Generate a combination of minimum coins that sums to a given value

Given an array arr[] of size N representing the available denominations and an integer X. The task is to find any combination of the minimum number of coins of the available denominations such that the sum of the coins is X. If the given sum cannot be obtained by the available denominations, print -1.

Examples:

Input: X = 21, arr[] = {2, 3, 4, 5}
Output: 2 4 5 5 5
Explanation:
One possible solution is {2, 4, 5, 5, 5} where 2 + 4 + 5 + 5 + 5 = 21. 
Another possible solution is {3, 3, 5, 5, 5}.

Input: X = 1, arr[] = {2, 4, 6, 9}
Output: -1
Explanation:
All coins are greater than 1. Hence, no solution exist.

Naive Approach: The simplest approach is to try all possible combinations of given denominations such that in each combination, the sum of coins is equal to X. From these combinations, choose the one having the minimum number of coins and print it. If the sum any combinations is not equal to X, print -1
Time Complexity: O(XN)
Auxiliary Space: O(N)

Efficient Approach: The above approach can be optimized using Dynamic Programming to find the minimum number of coins. While finding the minimum number of coins, backtracking can be used to track the coins needed to make their sum equals to X. Follow the below steps to solve the problem:



  1. Initialize an auxiliary array dp[], where dp[i] will store the minimum number of coins needed to make sum equals to i.
  2. Find the minimum number of coins needed to make their sum equals to X using the approach discussed in this article.
  3. After finding the minimum number of coins use the Backtracking Technique to track down the coins used, to make the sum equals to X.
  4. In backtracking, traverse the array and choose a coin which is smaller than the current sum such that dp[current_sum] equals to dp[current_sum – chosen_coin]+1. Store the chosen coin in an array.
  5. After completing the above step, backtrack again by passing the current sum as (current sum – chosen coin value).
  6. After finding the solution, print the array of chosen coins.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
#define MAX 100000
 
// dp array to memoize the results
int dp[MAX + 1];
 
// List to store the result
list<int> denomination;
 
// Function to find the minimum number of
// coins to make the sum equals to X
int countMinCoins(int n, int C[], int m)
{
    // Base case
    if (n == 0) {
        dp[0] = 0;
        return 0;
    }
 
    // If previously computed
    // subproblem occurred
    if (dp[n] != -1)
        return dp[n];
 
    // Initialize result
    int ret = INT_MAX;
 
    // Try every coin that has smaller
    // value than n
    for (int i = 0; i < m; i++) {
 
        if (C[i] <= n) {
 
            int x
                = countMinCoins(n - C[i],
                                C, m);
 
            // Check for INT_MAX to avoid
            // overflow and see if result
            // can be minimized
            if (x != INT_MAX)
                ret = min(ret, 1 + x);
        }
    }
 
    // Memoizing value of current state
    dp[n] = ret;
    return ret;
}
 
// Function to find the possible
// combination of coins to make
// the sum equal to X
void findSolution(int n, int C[], int m)
{
    // Base Case
    if (n == 0) {
 
        // Print Solutions
        for (auto it : denomination) {
            cout << it << ' ';
        }
 
        return;
    }
 
    for (int i = 0; i < m; i++) {
 
        // Try every coin that has
        // value smaller than n
        if (n - C[i] >= 0
            and dp[n - C[i]] + 1
                    == dp[n]) {
 
            // Add current denominations
            denomination.push_back(C[i]);
 
            // Backtrack
            findSolution(n - C[i], C, m);
            break;
        }
    }
}
 
// Function to find the minimum
// combinations of coins for value X
void countMinCoinsUtil(int X, int C[],
                       int N)
{
 
    // Initialize dp with -1
    memset(dp, -1, sizeof(dp));
 
    // Min coins
    int isPossible
        = countMinCoins(X, C,
                        N);
 
    // If no solution exists
    if (isPossible == INT_MAX) {
        cout << "-1";
    }
 
    // Backtrack to find the solution
    else {
        findSolution(X, C, N);
    }
}
 
