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 = [2], amount = 3
Output: -1

Example 3:

Input: coins = [1], amount = 0
Output: 0

Example 4:

Input: coins = [1], amount = 1
Output: 1

Example 5:

Input: coins = [1], 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] = 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. 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];
    }

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