Given an integer array of coins[ ] of size N representing different types of denominations and an integer sum, the task is to find the number of ways to make sum by using different denominations.
Note: Assume that you have an infinite supply of each type of coin.
Examples:
Input: sum = 4, coins[] = {1,2,3},
Output: 4
Explanation: there are four solutions: {1, 1, 1, 1}, {1, 1, 2}, {2, 2}, {1, 3}.Input: sum = 10, coins[] = {2, 5, 3, 6}
Output: 5
Explanation: There are five solutions:
{2,2,2,2,2}, {2,2,3,3}, {2,2,6}, {2,3,5} and {5,5}.
Coin Change Problem using Recursion:
Coin Change Using Recursion
Recurrence Relation:
count(coins,n,sum) = count(coins,n,sum-count[n-1]) + count(coins,n-1,sum)
For each coin, there are 2 options.
- Include the current coin: Subtract the current coin’s denomination from the target sum and call the count function recursively with the updated sum and the same set of coins i.e., count(coins, n, sum – coins[n-1] )
- Exclude the current coin: Call the count function recursively with the same sum and the remaining coins. i.e., count(coins, n-1,sum ).
The final result will be the sum of both cases.
Base case:
- If the target sum (sum) is 0, there is only one way to make the sum, which is by not selecting any coin. So, count(0, coins, n) = 1.
- If the target sum (sum) is negative or no coins are left to consider (n == 0), then there are no ways to make the sum, so count(sum, coins, 0) = 0.
Below is the Implementation of the above approach.
C++
// Recursive C++ program for// coin change problem.#include <bits/stdc++.h>using namespace std;// Returns the count of ways we can// sum coins[0...n-1] coins to get sum "sum"int count(int coins[], int n, int sum){ // If sum is 0 then there is 1 solution // (do not include any coin) if (sum == 0) return 1; // If sum is less than 0 then no // solution exists if (sum < 0) return 0; // If there are no coins and sum // is greater than 0, then no // solution exist if (n <= 0) return 0; // count is sum of solutions (i) // including coins[n-1] (ii) excluding coins[n-1] return count(coins, n, sum - coins[n - 1]) + count(coins, n - 1, sum);}// Driver codeint main(){ int i, j; int coins[] = { 1, 2, 3 }; int n = sizeof(coins) / sizeof(coins[0]); int sum = 5; cout << " " << count(coins, n, sum); return 0;} |
C
// Recursive C program for// coin change problem.#include <stdio.h>// Returns the count of ways we can// sum coins[0...n-1] coins to get sum "sum"int count(int coins[], int n, int sum){ // If sum is 0 then there is 1 solution // (do not include any coin) if (sum == 0) return 1; // If sum is less than 0 then no // solution exists if (sum < 0) return 0; // If there are no coins and sum // is greater than 0, then no // solution exist if (n <= 0) return 0; // count is sum of solutions (i) // including coins[n-1] (ii) excluding coins[n-1] return count(coins, n - 1, sum) + count(coins, n, sum - coins[n - 1]);}// Driver program to test above functionint main(){ int i, j; int coins[] = { 1, 2, 3 }; int n = sizeof(coins) / sizeof(coins[0]); printf("%d ", count(coins, n, 5)); getchar(); return 0;} |
Java
// Recursive JAVA program for// coin change problem.import java.util.*;class GFG { // Returns the count of ways we can // sum coins[0...n-1] coins to get sum "sum" static int count(int coins[], int n, int sum) { // If sum is 0 then there is 1 solution // (do not include any coin) if (sum == 0) return 1; // If sum is less than 0 then no // solution exists if (sum < 0) return 0; // If there are no coins and sum // is greater than 0, then no // solution exist if (n <= 0) return 0; // count is sum of solutions (i) // including coins[n-1] (ii) excluding coins[n-1] return count(coins, n - 1, sum) + count(coins, n, sum - coins[n - 1]); } // Driver code public static void main(String args[]) { int coins[] = { 1, 2, 3 }; int n = coins.length; System.out.println(count(coins, n, 5)); }}// This code is contributed by jyoti369 |
Python3
# Recursive Python3 program for# coin change problem.# Returns the count of ways we can sum# coins[0...n-1] coins to get sum "sum"def count(coins, n, sum): # If sum is 0 then there is 1 # solution (do not include any coin) if (sum == 0): return 1 # If sum is less than 0 then no # solution exists if (sum < 0): return 0 # If there are no coins and sum # is greater than 0, then no # solution exist if (n <= 0): return 0 # count is sum of solutions (i) # including coins[n-1] (ii) excluding coins[n-1] return count(coins, n - 1, sum) + count(coins, n, sum-coins[n-1])# Driver program to test above functioncoins = [1, 2, 3]n = len(coins)print(count(coins, n, 5))# This code is contributed by Smitha Dinesh Semwal |
C#
