Generate a combination of minimum coins that sums to a given value
Last Updated :
15 Jul, 2025
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:
- Initialize an auxiliary array dp[], where dp[i] will store the minimum number of coins needed to make sum equals to i.
- Find the minimum number of coins needed to make their sum equals to X using the approach discussed in this article.
- After finding the minimum number of coins use the Backtracking Technique to track down the coins used, to make the sum equals to X.
- 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.
- After completing the above step, backtrack again by passing the current sum as (current sum - chosen coin value).
- After finding the solution, print the array of chosen coins.
Below is the implementation of the above approach:
C++
// 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;
}
// 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
# 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
# can 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
// 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
<script>
// Javascript program for the above approach
var MAX = 100000;
// dp array to memoize the results
var dp = Array(MAX+1);
// List to store the result
var denomination = [];
// Function to find the minimum number of
// coins to make the sum equals to X
function 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
var ret = 1000000000;
// Try every coin that has smaller
// value than n
for (var i = 0; i < m; i++) {
if (C[i] <= n) {
var x
= countMinCoins(n - C[i],
C, m);
// Check for INT_MAX to avoid
// overflow and see if result
// can be minimized
if (x != 1000000000)
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
function findSolution(n, C, m)
{
// Base Case
if (n == 0) {
denomination.forEach(it => {
document.write( it + ' ');
});
return;
}
for (var 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.push(C[i]);
// Backtrack
findSolution(n - C[i], C, m);
break;
}
}
}
// Function to find the minimum
// combinations of coins for value X
function countMinCoinsUtil(X, C, N)
{
// Initialize dp with -1
dp = Array(MAX+1).fill(-1);
// Min coins
var isPossible
= countMinCoins(X, C,
N);
// If no solution exists
if (isPossible == 1000000000) {
document.write( "-1");
}
// Backtrack to find the solution
else {
findSolution(X, C, N);
}
}
// Driver code
var X = 21;
// Set of possible denominations
var arr = [2, 3, 4, 5];
var N = arr.length;
// Function Call
countMinCoinsUtil(X, arr, N);
</script>
Time Complexity: O(N*X), where N is the length of the given array and X is the given integer.
Auxiliary Space: O(N)
Efficient approach : Using DP Tabulation method ( Iterative approach )
The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.
Steps to solve this problem :
- Create a array DP to store the solution of the subproblems and initialize it with INT_MAX.
- Initialize the array DP with base case i.e. dp[0] = 0 .
- Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP
- Return the final solution stored in dp[X].
Implementation :
C++
// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
#define MAX 100000
// Function to find the minimum number of
// coins to make the sum equals to X
int countMinCoins(int X, int C[], int N)
{
// dp array to store the minimum number of coins for each value from 0 to X
int dp[X + 1];
// Initialize dp array
dp[0] = 0;
for (int i = 1; i <= X; i++) {
dp[i] = INT_MAX;
}
// Loop through all values from 1 to X and compute the minimum number of coins
for (int i = 1; i <= X; i++) {
for (int j = 0; j < N; j++) {
if (C[j] <= i && dp[i - C[j]] != INT_MAX) {
dp[i] = min(dp[i], 1 + dp[i - C[j]]);
}
}
}
// If no solution exists
if (dp[X] == INT_MAX) {
cout << "-1";
return -1;
}
// Print the solution
int i = X;
while (i > 0) {
for (int j = 0; j < N; j++) {
if (i >= C[j] && dp[i - C[j]] == dp[i] - 1) {
cout << C[j] << " ";
i -= C[j];
break;
}
}
}
return dp[X];
}
// 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
countMinCoins(X, arr, N);
return 0;
}
// this code is contributed by bhardwajji
import java.util.*;
public class Main {
// Function to find the minimum number of coins to make the sum equals to X
public static int countMinCoins(int X, int[] C, int N) {
