Coin Change - Minimum Coins to Make Sum
Last Updated :
14 Mar, 2025
Given an array of coins[] of size n and a target value sum, where coins[i] represent the coins of different denominations. You have an infinite supply of each of the coins. The task is to find the minimum number of coins required to make the given value sum. If it is not possible to form the sum using the given coins, return -1.
Examples:
Input: coins[] = [25, 10, 5], sum = 30
Output: 2
Explanation : Minimum 2 coins needed, 25 and 5
Input: coins[] = [9, 6, 5, 1], sum = 19
Output: 3
Explanation: 19 = 9 + 9 + 1
Input: coins[] = [5, 1], sum = 0
Output: 0
Explanation: For 0 sum, we do not need a coin
Input: coins[] = [4, 6, 2], sum = 5
Output: -1
Explanation: Not possible to make the given sum.
[Naive Approach ] Using Recursion – O(n^sum) Time and O(sum) Space
This problem is a variation of the problem Coin Change Problem. Here instead of finding the total number of possible solutions, we need to find the solution with the minimum number of coins.
The idea is to find the minimum number of coins required to reach the target sum by trying each coin denomination in the coins[] array. Starting from the target sum, for each coin coins[i], we can either include it or exclude it. If we include it, we subtract its value from sum and recursively try to make the remaining amount with the same coin denominations. If we exclude it, we move to the next coin in the list.
Mathematically the recurrence relation will look like the following:
minCoins(i, sum, coins) = min(1 + minCoins(i, sum-coins[i], coins), minCoins(i+1, sum, coins))
Base cases:
- minCoins(i, sum, coins) = 0, if sum = 0.
- minCoins(i, sum, coins) = INTEGER MAX, if sum < 0 or i == size of coins.
C++
// C++ program to find minimum of coins
// to make a given change sum
#include<bits/stdc++.h>
using namespace std;
int minCoinsRecur(int i, int sum, vector<int> &coins) {
// base case
if (sum == 0) return 0;
if (sum <0 || i == coins.size()) return INT_MAX;
int take = INT_MAX;
// take a coin only if its value
// is greater than 0.
if (coins[i]>0) {
take = minCoinsRecur(i, sum-coins[i], coins);
if (take != INT_MAX) take++;
}
// not take the coin
int noTake = minCoinsRecur(i+1, sum, coins);
return min(take, noTake);
}
int minCoins(vector<int> &coins, int sum) {
int res = minCoinsRecur(0, sum, coins);
return res!=INT_MAX?res:-1;
}
int main() {
vector<int> coins = {9, 6, 5, 1};
int sum = 19;
cout << minCoins(coins, sum);
return 0;
}
// Java program to find minimum of coins
// to make a given change sum
import java.util.Arrays;
class GfG {
static int minCoinsRecur(int i, int sum, int[] coins) {
// base case
if (sum == 0) return 0;
if (sum < 0 || i == coins.length) return Integer.MAX_VALUE;
int take = Integer.MAX_VALUE;
// take a coin only if its value
// is greater than 0.
if (coins[i] > 0) {
take = minCoinsRecur(i, sum - coins[i], coins);
if (take != Integer.MAX_VALUE) take++;
}
// not taking the coins
int noTake = minCoinsRecur(i + 1, sum, coins);
return Math.min(take, noTake);
}
static int minCoins(int[] coins, int sum) {
int ans = minCoinsRecur(0, sum, coins);
return ans != Integer.MAX_VALUE ? ans : -1;
}
public static void main(String[] args) {
int[] coins = {9, 6, 5, 1};
int sum = 19;
System.out.println(minCoins(coins, sum));
}
}
# Python program to find minimum of coins
# to make a given change sum
def minCoinsRecur(i, sum, coins):
# base case
if sum == 0:
return 0
if sum < 0 or i == len(coins):
return float('inf')
take = float('inf')
# take a coin only if its value
# is greater than 0.
if coins[i] > 0:
take = minCoinsRecur(i, sum - coins[i], coins)
if take != float('inf'):
take += 1
#not taking the coin
noTake = minCoinsRecur(i + 1, sum, coins)
return min(take, noTake)
def minCoins(coins, sum):
ans = minCoinsRecur(0, sum, coins)
return ans if ans != float('inf') else -1
if __name__ == "__main__":
coins = [9, 6, 5, 1]
sum = 19
print(minCoins(coins, sum))
// C# program to find minimum of coins
// to make a given change sum
using System;
class GfG {
static int minCoinsRecur(int i, int sum, int[] coins) {
// base case
if (sum == 0) return 0;
if (sum < 0 || i == coins.Length) return int.MaxValue;
int take = int.MaxValue;
// take a coin only if its value
// is greater than 0.
