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Count number of coins required to make a given value (Coin Change II)

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Given an integer array of coins[ ] of sizerepresenting different types of denominations and an integer sum, the task is to count the number of coins required to make a given value sum.  

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}.

Recommended Practice

Count number of coins required to make a given value using Recursion:

Coin-Change-With-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 code
int 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 function
int 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 function
coins = [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 function
var 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)

Count number of coins required to make a given value 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 coins
 
int 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 approach
import 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 coins
 
 
def 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 code
if __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 approach
using 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)

Count number of coins required to make a given value 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 to 1 which 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 (j from 0 to sum).
      • 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]].
  • 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 denominations
int 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 Code
int 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 denominations
def 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 Code
if __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));
    }
}

                    

Javascript

function count(coins, n, sum) {
    // Create a 2D array dp where n is the number of coin denominations
    // and sum is the target sum
    let dp = new Array(n + 1).fill(0).map(() => new Array(sum + 1).fill(0));
 
    // Set 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 (let i = 1; i <= n; i++) {
        for (let 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
let coins = [1, 2, 3];
let n = 3;
let sum = 5;
console.log(count(coins, n, sum));

                    

Output
5

Time complexity : O(N*sum)
Auxiliary Space : O(N*sum)

Count number of coins required to make a given value using Dynamic Programming (Space Optimized):

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 sum i using 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 in dp.
  • 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 Code
int 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 Coin
Change 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 problem
 
 
def 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 function
coins = [1, 2, 3]
n = len(coins)
sum = 5
x = count(coins, n, sum)
print(x)
 
# This code is contributed by Afzal Ansari

                    

C#

// Dynamic Programming C# implementation
// of Coin Change problem
using 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

<?php
function 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)



Last Updated : 20 Oct, 2023
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