8. Coin Change

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

Approach

• Bottom Up -> Check solution for the amount by subtracting the coin and add one
• Initialize array with 1 index and value amount + 1 so it can be used later too to return -1
• In simplest terms we d[i] will tell us the coins required to reach the i amount we fill all with amount + 1 to show that it is impossible to reach and we fill 0 index with zero saying it takes 0 coin to reach 0 amount
• The outer loop iterates over each possible amount from 1 to amount. For each amount i, we try to figure out how to make i using the available coin denominations.
• With inner loop we will check what was the solution if we remove the j coin and add one coin which is j, min condition is because initially we assigned a higher value to each index
• Time: O(n*t) Space: O(t)
• Where n is the length of the array coins and t is the given amount.

class Solution {
    public int coinChange(int[] coins, int amount) {
        int[] d = new int[amount + 1];
        Arrays.fill(d,amount + 1);
        d[0] = 0;
        for (int i = 1; i <= amount; i++){
            for (int j = 0; j < coins.length; j++) {
                if (coins[j] <= i) {
                    d[i] = Math.min(d[i], d[i - coins[j]] + 1);
                }
            }
        }
        return d[amount] > amount ? -1 : d[amount];
    }
}