if (coins[i] > 0) {
take = minCoinsRecur(i, sum - coins[i], coins);
if (take != int.MaxValue) take++;
}
// not taking the coin
int noTake = minCoinsRecur(i + 1, sum, coins);
return Math.Min(take, noTake);
}
static int minCoins(int[] coins, int sum) {
int ans = minCoinsRecur(0, sum, coins);
return ans != int.MaxValue ? ans : -1;
}
static void Main(string[] args) {
int[] coins = { 9, 6, 5, 1 };
int sum = 19;
Console.WriteLine(minCoins(coins, sum));
}
}
// JavaScript program to find minimum of coins
// to make a given change sum
function minCoinsRecur(i, sum, coins) {
// base case
if (sum === 0) return 0;
if (sum < 0 || i === coins.length) return Number.MAX_VALUE;
let take = Number.MAX_VALUE;
// take a coin only if its value
// is greater than 0.
if (coins[i] > 0) {
take = minCoinsRecur(i, sum - coins[i], coins);
if (take !== Number.MAX_VALUE) take++;
}
// not taking the coin
let noTake = minCoinsRecur(i + 1, sum, coins);
return Math.min(take, noTake);
}
function minCoins(coins, sum) {
let ans = minCoinsRecur(0, sum, coins);
return ans !== Number.MAX_VALUE ? ans : -1;
}
// Driver code
const coins = [9, 6, 5, 1];
const sum = 19;
console.log(minCoins(coins, sum));
[Better Approach 1] Using Top-Down DP (Memoization) - O(n*sum) Time and O(n*sum) Space
If we notice carefully, we can observe that the above recursive solution holds the following two properties of Dynamic Programming:
1. Optimal Substructure:
Minimum number of ways to make sum at index i, i.e., minCoins(i, sum, coins), depends on the optimal solutions of the subproblems minCoins(i, sum-coins[i], coins) , and minCoins(i+1, sum, coins). By comparing these optimal substructures, we can efficiently calculate the minimum number of coins to make target sum at index i.
2. Overlapping Subproblems:
While applying a recursive approach in this problem, we notice that certain subproblems are computed multiple times.
- There are only are two parameters: i and sum that changes in the recursive solution. So we create a 2D matrix of size n*(sum+1) for memoization.
- We initialize this matrix as -1 to indicate nothing is computed initially.
- Now we modify our recursive solution to first check if the value is -1, then only make recursive calls. This way, we avoid re-computations of the same subproblems.
C++
// C++ program to find minimum of coins
// to make a given change sum
#include<bits/stdc++.h>
using namespace std;
int minCoinsRecur(int i, int sum, vector<int> &coins, vector<vector<int>> &memo) {
// base case
if (sum == 0) return 0;
if (sum <0 || i == coins.size()) return INT_MAX;
if (memo[i][sum]!=-1) return memo[i][sum];
int take = INT_MAX;
// take a coin only if its value
// is greater than 0.
if (coins[i]>0) {
take = minCoinsRecur(i, sum-coins[i], coins, memo);
if (take != INT_MAX) take++;
}
// not take the coins
int noTake = minCoinsRecur(i+1, sum, coins, memo);
return memo[i][sum] = min(take, noTake);
}
int minCoins(vector<int> &coins, int sum) {
vector<vector<int>> memo(coins.size(), vector<int>(sum+1, -1));
int res = minCoinsRecur(0, sum, coins, memo);
return res!=INT_MAX?res:-1;
}
int main() {
vector<int> coins = {9, 6, 5, 1};
int sum = 19;
cout << minCoins(coins, sum);
return 0;
}
// Java program to find minimum of coins
// to make a given change sum
import java.util.Arrays;
class GfG {
static int minCoinsRecur(int i, int sum, int[] coins, int[][] memo) {
// base case
if (sum == 0) return 0;
if (sum < 0 || i == coins.length) return Integer.MAX_VALUE;
if (memo[i][sum] != -1) return memo[i][sum];
int take = Integer.MAX_VALUE;
// take a coin only if its value
// is greater than 0.