// Driver code
int main()
{
    int X = 21;
 
    // Set of possible denominations
    int arr[] = { 2, 3, 4, 5 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    countMinCoinsUtil(X, arr, N);
 
    return 0;
}

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Java

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// Java program for
// the above approach
import java.util.*;
class GFG{
   
static final int MAX = 100000;
 
// dp array to memoize the results
static int []dp = new int[MAX + 1];
 
// List to store the result
static List<Integer> denomination =
            new LinkedList<Integer>();
 
// Function to find the minimum
// number of coins to make the
// sum equals to X
static int countMinCoins(int n,
                         int C[], int m)
{
  // Base case
  if (n == 0)
  {
    dp[0] = 0;
    return 0;
  }
 
  // If previously computed
  // subproblem occurred
  if (dp[n] != -1)
    return dp[n];
 
  // Initialize result
  int ret = Integer.MAX_VALUE;
 
  // Try every coin that has smaller
  // value than n
  for (int i = 0; i < m; i++)
  {
    if (C[i] <= n)
    {
      int x = countMinCoins(n - C[i],
                            C, m);
 
      // Check for Integer.MAX_VALUE to avoid
      // overflow and see if result
      // can be minimized
      if (x != Integer.MAX_VALUE)
        ret = Math.min(ret, 1 + x);
    }
  }
 
  // Memoizing value of current state
  dp[n] = ret;
  return ret;
}
 
// Function to find the possible
// combination of coins to make
// the sum equal to X
static void findSolution(int n,
                         int C[], int m)
{
  // Base Case
  if (n == 0)
  {
    // Print Solutions
    for (int it : denomination)
    {
      System.out.print(it + " ");
    }
    return;
  }
 
  for (int i = 0; i < m; i++)
  {
    // Try every coin that has
    // value smaller than n
    if (n - C[i] >= 0 &&
        dp[n - C[i]] + 1 ==
        dp[n])
    {
      // Add current denominations
      denomination.add(C[i]);
 
      // Backtrack
      findSolution(n - C[i], C, m);
      break;
    }
  }
}
 
// Function to find the minimum
// combinations of coins for value X
static void countMinCoinsUtil(int X,
                              int C[], int N)
{
  // Initialize dp with -1
  for (int i = 0; i < dp.length; i++)
    dp[i] = -1;
 
  // Min coins
  int isPossible = countMinCoins(X, C, N);
 
  // If no solution exists
  if (isPossible == Integer.MAX_VALUE)
  {
    System.out.print("-1");
  }
 
  // Backtrack to find the solution
  else
  {
    findSolution(X, C, N);
  }
}
 
// Driver code
public static void main(String[] args)
{
  int X = 21;
 
  // Set of possible denominations
  int arr[] = {2, 3, 4, 5};
 
  int N = arr.length;
 
  // Function Call
  countMinCoinsUtil(X, arr, N);
}
}
 
// This code is contributed by Rajput-Ji

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Python3

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# Python3 program for the above approach
import sys
 
MAX = 100000
 
# dp array to memoize the results
dp = [-1] * (MAX + 1)
 
# List to store the result
denomination = []
 
# Function to find the minimum number of
# coins to make the sum equals to X
def countMinCoins(n, C, m):
     
    # Base case
    if (n == 0):
        dp[0] = 0
        return 0
 
    # If previously computed
    # subproblem occurred
    if (dp[n] != -1):
        return dp[n]
 
    # Initialize result
    ret = sys.maxsize
 
    # Try every coin that has smaller
    # value than n
    for i in range(m):
        if (C[i] <= n):
            x = countMinCoins(n - C[i], C, m)
 
            # Check for INT_MAX to avoid
            # overflow and see if result
            #. an be minimized
            if (x != sys.maxsize):
                ret = min(ret, 1 + x)
 
