Coin Change Problem

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

Example 4:

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

Example 5:

Input: coins = [1], amount = 2
Output: 2

Constraints:

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

Solution:

I solve this problem by using bottom-up approach. It is concept of dynamic programming where we identify the problem and if we see the overlap sub problem nature , we can use either memoization or tabulation approach.

In tabulation approach , we solve smaller problem and with the help of that smaller problem solution , we solve bigger problem until we reach to final solution.

public class Solution {
    public int CoinChange(int[] coins, int amount) {
        // Create an DP array where we will store the solution of  subproblem from  bottom-up  approach.
        // As it will have value for 0 also so we will have  Amount+1 length;
        int[] dp = new int[amount + 1];
       
      // Initialize the DP array with sum dummy value.
        for(int i = 0 ; i < dp.Length ;i++)
        {
            dp[i] = -99;  // You can initialize with any value greater than the "amount" value.
        }
        dp[0] = 0;
        for(int i = 1 ; i < amount+1 ; i++)
        {
            for(int  j = 0 ; j < coins.Length ; j++)
            {
                if( i-coins[j] >=0)
                {
                    dp[i] = Math.Min(dp[i], 1 + dp[i-coins[j]]);
                }
            }
        }
        return dp[amount] == -99 ? -1:dp[amount];
    }
}