if (coins[i] > 0) {
take = minCoinsRecur(i, sum - coins[i], coins, memo);
if (take != Integer.MAX_VALUE) take++;
}
// not take the coin
int noTake = minCoinsRecur(i + 1, sum, coins, memo);
return memo[i][sum] = Math.min(take, noTake);
}
static int minCoins(int[] coins, int sum) {
int[][] memo = new int[coins.length][sum + 1];
for (int[] row : memo) Arrays.fill(row, -1);
int ans = minCoinsRecur(0, sum, coins, memo);
return ans != Integer.MAX_VALUE ? ans : -1;
}
public static void main(String[] args) {
int[] coins = {9, 6, 5, 1};
int sum = 19;
System.out.println(minCoins(coins, sum));
}
}
# Python program to find minimum of coins
# to make a given change sum
def minCoinsRecur(i, sum, coins, memo):
# base case
if sum == 0:
return 0
if sum < 0 or i == len(coins):
return float('inf')
if memo[i][sum] != -1:
return memo[i][sum]
take = float('inf')
# take a coin only if its value
# is greater than 0.
if coins[i] > 0:
take = minCoinsRecur(i, sum - coins[i], coins, memo)
if take != float('inf'):
take += 1
#not take the coins
noTake = minCoinsRecur(i + 1, sum, coins, memo)
memo[i][sum] = min(take, noTake)
return memo[i][sum]
def minCoins(coins, sum):
memo = [[-1] * (sum + 1) for _ in range(len(coins))]
ans = minCoinsRecur(0, sum, coins, memo)
return ans if ans != float('inf') else -1
if __name__ == "__main__":
coins = [9, 6, 5, 1]
sum = 19
print(minCoins(coins, sum))
// C# program to find minimum of coins
// to make a given change sum
using System;
class GfG {
static int minCoinsRecur(int i, int sum, int[] coins, int[,] memo) {
// base case
if (sum == 0) return 0;
if (sum < 0 || i == coins.Length) return int.MaxValue;
if (memo[i, sum] != -1) return memo[i, sum];
int take = int.MaxValue;
// take a coin only if its value
// is greater than 0.
if (coins[i] > 0) {
take = minCoinsRecur(i, sum - coins[i], coins, memo);
if (take != int.MaxValue) take++;
}
// not take the coins
int noTake = minCoinsRecur(i + 1, sum, coins, memo);
memo[i, sum] = Math.Min(take, noTake);
return memo[i, sum];
}
static int minCoins(int[] coins, int sum) {
int[,] memo = new int[coins.Length, sum + 1];
for (int i = 0; i < coins.Length; i++)
for (int j = 0; j <= sum; j++)
memo[i, j] = -1;
int ans = minCoinsRecur(0, sum, coins, memo);
return ans != int.MaxValue ? ans : -1;
}
static void Main(string[] args) {
int[] coins = { 9, 6, 5, 1 };
int sum = 19;
Console.WriteLine(minCoins(coins, sum));
}
}
// JavaScript program to find minimum of coins
// to make a given change sum
function minCoinsRecur(i, sum, coins, memo) {
// base case
if (sum === 0) return 0;
if (sum < 0 || i === coins.length)
return Number.MAX_VALUE;
if (memo[i][sum] !== -1) return memo[i][sum];
let take = Number.MAX_VALUE;
// take a coin only if its value
// is greater than 0.
if (coins[i] > 0) {
take = minCoinsRecur(i, sum - coins[i], coins, memo);
if (take !== Number.MAX_VALUE) take++;
}
// not take the coin
let noTake = minCoinsRecur(i + 1, sum, coins, memo);
memo[i][sum] = Math.min(take, noTake);
return memo[i][sum];
}
function minCoins(coins, sum) {
const memo = Array.from({ length: coins.length }
, () => Array(sum + 1).fill(-1));
const ans = minCoinsRecur(0, sum, coins, memo);
return ans !== Number.MAX_VALUE ? ans : -1;
}
// Driver code
const coins = [9, 6, 5, 1];
const sum = 19;
console.log(minCoins(coins, sum));
[Better Approach 2] Using Bottom-Up DP (Tabulation) - O(n*sum) Time and O(n*sum) Space
The idea is to fill the DP table based on previous values. For each coin, we either include it or exclude it to compute the minimum number of coins needed for each sum. The table is filled in an iterative manner from i = n-1 to i = 0 and for each sum from 1 to sum.