// Recursive C# program for// coin change problem.using System;class GFG { // Returns the count of ways we can // sum coins[0...n-1] coins to get sum "sum" static int count(int[] coins, int n, int sum) { // If sum is 0 then there is 1 solution // (do not include any coin) if (sum == 0) return 1; // If sum is less than 0 then no // solution exists if (sum < 0) return 0; // If there are no coins and sum // is greater than 0, then no // solution exist if (n <= 0) return 0; // count is sum of solutions (i) // including coins[n-1] (ii) excluding coins[n-1] return count(coins, n - 1, sum) + count(coins, n, sum - coins[n - 1]); } // Driver program public static void Main() { int[] coins = { 1, 2, 3 }; int n = coins.Length; Console.Write(count(coins, n, 5)); }}// This code is contributed by Sam007 |
Javascript
<script>// Recursive javascript program for// coin change problem. // Returns the count of ways we can // sum coins[0...n-1] coins to get sum "sum"function count(coins , n , sum ){ // If sum is 0 then there is 1 solution // (do not include any coin) if (sum == 0) return 1; // If sum is less than 0 then no // solution exists if (sum < 0) return 0; // If there are no coins and sum // is greater than 0, then no // solution exist if (n <=0) return 0; // count is sum of solutions (i) // including coins[n-1] (ii) excluding coins[n-1] return count( coins, n - 1, sum ) + count( coins, n, sum - coins[n - 1] );}// Driver program to test above functionvar coins = [1, 2, 3];var n = coins.length;document.write( count(coins, n, 5));// This code is contributed by Amit Katiyar </script> |
PHP
<?php// Recursive PHP program for// coin change problem.// Returns the count of ways we can // sum coins[0...n-1] coins to get sum "sum"function coun($coins, $n, $sum){ // If sum is 0 then there is // 1 solution (do not include // any coin) if ($sum == 0) return 1; // If sum is less than 0 then no // solution exists if ($sum < 0) return 0; // If there are no coins and sum // is greater than 0, then no // solution exist if ($n <= 0) return 0; // count is sum of solutions (i) // including coins[n-1] (ii) excluding coins[n-1] return coun($coins, $n - 1,$sum ) + coun($coins, $n, $sum - $coins[$n - 1] );} // Driver Code $coins = array(1, 2, 3); $n = count($coins); echo coun($coins, $n, 5); // This code is contributed by Sam007?> |
Output
5
Time Complexity: O(2sum)
Auxiliary Space: O(sum)
Coin Change Problem using Dynamic Programming (Memoization) :
The above recursive solution has Optimal Substructure and Overlapping Subproblems so Dynamic programming (Memoization) can be used to solve the problem. So 2D array can be used to store results of previously solved subproblems.
Follow the below steps to Implement the idea:
- Create a 2D dp array to store the results of previously solved subproblems.
- dp[i][j] will represent the number of distinct ways to make the sum j by using the first i coins.
- During the recursion call, if the same state is called more than once, then we can directly return the answer stored for that state instead of calculating again.
Below is the implementation using the Memoization:
C++
#include <bits/stdc++.h>using namespace std;// Recursive function to count the numeber of distinct ways// to make the sum by using n coinsint count(vector<int>& coins, int n, int sum, vector<vector<int> >& dp){ // Base Case if (sum == 0) return dp[n][sum] = 1; // If number of coins is 0 or sum is less than 0 then // there is no way to make the sum. if (n == 0 || sum < 0) return 0; // If the subproblem is previously calculated then // simply return the result if (dp[n][sum] != -1) return dp[n][sum]; // Two options for the current coin return dp[n][sum] = count(coins, n, sum - coins[n - 1], dp) + count(coins, n - 1, sum, dp);}int32_t main(){ int tc = 1; // cin >> tc; while (tc--) { int n, sum; n = 3, sum = 5; vector<int> coins = { 1, 2, 3 }; // 2d dp array to store previously calculated // results vector<vector<int> > dp(n + 1, vector<int>(sum + 1, -1)); int res = count(coins, n, sum, dp); cout << res << endl; }} |
Java
