Java Program for Coin Change | DP-7
Given a value N, if we want to make change for N cents, and we have infinite supply of each of S = { S1, S2, .. , Sm} valued coins, how many ways can we make the change? The order of coins doesn\’t matter.
For example, for N = 4 and S = {1,2,3}, there are four solutions: {1,1,1,1},{1,1,2},{2,2},{1,3}. So output should be 4. For N = 10 and S = {2, 5, 3, 6}, there are five solutions: {2,2,2,2,2}, {2,2,3,3}, {2,2,6}, {2,3,5} and {5,5}. So the output should be 5.
Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.
Following is a simple recursive implementation of the Coin Change problem.
Java
// Recursive java program for// coin change problem.import java.io.*; class GFG { // Returns the count of ways we can // sum S[0...m-1] coins to get sum n static int count( int S[], int m, int n ) { // If n is 0 then there is 1 solution // (do not include any coin) if (n == 0) return 1; // If n is less than 0 then no // solution exists if (n < 0) return 0; // If there are no coins and n // is greater than 0, then no // solution exist if (m <=0 && n >= 1) return 0; // count is sum of solutions (i) // including S[m-1] (ii) excluding S[m-1] return count( S, m - 1, n ) + count( S, m, n-S[m-1] ); } // Driver program to test above function public static void main(String[] args) { int arr[] = {1, 2, 3}; int m = arr.length; System.out.println( count(arr, m, 4)); } } // This article is contributed by vt_m. |
Java
/* Dynamic Programming Java implementation of Coin Change problem */import java.util.Arrays; class CoinChange{ static long countWays(int S[], int m, int n) { //Time complexity of this function: O(mn) //Space Complexity of this function: O(n) // table[i] will be storing the number of solutions // for value i. We need n+1 rows as the table is // constructed in bottom up manner using the base // case (n = 0) long[] table = new long[n+1]; // Initialize all table values as 0 Arrays.fill(table, 0); //O(n) // Base case (If given value is 0) table[0] = 1; // Pick all coins one by one and update the table[] // values after the index greater than or equal to // the value of the picked coin for (int i=0; i<m; i++) for (int j=S[i]; j<=n; j++) table[j] += table[j-S[i]]; return table[n]; } // Driver Function to test above function public static void main(String args[]) { int arr[] = {1, 2, 3}; int m = arr.length; int n = 4; System.out.println(countWays(arr, m, n)); }}// This code is contributed by Pankaj Kumar |
Following is a simplified version of method 2. The auxiliary space required here is O(n) only.
Java
public static int count( int S[], int m, int n ){ // table[i] will be storing the number of solutions for // value i. We need n+1 rows as the table is constructed // in bottom up manner using the base case (n = 0) int table[]=new int[n+1]; // Base case (If given value is 0) table[0] = 1; // Pick all coins one by one and update the table[] values // after the index greater than or equal to the value of the // picked coin for(int i=0; i<m; i++) for(int j=S[i]; j<=n; j++) table[j] += table[j-S[i]]; return table[n];} |
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