322. Coin change

   // T(c^n) c coins.length, n =amount
    /*#Recursive Method:#  
    https://discuss.leetcode.com/topic/32489/java-both-iterative-and-recursive-solutions-with-explanations
    idea: dynamic programming: think of the last step we take. Suppose we have already found out the best way to sum up to amount a, then for the last step, we can choose any coin type which gives us a remainder r where r = a-coins[i] for all i's. For every remainder, go through exactly the same process as before until either the remainder is 0 or less than 0 (meaning not a valid solution). With this idea, the only remaining detail is to store the minimum number of coins needed to sum up to r so that we don't need to recompute it over and over again.*/
   
    public int coinChange(int[] coins, int amount) {
       if (amount < 1) return 0;
       return helper(coins, amount, new int[amount]);
    }
    
    public int helper(int[] coins, int rem, int[] count) {
        if (rem < 0) return -1; // not valid
        if (rem == 0) return 0;  // completed
        if (count[rem - 1] !=0) return count[rem - 1]; // already computed, 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];
    }
  // Iterative sol, we think in bottom0up manner. Suppose we have 
    // already computed all the minimum counts up to sum, what woule be the minimum count for sum+1?
    public int coinChange(int[] coins, int amount) {
        if (amount < 1) return 0;
        int[] dp = new int[amount +1];
        int sum = 0;
        while (++sum <= amount) {
            int min = -1;
            for (int coin : coins) {
                if (sum >= coin && dp[sum-coin] != -1) {
                    int temp = dp[sum-coin] + 1;
                    min = min<0 ? temp : (temp < min ? temp : min);
                }
            }
            dp[sum] = min;
        }
        return dp[amount];
    }**