Coin Change – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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

C++ Coin Change LeetCode Solution

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        int Max = amount + 1;
        vector<int> dp(amount + 1, Max);
        dp[0] = 0;
        for (int i = 1; i <= amount; i++) {
            for (int j = 0; j < coins.size(); j++) {
                if (coins[j] <= i) {
                    dp[i] = min(dp[i], dp[i - coins[j]] + 1);
                }
            }
        }
        return dp[amount] > amount ? -1 : dp[amount];
    }
};

Java Coin Change LeetCode Solution

#Recursive Method:#
public class Solution {
public int coinChange(int[] coins, int amount) {
    if(amount<1) return 0;
    return helper(coins, amount, new int[amount]);
}

private int helper(int[] coins, int rem, int[] count) { // rem: remaining coins after the last step; count[rem]: minimum number of coins to sum up to rem
    if(rem<0) return -1; // not valid
    if(rem==0) return 0; // completed
    if(count[rem-1] != 0) return count[rem-1]; // already computed, so reuse
    int min = Integer.MAX_VALUE;
    for(int coin : coins) {
        int res = helper(coins, rem-coin, count);
        if(res>=0 && res < min)
            min = 1+res;
    }
    count[rem-1] = (min==Integer.MAX_VALUE) ? -1 : min;
    return count[rem-1];
}
}
 

Python 3 Coin Change LeetCode Solution

class Solution(object):
    def coinChange(self, coins, amount):
        MAX = float('inf')
        dp = [0] + [MAX] * amount

        for i in xrange(1, amount + 1):
            dp[i] = min([dp[i - c] if i - c >= 0 else MAX for c in coins]) + 1

        return [dp[amount], -1][dp[amount] == MAX]

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