# LeetCode 322. Coin Change

## Description

https://leetcode.com/problems/coin-change/

You are given coins of different denominations and a total amount of money amount. Write a function to compute 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 = , amount = 3
Output: -1
```

Example 3:

```Input: coins = , amount = 0
Output: 0
```

Example 4:

```Input: coins = , amount = 1
Output: 1
```

Example 5:

```Input: coins = , amount = 2
Output: 2
```

Constraints:

• `1 <= coins.length <= 12`
• `1 <= coins[i] <= 231 - 1`
• `0 <= amount <= 104`

## Explanation

dynamic programming array to indicate how many coins needed to make up every amount

## Python Solution

``````class Solution:
def coinChange(self, coins: List[int], amount: int) -> int:

MAX = sys.maxsize
result = [MAX for i in range(0, amount + 1)]

result = 0

for i in range(1, amount + 1):
for coin in coins:
if i - coin < 0:
continue

result[i] = min(result[i], result[i - coin] + 1)

if result[amount] == MAX:
return -1

return result[amount]
``````
• Time complexity: ~N*M, M is amount
• Space complexity: ~M, M is amount

## 2 Thoughts to “LeetCode 322. Coin Change”

1. vickey says:

public int coinChange(int[] coins, int amount) {
int n = coins.length;
int[] arr = new int[amount + 1];
for (int i = 1; i <= amount; i++) {
arr[i] = Integer.MAX_VALUE – 1;
for (int j = 0; j < n; j++) {
if (i – coins[j] < 0) {
continue;
}
arr[i] = Math.min(arr[i – coins[j]] + 1,arr[i]);
}
}
return arr[amount] == Integer.MAX_VALUE – 1 ? -1 : arr[amount];
}