Coin Distribution Problem

Here on this page we have provided a previous year coding problem that was asked in TCS CodeVita Coding Competition i.e; Coin Distribution Problem, in this problem we have to find out the minimum number of coins needed, for generating a specific value. Let’s see how to code a C++ Program for Coin Distribution Problem

Problem Statement

Find the minimum number of coins required to form any value between 1 to N,both inclusive.Cumulative value of coins should not exceed N. Coin denominations are 1 Rupee, 2 Rupee and 5 Rupee.Let’s Understand the problem using the following example. Consider the value of N is 13, then the minimum number of coins required to formulate any value between 1 and 13, is 6. One 5 Rupee, three 2 Rupee and two 1 Rupee coins are required to realize any value between 1 and 13. Hence this is the answer.However, if one takes two 5 Rupee coins, one 2 rupee coin and two 1 rupee coin, then too all values between 1 and 13 are achieved. But since the cumulative value of all coins equals 14, i.e., exceeds 13, this is not the answer.

Input Format:

A single integer value.

Output Format:

Four space separated integer values.
- - 1st – Total number of coins.
  - 2nd – number of 5 Rupee coins.
  - 3rd – number of 2 Rupee coins.
  - 4th – number of 1 Rupee coins.

Constraints:

0 < n < 1000

Refer the sample output for formatting

Sample Input:

Sample Output:

6 1 3 2

Explanation:

The minimum number of coins required is 6 with in it:
- minimum number of 5 Rupee coins = 1
- minimum number of 2 Rupee coins = 3
- minimum number of 1 Rupee coins = 2

Using these coins, we can form any value with in the given value and itself, like below:

Here the given value is 13

For 1 = one 1 Rupee coin
For 2 = one 2 Rupee coin
For 3 = one 1 Rupee coin and one 2 Rupee coins
For 4 = two 2 Rupee coins
For 5 = one 5 Rupee coin
For 6 = one 5 Rupee and one 1 Rupee coins
For 7 = one 5 Rupee and one 2 Rupee coins
For 8 = one 5 Rupee, one 2 Rupee and one 1 Rupee coins
For 9 = one 5 Rupee and two 2 Rupee coins
For 10 = one 5 Rupee, two 2 Rupee and one 1 Rupee coins
For 11 = one 5 Rupee, two 2 Rupee and two 1 Rupee coins
For 12 = one 5 Rupee, three 2 Rupee and one 1 Rupee coins
For 13 = one 5 Rupee, three 2 Rupee and two 1 Rupee coins

Approach 1 – DP Approach

Time Complexity : O(1)

The code to implement this:

#include <stdio.h>

int main()
{
	int n;
	scanf("%d",&n);
	if(!n)
        {
		printf("0 0 0 0");
        }
    int t=n/5;
    int tr=n%5;
    if(n>3 && tr<4)
    {
        tr+=5;
        t-=1;
    }
    int dp[10][3]={{0,0,0},{1,0,0},{2,0,0},{1,1,0},{2,1,0},{1,2,0},{2,2,0},{1,3,0},{2,3,0},{2,1,1}};
    int total=t+ dp[tr][0]+dp[tr][1];
	printf("%d %d %d %d",total,t,dp[tr][1],dp[tr][0]);
}

Approach 2 – Static Approach

Time Complexity : O(1)

The code to implement this:

#include <stdio.h>

int main()
{
	int n;
	scanf("%d",&n);
	int f = (n-4)/5;
	int o;
	if( (n-5*f)%2 == 0) o=2;
	else o=1;
	int t=(n-5*f -o)/2;
	printf("%d %d %d %d",o+t+f,f,t,o);
}

Coin Distribution Problem in few other Coding Languages

C++

To find the solution of Coin Distribution Problem in C++ Programming language click on the button below:

C++

Java

To find the solution of Coin Distribution Problem in Java Programming language click on the button below:

Java

Python

To find the solution of Coin Distribution Problem in Python Programming language click on the button below:

Python

Article is contributed by Rahit Saha

One comment on “Coin Distribution Problem | C Solution”

mevinodrao
anotherway in c language
#include
int i,j,k,x,a,b,c,sum=0,n;
int main()
{
printf(“enter the number”);
scanf(“%d”,&n);

signed x=n-5;
if(x>=0)
{
a=x/5;
b=(x%5)/2;
c=x-(a*5+b*2);

printf(“the total no of coins are %d\n”,3+a+b+c);
printf(“the minimum no of 5rs coins are %d\n”,a);
printf(“the minimum no of 2rs coins are %d\n”,b+2);
printf(“the minimum no of 1rs coins are %d\n”,c+1);
}
else
{
if(n==1)
{
printf(“the total no of coins are %d\n”,1);
printf(“the minimum no of 5rs coins are %d\n”,0);
printf(“the minimum no of 2rs coins are %d\n”,0);
printf(“the minimum no of 1rs coins are %d\n”,1);
}
else if(n==2)
{
printf(“the total no of coins are %d\n”,2);
printf(“the minimum no of 5rs coins are %d\n”,0);
printf(“the minimum no of 2rs coins are %d\n”,0);
printf(“the minimum no of 1rs coins are %d\n”,2);
}
else if(n==3)
{
printf(“the total no of coins are %d\n”,2);
printf(“the minimum no of 5rs coins are %d\n”,0);
printf(“the minimum no of 2rs coins are %d\n”,1);
printf(“the minimum no of 1rs coins are %d\n”,1);
}
else if(n==4)
{
printf(“the total no of coins are %d\n”,3);
printf(“the minimum no of 5rs coins are %d\n”,0);
printf(“the minimum no of 2rs coins are %d\n”,1);
printf(“the minimum no of 1rs coins are %d\n”,2);
}
}
return 0;
}

1

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