// dp array to store the minimum number of coins for each value from 0 to X
int[] dp = new int[X + 1];
// Initialize dp array
dp[0] = 0;
for (int i = 1; i <= X; i++) {
dp[i] = Integer.MAX_VALUE;
}
// Loop through all values from 1 to X and compute
// the minimum number of coins
for (int i = 1; i <= X; i++) {
for (int j = 0; j < N; j++) {
if (C[j] <= i && dp[i - C[j]] != Integer.MAX_VALUE) {
dp[i] = Math.min(dp[i], 1 + dp[i - C[j]]);
}
}
}
// If no solution exists
if (dp[X] == Integer.MAX_VALUE) {
System.out.println("-1");
return -1;
}
// Print the solution
int i = X;
while (i > 0) {
for (int j = 0; j < N; j++) {
if (i >= C[j] && dp[i - C[j]] == dp[i] - 1) {
System.out.print(C[j] + " ");
i -= C[j];
break;
}
}
}
System.out.println();
return dp[X];
}
// 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
countMinCoins(X, arr, N);
}
}
import sys
# Function to find the minimum number of coins to make the sum equals to X
def countMinCoins(X, C, N):
# dp array to store the minimum number of coins for each value from 0 to X
dp = [sys.maxsize] * (X + 1)
# Initialize dp array
dp[0] = 0
# Loop through all values from 1 to X and compute the minimum number of coins
for i in range(1, X + 1):
for j in range(N):
if C[j] <= i and dp[i - C[j]] != sys.maxsize:
dp[i] = min(dp[i], 1 + dp[i - C[j]])
# If no solution exists
if dp[X] == sys.maxsize:
print("-1")
return -1
# Print the solution
i = X
while i > 0:
for j in range(N):
if i >= C[j] and dp[i - C[j]] == dp[i] - 1:
print(C[j], end=" ")
i -= C[j]
break
return dp[X]
# Driver code
if __name__ == "__main__":
X = 21
# Set of possible denominations
arr = [2, 3, 4, 5]
N = len(arr)
# Function Call
countMinCoins(X, arr, N)
using System;
class Program
{
// Function to find the minimum number of coins to make the sum equals to X
static int CountMinCoins(int X, int[] C, int N)
{
// dp array to store the minimum number of coins for each value from 0 to X
int[] dp = new int[X + 1];
// Initialize dp array
dp[0] = 0;
for (int i = 1; i <= X; i++)
{
dp[i] = int.MaxValue;
}
// Loop through all values from 1 to X and compute the minimum number of coins
for (int i = 1; i <= X; i++)
{
for (int j = 0; j < N; j++)
{
if (C[j] <= i && dp[i - C[j]] != int.MaxValue)
{
dp[i] = Math.Min(dp[i], 1 + dp[i - C[j]]);
}
}
}
// If no solution exists
if (dp[X] == int.MaxValue)
{
Console.WriteLine("-1");
return -1;
}
// Print the solution
int k = X;
while (k > 0)
{
for (int j = 0; j < N; j++)
{
if (k >= C[j] && dp[k - C[j]] == dp[k] - 1)
{
Console.Write(C[j] + " ");
k -= C[j];
break;
}
}
}
return dp[X];
}
// Driver code
static void Main(string[] args)
{
int X = 21;
// Set of possible denominations
int[] arr = { 2, 3, 4, 5 };
int N = arr.Length;
// Function Call
CountMinCoins(X, arr, N);
}
}
// javascript code addition
// Function to find the minimum number of coins to make the sum equals to X
function countMinCoins(X, C, N) {
// dp array to store the minimum number of coins for each value from 0 to X
const dp = new Array(X + 1).fill(Infinity);
// Initialize dp array
dp[0] = 0;
// Loop through all values from 1 to X and compute the minimum number of coins
for (let i = 1; i <= X; i++) {
for (let j = 0; j < N; j++) {
if (C[j] <= i && dp[i - C[j]] !== Infinity) {
dp[i] = Math.min(dp[i], 1 + dp[i - C[j]]);
}
}
}
// If no solution exists
if (dp[X] === Infinity) {
console.log("-1");
return -1;
}
// Print the solution
let i = X;
const result = [];
while (i > 0) {
for (let j = 0; j < N; j++) {
if (i >= C[j] && dp[i - C[j]] === dp[i] - 1) {
result.push(C[j]);
i -= C[j];
break;
}
}
}
console.log(result.join(' '));
return dp[X];
}
// Driver code
const X = 21;
// Set of possible denominations
const arr = [2, 3, 4, 5];
const N = arr.length;
// Function Call
countMinCoins(X, arr, N);
// The code is contributed by Nidhi goel.
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(X)
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Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
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Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
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GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
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