// Java program for the above approachimport java.util.*;class GFG { // Recursive function to count the numeber of distinct // ways to make the sum by using n coins static int count(int[] coins, int sum, int n, int[][] dp) { // Base Case if (sum == 0) return dp[n][sum] = 1; // If number of coins is 0 or sum is less than 0 then // there is no way to make the sum. if (n == 0 || sum<0) return 0; // If the subproblem is previously calculated then // simply return the result if (dp[n][sum] != -1) return dp[n][sum]; // Two options for the current coin return dp[n][sum] = count(coins, sum - coins[n - 1], n, dp) + count(coins, sum, n - 1, dp); } // Driver code public static void main(String[] args) { int tc = 1; while (tc != 0) { int n, sum; n = 3; sum = 5; int[] coins = { 1, 2, 3 }; int[][] dp = new int[n + 1][sum + 1]; for (int[] row : dp) Arrays.fill(row, -1); int res = count(coins, sum, n, dp); System.out.println(res); tc--; } }} |
Python3
# Python program for the above approach# Recursive function to count the numeber of distinct ways# to make the sum by using n coinsdef count(coins, sum, n, dp): # Base Case if (sum == 0): dp[n][sum] = 1 return dp[n][sum] # If number of coins is 0 or sum is less than 0 then there is no way to make the sum. if (n == 0 or sum < 0): return 0 # If the subproblem is previously calculated then simply return the result if (dp[n][sum] != -1): return dp[n][sum] # Two options for the current coin dp[n][sum] = count(coins, sum - coins[n - 1], n, dp) + \ count(coins, sum, n - 1, dp) return dp[n][sum]# Driver codeif __name__ == '__main__': tc = 1 while (tc != 0): n = 3 sum = 5 coins = [1, 2, 3] dp = [[-1 for i in range(sum+1)] for j in range(n+1)] res = count(coins, sum, n, dp) print(res) tc -= 1 |
C#
// C# program for the above approachusing System;public class GFG { // Recursive function to count the numeber of distinct // ways to make the sum by using n coins static int count(int[] coins, int sum, int n, int[, ] dp) { // Base Case if (sum == 0) return dp[n, sum] = 1; // If number of coins is 0 or sum is less than 0 then // there is no way to make the sum. if (n == 0 || sum < 0) return 0; // If the subproblem is previously calculated then // simply return the result if (dp[n, sum] != -1) return dp[n, sum]; // Two options for the current coin return dp[n, sum] = count(coins, sum - coins[n - 1], n, dp) + count(coins, sum, n - 1, dp); } // Driver code public static void Main(String[] args) { int tc = 1; while (tc != 0) { int n, sum; n = 3; sum = 5; int[] coins = { 1, 2, 3 }; int[, ] dp = new int[n + 1, sum + 1]; for (int j = 0; j < n + 1; j++) { for (int l = 0; l < sum + 1; l++) dp[j, l] = -1; } int res = count(coins, sum, n, dp); Console.WriteLine(res); tc--; } }} |
Javascript
// javascript program for the above approach // Recursive function to count the numeber of distinct ways // to make the sum by using n coins function count(coins , sum , n, dp) { // Base Case if (sum == 0) return dp[n][sum] = 1; // If number of coins is 0 or sum is less than 0 then // there is no way to make the sum. if (n == 0 || sum<0) return 0; // If the subproblem is previously calculated then // simply return the result if (dp[n][sum] != -1) return dp[n][sum]; // Two options for the current coin return dp[n][sum] = count(coins, sum - coins[n - 1], n, dp) + count(coins, sum, n - 1, dp); } // Driver code var tc = 1; while (tc != 0) { var n, sum; n = 3; sum = 5; var coins = [ 1, 2, 3 ]; var dp = Array(n+1).fill().map(() => Array(sum+1).fill(-1)); var res = count(coins, sum, n, dp); console.log(res); tc--; } |
Output
5
Time Complexity: O(N*sum), where N is the number of coins and sum is the target sum.
Auxiliary Space: O(N*sum)
Coin Change Problem using Dynamic Programming (Tabulation):
We can use the following steps to implement the dynamic programming(tabulation) approach for Coin Change.
- Create a 2D dp array with rows and columns equal to the number of coin denominations and target sum.
dp[0][0]will be set to1which represents the base case where the target sum is 0, and there is only one way to make the change by not selecting any coin.- Iterate through the rows of the dp array (i from 1 to
n), representing the current coin being considered.- The inner loop iterates over the target sums (
jfrom 0 tosum).- Add the number of ways to make change without using the current coin, i.e.,
dp[i][j] += dp[i-1][j]. - Add the number of ways to make change using the current coin, i.e.,
dp[i][j] += dp[i][j-coins[i-1]].
