Algorithm
Problem Name: 322. Coin Change
Problem Link: https://leetcode.com/problems/coin-change/
You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.
Return the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.
You may assume that you have an infinite number of each kind of coin.
Example 1:
Input: coins = [1,2,5], amount = 11 Output: 3 Explanation: 11 = 5 + 5 + 1
Example 2:
Input: coins = [2], amount = 3 Output: -1
Example 3:
Input: coins = [1], amount = 0 Output: 0
Constraints:
1 <= coins.length <= 121 <= coins[i] <= 231 - 10 <= amount <= 104
Code Examples
#1 Code Example with C Programming
Code -
C Programming
int coinChange(int* coins, int coinsSize, int amount) {
int *dp, i, j, d, n;
dp = malloc((amount + 1) * sizeof(int));
//assert(dp);
dp[0] = 0;
for (i = 1; i < = amount; i ++) {
n = -1;
for (j = 0; j < coinsSize; j ++) {
if (i >= coins[j]) {
d = i - coins[j];
if (dp[d] != -1 && (n == -1 || n > dp[d] + 1)) {
n = dp[d] + 1; // plus one of this coin
}
}
}
dp[i] = n;
}
n = dp[amount];
free(dp);
return n;
}
Input
Output
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int coinChange(int[] coins, int amount) {
Integer[] dp = new Integer[amount + 1];
Arrays.fill(dp, amount + 1);
dp[0] = 0;
for (int i = 1; i < = amount; i++) {
for (int coin : coins) {
if (i - coin < 0) {
continue;
}
dp[i] = Math.min(dp[i], dp[i - coin] + 1);
}
}
return dp[amount] == (amount + 1) ? -1 : dp[amount];
}
private int coinChangeMemoization(int[] coins, int amount, Integer[] memo) {
if (amount < 0) {
return -1;
}
if (amount == 0) {
return 0;
}
if (memo[amount] != null) {
return memo[amount];
}
int minCount = Integer.MAX_VALUE;
for (int coin : coins) {
int count = coinChangeMemoization(coins, amount - coin, memo);
if (count == -1) {
continue;
}
minCount = Math.min(minCount, count + 1);
}
memo[amount] = minCount == Integer.MAX_VALUE ? -1 : minCount;
return memo[amount];
}
// This approach times out due to overlapping subproblems
private int coinChangeRecursive(int[] coins, int amount) {
if (amount < 0) {
return -1;
}
if (amount == 0) {
return 0;
}
int minCount = Integer.MAX_VALUE;
for (int coin : coins) {
int count = coinChangeRecursive(coins, amount - coin);
if (count == -1) {
continue;
}
minCount = Math.min(minCount, count);
}
return minCount == Integer.MAX_VALUE ? -1 : minCount;
}
}
Input
Output
#3 Code Example with Javascript Programming
Code -
Javascript Programming
const coinChange = function(coins, amount) {
const dp = new Array(amount + 1).fill(amount + 1)
dp[0] = 0
for (let i = 1; i < = amount; i++) {
for (let coin of coins) {
if (coin <= i) dp[i] = Math.min(dp[i], dp[i - coin] + 1)
}
}
return dp[amount] === amount + 1 ? -1 : dp[amount]
}
Input
Output
#4 Code Example with Python Programming
Code -
Python Programming
class Solution:
def coinChange(self, coins, amount):
"""
:type coins: List[int]
:type amount: int
:rtype: int
"""
dp = [0] + [float('inf')] * amount
for i in range(1, amount + 1): dp[i] = min([dp[i - c] if i - c >= 0 else float('inf') for c in coins]) + 1
return dp[amount] if dp[amount] != float('inf') else -1
Input
Output
#5 Code Example with C# Programming
Code -
C# Programming
using System;
namespace LeetCode
{
public class _0322_CoinChange
{
public int CoinChange(int[] coins, int amount)
{
var dp = new long[amount + 1];
for (int i = 0; i < = amount; i++)
dp[i] = int.MaxValue;
dp[0] = 0;
for (int i = 1; i < = amount; i++)
foreach (var coin in coins)
if (coin <= i)
dp[i] = Math.Min(dp[i], dp[i - coin] + 1);
return dp[amount] > amount ? -1 : (int)dp[amount];
}
}
}
Input