The dynamic programming relation is as follows:
- if (sum-coins[i]) is greater than 0, then dp[i][sum] = min(1+dp[i][sum-coins[i]], dp[i+1][sum])
- else dp[i][sum] = dp[i+1][sum].
C++
// C++ program to find minimum of coins
// to make a given change sum
#include<bits/stdc++.h>
using namespace std;
int minCoins(vector<int> &coins, int sum) {
vector<vector<int>> dp(coins.size(), vector<int>(sum+1, 0));
for (int i=coins.size()-1; i>=0; i--) {
for (int j=1; j<=sum; j++) {
dp[i][j] = INT_MAX;
int take = INT_MAX, noTake = INT_MAX;
// If we take coins[i] coin
if (j-coins[i]>=0) {
take = dp[i][j-coins[i]];
if (take != INT_MAX) take++;
}
if (i+1<coins.size())
// not take the coin
noTake = dp[i+1][j];
dp[i][j] = min(take, noTake);
}
}
if (dp[0][sum]!=INT_MAX) return dp[0][sum];
return -1;
}
int main() {
vector<int> coins = {9, 6, 5, 1};
int sum = 19;
cout << minCoins(coins, sum);
return 0;
}
// Java program to find minimum of coins
// to make a given change sum
import java.util.Arrays;
class GfG {
static int minCoins(int[] coins, int sum) {
int[][] dp = new int[coins.length][sum + 1];
for (int i = coins.length - 1; i >= 0; i--) {
for (int j = 1; j <= sum; j++) {
dp[i][j] = Integer.MAX_VALUE;
int take = Integer.MAX_VALUE, noTake = Integer.MAX_VALUE;
// If we take coins[i] coin
if (j - coins[i] >= 0) {
take = dp[i][j - coins[i]];
if (take != Integer.MAX_VALUE) take++;
}
// not take the coin
if (i + 1 < coins.length) noTake = dp[i + 1][j];
dp[i][j] = Math.min(take, noTake);
}
}
if (dp[0][sum] != Integer.MAX_VALUE) return dp[0][sum];
return -1;
}
public static void main(String[] args) {
int[] coins = {9, 6, 5, 1};
int sum = 19;
System.out.println(minCoins(coins, sum));
}
}
# Python program to find minimum of coins
# to make a given change sum
def minCoins(coins, sum):
dp = [[0] * (sum + 1) for _ in range(len(coins))]
for i in range(len(coins) - 1, -1, -1):
for j in range(1, sum + 1):
dp[i][j] = float('inf')
take = float('inf')
noTake = float('inf')
# If we take coins[i] coin
if j - coins[i] >= 0:
take = dp[i][j - coins[i]]
if take != float('inf'):
take += 1
if i + 1 < len(coins):
#not take the coins
noTake = dp[i + 1][j]
dp[i][j] = min(take, noTake)
if dp[0][sum] != float('inf'):
return dp[0][sum]
return -1
if __name__ == "__main__":
coins = [9, 6, 5, 1]
sum = 19
print(minCoins(coins, sum))
// C# program to find minimum of coins
// to make a given change sum
using System;
class GfG {
static int minCoins(int[] coins, int sum) {
int[,] dp = new int[coins.Length, sum + 1];
for (int i = coins.Length - 1; i >= 0; i--) {
for (int j = 1; j <= sum; j++) {
dp[i, j] = int.MaxValue;
int take = int.MaxValue, noTake = int.MaxValue;
// If we take coins[i] coin
if (j - coins[i] >= 0) {
take = dp[i, j - coins[i]];
if (take != int.MaxValue) take++;
}
if (i + 1 < coins.Length)
// not take the coins
noTake = dp[i + 1, j];
dp[i, j] = Math.Min(take, noTake);
}
}
if (dp[0, sum] != int.MaxValue) return dp[0, sum];
return -1;
}
static void Main(string[] args) {
int[] coins = { 9, 6, 5, 1 };
int sum = 19;
Console.WriteLine(minCoins(coins, sum));
}
}
// JavaScript program to find minimum of coins
// to make a given change sum
function minCoins(coins, sum) {
let dp = Array.from({ length: coins.length }, () => Array(sum + 1).fill(0));
for (let i = coins.length - 1; i >= 0; i--) {
for (let j = 1; j <= sum; j++) {
dp[i][j] = Number.MAX_VALUE;
let take = Number.MAX_VALUE, noTake = Number.MAX_VALUE;
// If we take coins[i] coin
if (j - coins[i] >= 0) {
take = dp[i][j - coins[i]];
if (take !== Number.MAX_VALUE) take++;
}
if (i + 1 < coins.length)
