i;k>=1;k--){
            printf("%d",k);
        }
        printf("\n");
    }
    return 0;
}

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