Given an infinite supply of 25 cents, 10 cents, 5 cents, and 1 cent. Find the number of ways of representing n cents. (The order doesn’t matter). This is a coin change problem and will require the implementation of Dynamic Programming.

Ex: n = 10 => {1,1,1,1,1,1,1,1,1,1} {1,1,1,1,1,5} {5,5} {10}

### Algorithm:

We are given a set of 4 Coins of type 1 cents, 5 cents, 10 cents, 25 cents. To find the number of ways of making n cents using these 4 cents, we will consider 2 conditions:

- Try to make n cents by including the i
*th*cent from the set of 4 coins. number_of_ways(n-coin_arr[m], coin_arr m);*here m in the number of coins in given set(here m=4).* - Try to make n cents by not including the i
*th*cent from the given set of the 4 coins. number_of_ways(n, coin_arr, m-1);

This problem involves the repetition of subproblems, so we will use Dynamic Programming.

#### Implementation of the above Algorithm

1. **Iterative Approach**

#include <bits/stdc++.h> using namespace std; int number_of_ways(int cents, vector<int>coin_arr, int n){ vector<vector<int> >dp(n+1, vector<int>(cents+1,0)); for(int j=0; j<n; j++) dp[j][0] = 1; for(int i=1; i<=cents; i++){ for(int j=0; j<n; j++){ if(i-coin_arr[j]>=0) dp[j][i] += dp[j][i-coin_arr[j]]; if(j>0) dp[j][i] += dp[j-1][i]; } } return dp[n-1][cents]; } int main() { int cents; vector<int> coin_arr{1,5,10,25}; cin>>cents; cout<<number_of_ways(cents, coin_arr, coin_arr.size())<<endl; return 0; }

2.** Recursive Approach**

#include <bits/stdc++.h> using namespace std; int number_of_ways(int cents, vector<int>coin_arr, int n, vector<vector<int> > &dp){ if(cents<0 || n<0) return 0; if(cents == 0) return 1; if(dp[cents][n]==-1) dp[cents][n] = number_of_ways(cents-coin_arr[n], coin_arr, n, dp) +number_of_ways(cents, coin_arr, n-1, dp); return dp[cents][n]; } int main() { int cents; vector<int> coin_arr{1,5,10,25}; cin>>cents; //creating a 2D array for storing subproblems //of size= dp[cents+1][coin_arr.size()+1] vector<vector<int> > dp(cents+1, vector<int>(coin_arr.size()+1,-1)); cout<<number_of_ways(cents, coin_arr, coin_arr.size()-1, dp)<<endl; return 0; }

#### Output

Coins Output n = 5 2 n = 26 13 n = 1000 142511

#### Time Complexity

Since we have used Dynamic Programming here, hence the time complexity of the above program for coin change problem is **O(NM)**, N the total cents to make and M is the number of the coin in a given set.