    # Memoizing value of current state
    dp[n] = ret
    return ret
 
# Function to find the possible
# combination of coins to make
# the sum equal to X
def findSolution(n, C, m):
     
    # Base Case
    if (n == 0):
 
        # Print Solutions
        for it in denomination:
            print(it, end = " ")
 
        return
 
    for i in range(m):
 
        # Try every coin that has
        # value smaller than n
        if (n - C[i] >= 0 and
         dp[n - C[i]] + 1 == dp[n]):
 
            # Add current denominations
            denomination.append(C[i])
 
            # Backtrack
            findSolution(n - C[i], C, m)
            break
 
# Function to find the minimum
# combinations of coins for value X
def countMinCoinsUtil(X, C,N):
 
    # Initialize dp with -1
    # memset(dp, -1, sizeof(dp))
 
    # Min coins
    isPossible = countMinCoins(X, C,N)
 
    # If no solution exists
    if (isPossible == sys.maxsize):
        print("-1")
 
    # Backtrack to find the solution
    else:
        findSolution(X, C, N)
 
# Driver code
if __name__ == '__main__':
     
    X = 21
 
    # Set of possible denominations
    arr = [ 2, 3, 4, 5 ]
 
    N = len(arr)
 
    # Function call
    countMinCoinsUtil(X, arr, N)
 
# This code is contributed by mohit kumar 29

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C#

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// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
   
static readonly int MAX = 100000;
 
// dp array to memoize the results
static int []dp = new int[MAX + 1];
 
// List to store the result
static List<int> denomination =
            new List<int>();
 
// Function to find the minimum
// number of coins to make the
// sum equals to X
static int countMinCoins(int n,
                         int []C,
                         int m)
{
  // Base case
  if (n == 0)
  {
    dp[0] = 0;
    return 0;
  }
 
  // If previously computed
  // subproblem occurred
  if (dp[n] != -1)
    return dp[n];
 
  // Initialize result
  int ret = int.MaxValue;
 
  // Try every coin that has smaller
  // value than n
  for (int i = 0; i < m; i++)
  {
    if (C[i] <= n)
    {
      int x = countMinCoins(n - C[i],
                            C, m);
 
      // Check for int.MaxValue to avoid
      // overflow and see if result
      // can be minimized
      if (x != int.MaxValue)
        ret = Math.Min(ret, 1 + x);
    }
  }
 
  // Memoizing value of current state
  dp[n] = ret;
  return ret;
}
 
// Function to find the possible
// combination of coins to make
// the sum equal to X
static void findSolution(int n,
                         int []C,
                         int m)
{
  // Base Case
  if (n == 0)
  {
    // Print Solutions
    foreach (int it in denomination)
    {
      Console.Write(it + " ");
    }
    return;
  }
 
  for (int i = 0; i < m; i++)
  {
    // Try every coin that has
    // value smaller than n
    if (n - C[i] >= 0 &&
        dp[n - C[i]] + 1 ==
        dp[n])
    {
      // Add current denominations
      denomination.Add(C[i]);
 
      // Backtrack
      findSolution(n - C[i], C, m);
      break;
    }
  }
}
 
// Function to find the minimum
// combinations of coins for value X
static void countMinCoinsUtil(int X,
                              int []C,
                              int N)
{
  // Initialize dp with -1
  for (int i = 0; i < dp.Length; i++)
    dp[i] = -1;
 
  // Min coins
  int isPossible = countMinCoins(X, C, N);
 
  // If no solution exists
  if (isPossible == int.MaxValue)
  {
    Console.Write("-1");
  }
 
  // Backtrack to find the solution
  else
  {
    findSolution(X, C, N);
  }
}
 
// Driver code
public static void Main(String[] args)
{
  int X = 21;
 
  // Set of possible denominations
  int []arr = {2, 3, 4, 5};
 
  int N = arr.Length;
 
  // Function Call
  countMinCoinsUtil(X, arr, N);
}
}
 
// This code is contributed by shikhasingrajput

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Output: 

2 4 5 5 5





 

Time Complexity: O(N*X), where N is the length of the given array and X is the given integer.
Auxiliary Space: O(N)

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