322 coin change of leetcode

1. Description

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.

2. Test Cases

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

3. Solution

3.1 Recursion with memo

class Solution {
    
    public int coinChange(int[] coins, int amount) {
    
        Integer[] dp = new Integer[amount +1];
        dp[0] = 0;
        
        find(coins, amount, dp);
        
        return dp[amount];
    }
    
    private int find(int[] coins, int amount, Integer[] dp) {
    
        if (amount == 0) return 0;
        
        if(dp[amount] != null )
            return dp[amount];
        
        int min =Integer.MAX_VALUE;
        
        for (int c : coins) {
    
            if (amount -c >= 0) {
    
                int r = find(coins, amount-c, dp);
                if (r != -1)
                    min = Math.min(min, r + 1);
            }
        }
        
        return dp[amount] = (min == Integer.MAX_VALUE ? -1 : min);
    }
    
}

3.2 iterative

class Solution {
    
    public int coinChange(int[] coins, int amount) {
    
        Integer[] dp = new Integer[amount +1];
        dp[0] = 0;
        
        for (int i = 1; i <= amount; i++) {
    
            int min = Integer.MAX_VALUE;
            
            for (int c : coins) {
    
                if (i - c >= 0 ) {
    
                    int r = dp[i - c];
                    if (r != -1)
                        min = Math.min(min, r + 1);
                }
            }
            
            dp[i] = min == Integer.MAX_VALUE ? -1 : min;
        }
        
        return dp[amount];
    }
}

4. reflection

It seems that the essence of dynamic planning is exhaustion , Just how to enumerate fewer times , Get results faster .