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Commit 1509b3d

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Merge pull request #242 from nandikeshsingh/master
Coin change problem
2 parents 4505b18 + df3c086 commit 1509b3d

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Problem :-
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===
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Given a value N and the values of different denomination,find the number of ways to make changes for N cents,if we have infinite supply of all denominations.
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Input:-
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---
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Size of an array and all the values in the array and the number of cents.
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Output :-
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---
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An integer that denotes the number of ways to make change.
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#### Language : `C++`
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#### Algorithm Paradigm : `Dynamic Programming`
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#### Time Complexity : `O(N*M)`
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#### Space Complexity : `O(N*M)`
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Working :-
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---
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An arr array stores the different currency values.
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Another array temp stores the best values for sub-problems.For eg: `temp[3][4]` stores the optimal value for the number of ways
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to make change if the number of cents(N) were 4 and we only had `arr[0]`,`arr[1]` and `arr[2]` as the different values of currency.
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By using dynamic programming we are bringing the time complexity down from exponentional(in case of brute force solution)
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to polynomial.
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//Coin change problem using dynamic programming in C++.
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#include<bits/stdc++.h>
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using namespace std;
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int main()
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{
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int M;//Variable to store the number of different currency values.
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cout<<"Enter the number of denominations : ";
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cin>>M;//Inputting the number of different currency value from user.
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cout<<endl;
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int arr[M];//Array to store different currency values.
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for(int i=0;i<M;i++)
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{
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cout<<"Enter the value of denominaion "<<i+1<<" : ";
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cin>>arr[i];//Inputting the value of each currency from user.
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}
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cout<<endl;
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int N;//Variable to store the number of cents whose number of ways to make change is to be found.
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cout<<"Enter the number of cents : ";
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cin>>N;//Inputting the number of cents from user.
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cout<<endl;
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/*2D array to store the optimal value for sub-cases.*/
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int temp[M+1][N+1];
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for(int i=0;i<M+1;i++)//Implementing the algorithm.
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{
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for(int j=0;j<N+1;j++)
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{
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if(i==0 || j==0)
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temp[i][j]=0;
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else if(j<arr[i-1])
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temp[i][j]=temp[i-1][j];
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else if(j==arr[i-1])
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temp[i][j]=1+temp[i-1][j];
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else
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temp[i][j]=temp[i-1][j]+temp[i][j-arr[i-1]];
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}
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}
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cout<<"Number of ways to make change for "<<N<<" cent is "<<temp[M][N]<<endl;//Printing the final answer.
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return 0;
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}

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