- Add the number of ways to make change without using the current coin, i.e.,
- The inner loop iterates over the target sums (
dp[n][sum]will contain the total number of ways to make change for the given target sum using the available coin denominations.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>using namespace std;// Returns total distinct ways to make sum using n coins of// different denominationsint count(vector<int>& coins, int n, int sum){ // 2d dp array where n is the number of coin // denominations and sum is the target sum vector<vector<int> > dp(n + 1, vector<int>(sum + 1, 0)); // Represents the base case where the target sum is 0, // and there is only one way to make change: by not // selecting any coin dp[0][0] = 1; for (int i = 1; i <= n; i++) { for (int j = 0; j <= sum; j++) { // Add the number of ways to make change without // using the current coin, dp[i][j] += dp[i - 1][j]; if ((j - coins[i - 1]) >= 0) { // Add the number of ways to make change // using the current coin dp[i][j] += dp[i][j - coins[i - 1]]; } } } return dp[n][sum];}// Driver Codeint main(){ vector<int> coins{ 1, 2, 3 }; int n = 3; int sum = 5; cout << count(coins, n, sum); return 0;} |
Java
import java.util.*;public class CoinChangeWays { // Returns total distinct ways to make sum using n coins of // different denominations static int count(List<Integer> coins, int n, int sum) { // 2D dp array where n is the number of coin // denominations and sum is the target sum int[][] dp = new int[n + 1][sum + 1]; // Represents the base case where the target sum is 0, // and there is only one way to make change: by not // selecting any coin dp[0][0] = 1; for (int i = 1; i <= n; i++) { for (int j = 0; j <= sum; j++) { // Add the number of ways to make change without // using the current coin dp[i][j] += dp[i - 1][j]; if ((j - coins.get(i - 1)) >= 0) { // Add the number of ways to make change // using the current coin dp[i][j] += dp[i][j - coins.get(i - 1)]; } } } return dp[n][sum]; } // Driver Code public static void main(String[] args) { List<Integer> coins = Arrays.asList(1, 2, 3); int n = 3; int sum = 5; System.out.println(count(coins, n, sum)); }}// This code is contributed by Veerendra_Singh_Rajpoot |
Python3
# Function to calculate the total distinct ways to make a sum using n coins of different denominationsdef count(coins, n, target_sum): # 2D dp array where n is the number of coin denominations and target_sum is the target sum dp = [[0 for j in range(target_sum + 1)] for i in range(n + 1)] # Represents the base case where the target sum is 0, and there is only one way to make change: by not selecting any coin dp[0][0] = 1 for i in range(1, n + 1): for j in range(target_sum + 1): # Add the number of ways to make change without using the current coin dp[i][j] += dp[i - 1][j] if j - coins[i - 1] >= 0: # Add the number of ways to make change using the current coin dp[i][j] += dp[i][j - coins[i - 1]] return dp[n][target_sum]# Driver Codeif __name__ == "__main__": coins = [1, 2, 3] n = 3 target_sum = 5 print(count(coins, n, target_sum)) |
C#
using System;using System.Collections.Generic;class Program { // Returns total distinct ways to make sum using n coins // of different denominations static int Count(List<int> coins, int n, int sum) { // 2d dp array where n is the number of coin // denominations and sum is the target sum int[, ] dp = new int[n + 1, sum + 1]; // Represents the base case where the target sum is // 0, and there is only one way to make change: by // not selecting any coin dp[0, 0] = 1; for (int i = 1; i <= n; i++) { for (int j = 0; j <= sum; j++) { // Add the number of ways to make change // without using the current coin dp[i, j] += dp[i - 1, j]; if ((j - coins[i - 1]) >= 0) { // Add the number of ways to make change // using the current coin dp[i, j] += dp[i, j - coins[i - 1]]; } } } return dp[n, sum]; } // Driver Code static void Main(string[] args) { List<int> coins = new List<int>{ 1, 2, 3 }; int n = 3; int sum = 5; Console.WriteLine(Count(coins, n, sum)); }} |
Output
5
Time complexity : O(N*sum)
Auxiliary Space : O(N*sum)
Coin change Problem using the Space Optimized 1D array:
In the above tabulation approach we are only using dp[i-1][j] and dp[i][j] etc, so we can do space optimization by only using a 1d dp array.
