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518. Coin Change II

You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.

Return the number of combinations that make up that amount. If that amount of money cannot be made up by any combination of the coins, return 0.

You may assume that you have an infinite number of each kind of coin.

The answer is guaranteed to fit into a signed 32-bit integer.

Example 1:

Input: amount = 5, coins = [1,2,5]
Output: 4
Explanation: there are four ways to make up the amount:
5=5
5=2+2+1
5=2+1+1+1
5=1+1+1+1+1

Example 2:

Input: amount = 3, coins = [2]
Output: 0
Explanation: the amount of 3 cannot be made up just with coins of 2.

Example 3:

Input: amount = 10, coins = [10]
Output: 1

Constraints:

  • 1 <= coins.length <= 300
  • 1 <= coins[i] <= 5000
  • All the values of coins are unique.
  • 0 <= amount <= 5000

Solution: 

class Solution {
    int[][] dp;
    public int change(int amount, int[] coins) {
        dp = new int[coins.length][amount + 1];
// dp[i][j] represents the number of ways to make up amount j using the first i coin
        /*
        a/c  0  1  2  3  4  5  6
          0  -1 -1 -1 -1 -1 -1 -1
          1  -1 -1 -1 -1 -1 -1 -1 
          2  -1 -1 -1 -1 -1 -1 -1 
        */
        for (int i = 0; i < coins.length; i++){
            for (int j = 0; j < amount + 1; j++){
                dp[i][j] = -1;
            }
        }

        int index = coins.length - 1;
        return helper(amount, coins, index);



    }

    private int helper(int amount, int[] coins, int index){
        if (amount == 0){
            return 1;
        }
        //// If amount is 0, there is one way to achieve it (use no coins)
        if (index == -1){
            return 0;
        }
        //// If no coins left and amount is not 0, there is no way to achieve the amount

        if (dp[index][amount] != -1){
            return dp[index][amount];
        }
 // Include the coin if its value is less than or equal to amount, and exclude it
        if (coins[index] <= amount){
            dp[index][amount] = helper(amount - coins[index], coins, index) + helper(amount, coins, index - 1);

            return dp[index][amount];
        }else{
            dp[index][amount] = helper(amount, coins, index - 1);
            return dp[index][amount];
        }
    }
}


// TC: O(n*m)
// SC: O(n*n)