# C Programming – HackerRank Solution | STACKS – DENOMINATIONS |

### (HackerRank) Write a Modular C Programming code to solve STACKS – DENOMINATIONS, Frankinson has N number of coins in his pocket. Coins are of denominations of 1, 2, 5 and 10.

Frankinson has N number of coins in his pocket. Coins are of denominations of 1, 2, 5 and 10.

He keeps all the coins on the table one above the other. Then he picks out one coin at a time and counts them of specific denomination.

Help Frankinson to count the coins of all the different denominations.

Input Format

First line is N number of coins.

Second line is coins of different denominations.

Constraints

Denomination of each coin should be either 1,2,5 or 10.

Output Format

Number of Coins of Each denomination in separate line.

Sample Input 0

5
1 2 5 10 1

Sample Output 0

Coins of 1 re = 2
Coins of 2 rs = 1
Coins of 5 rs = 1
Coins of 10 rs = 1
Coins of invalid denominations = 0

Sample Input 1

2
1 3

Sample Output 1

Coins of 1 re = 1
Coins of 2 rs = 0
Coins of 5 rs = 0
Coins of 10 rs = 0
Coins of invalid denominations = 1

Sample Input 2

-5

Sample Output 2

Invalid number of coins

CODE:

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define SIZE 100
struct stack
{
int coin[SIZE];
int top;
};
void push(struct stack *sptr, int num);
void pop_count_display(struct stack *sptr);
int main()
{
struct stack * sptr;
struct stack s;
sptr=&s;
sptr->top=-1;
int c,n,i;
scanf("%d",&n);
if(n<0)
{
printf("Invalid number of coins");
exit(0);
}
for(i=0;i<n;i++)
{
scanf("%d",&c);
push(sptr,c);
}
pop_count_display(sptr);
}
void push(struct stack *sptr, int num)
{
if(sptr->top==SIZE-1)
{
printf("Stack Overflow\n");
}
else
{
sptr->top++;
sptr->coin[sptr->top]=num;
}
}
void pop_count_display(struct stack *sptr)
{
int i,c1=0,c2=0,c5=0,c10=0,c=0;
for(i=sptr->top;i>=0;i--)
{
if(sptr->coin[i]==1)
{
c1++;
}

else if(sptr->coin[i]==2)
{
c2++;
}
else if(sptr->coin[i]==5)
{
c5++;
}
else if(sptr->coin[i]==10)
{
c10++;
}
else
{
c++;
}
sptr->top--;
}
printf("Coins of 1 re = %d\n",c1);
printf("Coins of 2 rs = %d\n",c2);
printf("Coins of 5 rs = %d\n",c5);
printf("Coins of 10 rs = %d\n",c10);
printf("Coins of invalid denominations = %d",c);
}

OUTPUT

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Input (stdin)

5
1 2 5 10 1

Coins of 1 re = 2
Coins of 2 rs = 1
Coins of 5 rs = 1
Coins of 10 rs = 1
Coins of invalid denominations = 0
Expected Output

Coins of 1 re = 2
Coins of 2 rs = 1
Coins of 5 rs = 1
Coins of 10 rs = 1
Coins of invalid denominations = 0

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