Follow the below steps to Implement the idea:
- Create a 1D dp array,
dp[i]represents the number of ways to make the sumiusing the given coin denominations. - The outer loop iterates over the coins, and the inner loop iterates over the target sums. For each
dp[j], it calculates the number of ways to make change using the current coin denomination and the previous results stored indp. dp[sum]contains the total number of ways to make change for the given target sum using the available coin denominations. This approach optimizes space by using a 1D array instead of a 2D DP table.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>using namespace std;// This code is int count(int coins[], int n, int sum){ // table[i] will be storing the number of solutions for // value i. We need sum+1 rows as the dp is // constructed in bottom up manner using the base case // (sum = 0) int dp[sum + 1]; // Initialize all table values as 0 memset(dp, 0, sizeof(dp)); // Base case (If given value is 0) dp[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 < n; i++) for (int j = coins[i]; j <= sum; j++) dp[j] += dp[j - coins[i]]; return dp[sum];}// Driver Codeint main(){ int coins[] = { 1, 2, 3 }; int n = sizeof(coins) / sizeof(coins[0]); int sum = 5; cout << count(coins, n, sum); return 0;} |
Java
/* Dynamic Programming Java implementation of CoinChange problem */import java.util.Arrays;class CoinChange { static long count(int coins[], int n, int sum){ // dp[i] will be storing the number of solutions for // value i. We need sum+1 rows as the dp is // constructed in bottom up manner using the base case // (sum = 0) int dp[] = new int[sum + 1]; // Base case (If given value is 0) dp[0] = 1; // Pick all coins one by one and update the dp[] // values after the index greater than or equal to the // value of the picked coin for (int i = 0; i < n; i++) for (int j = coins[i]; j <= sum; j++) dp[j] += dp[j - coins[i]]; return dp[sum];} // Driver Function to test above function public static void main(String args[]) { int coins[] = { 1, 2, 3 }; int n = coins.length; int sum = 5; System.out.println(count(coins, n, sum)); }}// This code is contributed by Pankaj Kumar |
Python
# Dynamic Programming Python implementation of Coin# Change problemdef count(coins, n, sum): # dp[i] will be storing the number of solutions for # value i. We need sum+1 rows as the dp is constructed # in bottom up manner using the base case (sum = 0) # Initialize all table values as 0 dp = [0 for k in range(sum+1)] # Base case (If given value is 0) dp[0] = 1 # Pick all coins one by one and update the dp[] values # after the index greater than or equal to the value of the # picked coin for i in range(0, n): for j in range(coins[i], sum+1): dp[j] += dp[j-coins[i]] return dp[sum]# Driver program to test above functioncoins = [1, 2, 3]n = len(coins)sum = 5x = count(coins, n, sum)print(x)# This code is contributed by Afzal Ansari |
C#
// Dynamic Programming C# implementation// of Coin Change problemusing System;class GFG { static int count(int[] coins, int n, int sum) { // dp[i] will be storing the // number of solutions for value i. // We need sum+1 rows as the dp // is constructed in bottom up manner // using the base case (sum = 0) int[] dp = new int[sum + 1]; // Base case (If given value is 0) dp[0] = 1; // Pick all coins one by one and // update the dp[] values after // the index greater than or equal // to the value of the picked coin for (int i = 0; i < n; i++) for (int j = coins[i]; j <= sum; j++) dp[j] += dp[j - coins[i]]; return dp[sum]; } // Driver Code public static void Main() { int[] coins = { 1, 2, 3 }; int n = coins.Length; int sum = 5; Console.Write(count(coins, n, sum)); }}// This code is contributed by Raj |
Javascript
<script> // Dynamic Programming Javascript implementation // of Coin Change problem function count(coins, n, sum) { // dp[i] will be storing the // number of solutions for value i. // We need n+1 rows as the dp // is constructed in bottom up manner // using the base case (sum = 0) let dp = new Array(sum + 1); dp.fill(0); // Base case (If given value is 0) dp[0] = 1; // Pick all coins one by one and // update the dp[] values after // the index greater than or equal // to the value of the picked coin for(let i = 0; i < n; i++) for(let j = coins[i]; j <= sum; j++) dp[j] += dp[j - coins[i]]; return dp[sum]; } let coins = [1, 2, 3]; let n = coins.length; let sum = 4; document.write(count(coins, n, sum));</script> |
PHP
<?phpfunction count_1( &$coins, $n, $sum ){ // table[i] will be storing the number // of solutions for value i. We need sum+1 // rows as the table is constructed in // bottom up manner using the base case (sum = 0) $table = array_fill(0, $sum + 1, NULl); // Base case (If given value is 0) $table[0] = 1; // Pick all coins one by one and update // the table[] values after the index // greater than or equal to the value // of the picked coin for($i = 0; $i < $n; $i++) for($j = $coins[$i]; $j <= $sum; $j++) $table[$j] += $table[$j - $coins[$i]]; return $table[$sum];}// Driver Code$coins = array(1, 2, 3);$n = sizeof($coins);$sum = 4;$x = count_1($coins, $n, $sum);echo $x;// This code is contributed// by ChitraNayal?> |
Output
5
Time complexity : O(N*sum)
Auxiliary Space : O(sum)
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