class Solution {
    /** Recursion */
    public int coinChange(int[] coins, int amount) {
        int n = coins.length;
 
        int ans = findMinCoins(coins, amount, n);
        if(ans == Integer.MAX_VALUE-1)
            return -1;
        return ans;
    }
 
    public int findMinCoins(int[] coins, int amount, int n)
    {
        // base condition: No coins left
        // At this point, if amount > 0 (positive), we cannot form the remaining amount, because there are no coins left to use.
        // Returning 0 here would mean 0 coins are needed, which is incorrect, because we cannot form the positive amount with no coins.
        if(n == 0)
            return Integer.MAX_VALUE-1;
 
        // 0 coins needed to make amount as 0
        if(amount == 0)
            return 0;
 
        if(coins[n-1] <= amount)
        {
            // we can pick same coin multiple times
            int pick = 1 + findMinCoins(coins, amount-coins[n-1], n);
            int notPick = findMinCoins(coins, amount, n-1);
            return Math.min(pick, notPick);
        }
        return findMinCoins(coins, amount, n-1);
    }
 
    /** Memoization */
    public int coinChange(int[] coins, int amount) {
        int n = coins.length;
 
        int[][] t = new int[n+1][amount+1];
 
        // when we have coins, but amount is always 0, so 0 coins needed to make amount as 0
        for(int i=0; i<n+1; i++)
        {
            Arrays.fill(t[i], 0);
        }
 
        // when we do not have coins, but amount  > 0
        for(int j=0; j<n+1; j++)
        {
            Arrays.fill(t[j], Integer.MAX_VALUE-1);
        }
 
        int ans = findMinCoins(coins, amount, n, t);
        if(ans == Integer.MAX_VALUE-1)
            return -1;
        return ans;
    }
 
    public int findMinCoins(int[] coins, int amount, int n, int[][] t)
    {
        // base condition: No coins left
        // At this point, if amount > 0 (positive), we cannot form the remaining amount, because there are no coins left to use.
        // Returning 0 here would mean 0 coins are needed, which is incorrect, because we cannot form the positive amount with no coins.
        if(n == 0)
            return Integer.MAX_VALUE-1;
 
        // 0 coins needed to make amount as 0
        if(amount == 0)
            return 0;
 
        if(t[n][amount] != Integer.MAX_VALUE - 1)
            return t[n][amount];
 
        if(coins[n-1] <= amount)
        {
            // we can pick same coin multiple times
            int pick = 1 + findMinCoins(coins, amount-coins[n-1], n, t);
            int notPick = findMinCoins(coins, amount, n-1, t);
            t[n][amount] = Math.min(pick, notPick);
            return t[n][amount];
        }
        t[n][amount] = findMinCoins(coins, amount, n-1, t);
        return t[n][amount];
    }
 
    /** Top-down approach */
    public int coinChange(int[] coins, int amount) {
        int n = coins.length;
 
        int[][] t = new int[n+1][amount+1];
 
        for(int i=0; i<n+1; i++)
        {
            for(int j=0; j<amount+1; j++)
            {
                // when coins are 0, we cannot form the amount
                if(i == 0)
                    t[i][j] = Integer.MAX_VALUE - 1;
 
                // when amount to be formed is 0, then 0 coins needed
                if(j == 0)
                    t[i][j] = 0;
            }
        }
 
        for(int i=1; i<n+1; i++)
        {
            for(int j=0; j<amount+1; j++)
            {
                if(coins[i-1] <= j)
                {
                    // we can pick same coin multiple times
                    int pick = 1 + t[i][j - coins[i-1]];
                    int notPick = t[i-1][j];
                    t[i][j] = Math.min(pick, notPick);
                }
                else
                    t[i][j] = t[i-1][j];
            }
        }
 
        int ans = t[n][amount];
        if(ans == Integer.MAX_VALUE-1)
            return -1;
        return ans;
    }
}