Coin change(DP) in C++

Author: raunakpandey1

Dynamic Programming/Coin_Change.cpp

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+//Link : https://practice.geeksforgeeks.org/problems/coin-change2448/1
+
+//Application of Unbounded Knapsack
+// Coin change (1. Total no of ways)
+//Note : This problem is similar to count of subset sum but only the difference is that the repetition of same elements are allowed.
+
+// Given a value N, find the number of ways to make change for N cents, if we have infinite supply of each of S = { S1, S2, .. , SM } valued coins.
+
+
+// Example 1:
+
+// Input:
+// n = 4 , m = 3
+// S[] = {1,2,3}
+// Output: 4
+// Explanation: Four Possible ways are:
+// {1,1,1,1},{1,1,2},{2,2},{1,3}.
+// Example 2:
+
+// Input:
+// n = 10 , m = 4
+// S[] ={2,5,3,6}
+// Output: 5
+// Explanation: Five Possible ways are:
+// {2,2,2,2,2}, {2,2,3,3}, {2,2,6}, {2,3,5} 
+// and {5,5}.
+
+// Your Task:
+// You don't need to read input or print anything. Your task is to complete the function count() which accepts an array S[] its size m and n as input parameter and returns the number of ways to make change for N cents.
+
+
+// Expected Time Complexity: O(m*n).
+// Expected Auxiliary Space: O(n).
+
+
+// Constraints:
+// 1 <= n,m <= 103
+
+#include<bits/stdc++.h>
+using namespace std;
+
+ // } Driver Code Ends
+class Solution
+{
+  public:
+    long long int count( int arr[], int m, int n )
+    {
+       
+        //code here.
+        long long int dp[m+1][n+1];
+        
+        for(int i = 0; i <= m; i++) dp[i][0] = 1;
+        for(int i = 0; i <= n; i++)  dp[0][i] = 0;
+        
+        for(int i = 1; i <= m; i++){
+            for(int j = 1; j <= n; j++){
+                if(arr[i-1] <= j){
+                    dp[i][j] = dp[i][j - arr[i - 1]] + dp[i-1][j];
+                }else if(arr[i-1] > j){
+                    dp[i][j] = dp[i-1][j];
+                }
+            }
+        }
+        
+        return dp[m][n];
+    }
+};
+
+// { Driver Code Starts.
+int main()
+{
+    int t;
+    cin>>t;
+	while(t--)
+	{
+		int n,m;
+		cin>>n>>m;
+		int arr[m];
+		for(int i=0;i<m;i++)
+		    cin>>arr[i];
+	    Solution ob;
+		cout<<ob.count(arr,m,n)<<endl;
+	}
+    
+    
+    return 0;
+}  // } Driver Code Ends