Coin Distribution Problem | C++ Solution

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

Coin exchange Problem C++

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:

13

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<bits/stdc++.h>
using namespace std;
int main() { int n,total,five,two,one; cin>>n; if(!n) {cout<<0<<" "<<0<<" "<<0<<" "<<0<<endl; } 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}}; total=t+ dp[tr][0]+dp[tr][1]; cout<<total<<" "<<t<<" "<<dp[tr][1]<<" "<<dp[tr][0]<<endl; }

Approach 2 –  Static Approach

Time Complexity : O(1)

The code to implement this:

#include<bits/stdc++.h>
using namespace std;
int main() { int n; cin>>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; cout<<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”


  • Maths

    num=int (input())
    five=int((num-4)/5)
    if((num-5*five)%2)==0:
    one=2
    else:
    one=1
    two=(num-5*five-one)//2
    if(five==0):
    print(five+1,two,one)
    else:
    print(five,two,one)