// not take the coins
noTake = dp[i + 1][j];
dp[i][j] = Math.min(take, noTake);
}
}
if (dp[0][sum] !== Number.MAX_VALUE) return dp[0][sum];
return -1;
}
// Driver code
const coins = [9, 6, 5, 1];
const sum = 19;
console.log(minCoins(coins, sum));
[Expected Approach] Using Space Optimized DP – O(n*sum) Time and O(sum) Space
In previous approach of dynamic programming we have derive the relation between states as given below:
- if (sum-coins[i]) is greater than 0, then dp[i][sum] = min(1+dp[i][sum-coins[i]], dp[i+1][sum])
- else dp[i][sum] = dp[i+1][sum].
If we observe that for calculating current dp[i][sum] state we only need previous row dp[i-1][sum] or current row dp[i][sum-coins[i]]. There is no need to store all the previous states just one previous state is used to compute result.
C++
// C++ program to find minimum of coins
// to make a given change sum
#include<bits/stdc++.h>
using namespace std;
// Function to find the minimum number of
// coins needed to make the given sum
int minCoins(vector<int> &coins, int sum) {
// Create a DP array to store the minimum coins
// for each value up to sum
vector<int> dp(sum+1, INT_MAX);
// Base case: 0 coins are needed to make sum 0
dp[0] = 0;
// Iterate over each coin in the coins array
for (int i=coins.size()-1; i>=0; i--) {
// Iterate through all sums from 1 to the given sum
for (int j=1; j<=sum; j++) {
// Initialize take and noTake variables
int take = INT_MAX, noTake = INT_MAX;
// Check if we can take the current coin
if (j-coins[i]>=0 && coins[i]>0) {
// Get the minimum coins needed for
// the remaining sum
take = dp[j-coins[i]];
// Increment take if it's a valid option
if (take != INT_MAX) take++;
}
// If we have more coins available,
// consider not taking the current coin
if (i+1<coins.size())
// Get the minimum coins needed without
// taking the current coi
noTake = dp[j];
// Update the DP array with the minimum of
// taking or not taking the coin
dp[j] = min(take, noTake);
}
}
// found a solution, return the minimum
// coins needed, otherwise return -1
if (dp[sum]!=INT_MAX) return dp[sum];
return -1;
}
int main() {
vector<int> coins = {9, 6, 5, 1};
int sum = 19;
cout << minCoins(coins, sum);
return 0;
}
// Java program to find minimum of coins
import java.util.Arrays;
class GfG {
// Function to find the minimum number
// of coins required to make a given sum
static int minCoins(int[] coins, int sum) {
// Array to store the minimum coins needed for
// each value up to sum
int[] dp = new int[sum + 1];
// Initialize the dp array with a large value (infinity)
Arrays.fill(dp, Integer.MAX_VALUE);
// Base case: 0 coins are needed to make the sum of 0
dp[0] = 0;
// Iterate through each coin in reverse order
for (int i = coins.length - 1; i >= 0; i--) {
// Iterate through all sums from 1 to the target sum
for (int j = 1; j <= sum; j++) {
// Initialize variables to track taking
// or not taking the coin
int take = Integer.MAX_VALUE, noTake = Integer.MAX_VALUE;
// If we can take the current coin
if (j - coins[i] >= 0 && coins[i] > 0) {
take = dp[j - coins[i]];
// If taking the coin is possible,
// increment the count
if (take != Integer.MAX_VALUE) take++;
}
// If there are more coins available,
// consider not taking the current coin
if (i + 1 < coins.length)
noTake = dp[j];
// Store the minimum of taking or
// not taking the coin
dp[j] = Math.min(take, noTake);
}
}
// Return the minimum coins needed for
// the target sum, or -1 if not possible
return dp[sum] != Integer.MAX_VALUE ? dp[sum] : -1;
}
public static void main(String[] args) {
int[] coins = {9, 6, 5, 1};
