Google - Find minimum number of coins that make a given value

Posted 2021-01-23 incrediblechangshuo

Given a value V, if we want to make change for V cents, and we have infinite supply of each of C = { C1, C2, .. , Cm} valued coins, what is the minimum number of coins to make the change?
Examples:

Input: coins[] = {25, 10, 5}, V = 30
Output: Minimum 2 coins required
We can use one coin of 25 cents and one of 5 cents 

Input: coins[] = {9, 6, 5, 1}, V = 11
Output: Minimum 2 coins required
We can use one coin of 6 cents and 1 coin of 5 cents

// "static void main" must be defined in a public class.
public class Main {
    public static void main(String[] args) {
        int[] coins = {1,3,5};
        System.out.println(new Solution().minCoinChangeRecursiveWithMemo(coins, 3, 50));
        System.out.println(new Solution().minCoinChangeDP(coins, 3, 50));
    }
}

class Solution {
    public int minCoinChangeRecursiveWithMemo(int[] coins, int m, int v){
        HashMap<Integer, Integer> map = new HashMap<>();
        return helper(coins, m, v, map);
    }
    
    public int helper(int[] coins, int m, int v, HashMap<Integer, Integer> map){
        if(v == 0){
            return 0;
        }
        if(map.containsKey(v)){
            return map.get(v);
        }
        int min = Integer.MAX_VALUE;
        for(int i = 0; i < m; i++){
            if(coins[i] <= v){
                int sub_res = helper(coins, m, v-coins[i], map);
                if(sub_res !=  Integer.MAX_VALUE && sub_res + 1 < min){
                    min = sub_res+1;
                }
            }
        }
        map.put(v, min);
        return min;
    }
    
    public int minCoinChangeDP (int[] coins, int m, int v){
        int[] dp = new int[v+1];
        dp[0] = 0;
        for(int i= 1; i <= v; i++){
            dp[i] = Integer.MAX_VALUE;
            for(int j = 0; j < m; j++){
                if(coins[j] <= i){
                    if(dp[i - coins[j]] + 1 < dp[i]) {
                        dp[i] = dp[i - coins[j]] + 1;
                    }
                }
            }
        }
        return dp[v];
    }
    
}