int sum = 19;
System.out.println(minCoins(coins, sum));
}
}
# Python program to find minimum of coins
def minCoins(coins, sum):
# Initialize a list to store the minimum
# number of coins for each amount
dp = [float('inf')] * (sum + 1)
# Base case: 0 coins are needed to make the sum of 0
dp[0] = 0
# Iterate over each coin in reverse order
for i in range(len(coins) - 1, -1, -1):
# Iterate through all amounts from 1 to sum
for j in range(1, sum + 1):
# take variable for the current coin
take = float('inf')
# noTake variable for the current amount
noTake = float('inf')
# If we can take the current coin
if j - coins[i] >= 0 and coins[i] > 0:
# Get the minimum coins needed
# for the remaining amount
take = dp[j - coins[i]]
# Increment the count if it's a valid take
if take != float('inf'):
take += 1
# If there are coins left to consider
if i + 1 < len(coins):
# Get the minimum coins needed without
# taking the current coin
noTake = dp[j]
# Store the minimum of taking or not
# taking the current coin
dp[j] = min(take, noTake)
# Return the result for the given sum,
# or -1 if it's not possible
return dp[sum] if dp[sum] != float('inf') else -1
if __name__ == "__main__":
coins = [9, 6, 5, 1]
sum = 19
print(minCoins(coins, sum))
// C# program to find minimum of coins
using System;
using System.Linq;
class GfG {
// Function to find the minimum number of
// coins required to make a given sum
static int minCoinsFunc(int[] coins, int sum) {
// Create a DP array initialized to maximum value
int[] dp = Enumerable.Repeat(int.MaxValue, sum + 1).ToArray();
dp[0] = 0;
// Iterate through each coin
for (int i = coins.Length - 1; i >= 0; i--) {
// Iterate through all sums from 1 to the target sum
for (int j = 1; j <= sum; j++) {
int take = int.MaxValue, noTake = int.MaxValue;
// If we take coins[i] coin
if (j - coins[i] >= 0 && coins[i] > 0) {
// Get the value if we take this coin
take = dp[j - coins[i]];
// Increment the coin count
if (take != int.MaxValue) take++;
}
// Get the value if we do not take this coin
if (i + 1 < coins.Length)
noTake = dp[j];
// Store the minimum of taking or not taking the coin
dp[j] = Math.Min(take, noTake);
}
}
// Return the result: if no combination found, return -1
return dp[sum] != int.MaxValue ? dp[sum] : -1;
}
// Main function to test the MinCoinsFunc
public static void Main() {
int[] coins = {9, 6, 5, 1};
int sum = 19;
Console.WriteLine(minCoinsFunc(coins, sum));
}
}
// JavaScript program to find minimum of coins
function minCoins(coins, sum) {
// Initialize an array to store the
// minimum coins needed for each amount
let dp = new Array(sum + 1).fill(Infinity);
dp[0] = 0;
// Iterate through each coin in the coins array
for (let i = coins.length - 1; i >= 0; i--) {
// Iterate through all sums from 1
// to the target sum
for (let j = 1; j <= sum; j++) {
// Initialize take and noTake variables
let take = Infinity, noTake = Infinity;
// If we can take the current coin
if (j - coins[i] >= 0 && coins[i] > 0) {
// Get the minimum coins for the remaining sum
take = dp[j - coins[i]];
// Increment the coins count if valid
if (take !== Infinity) take++;
}
// If not taking the current coin,
// just take the previous value
if (i + 1 < coins.length)
noTake = dp[j];
// Store the minimum of taking or not taking the coin
dp[j] = Math.min(take, noTake);
}
}
// Return the result for the target sum,
// or -1 if it's not possible
return dp[sum] !== Infinity ? dp[sum] : -1;
}
// Driver code
let coins = [9, 6, 5, 1];
let sum = 19;
console.log(minCoins(coins, sum));
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