Strengths:

• You correctly identified that the recursive solution would lead to TLE and implemented an iterative DP solution.
• The DP solution is correct and efficient for the given constraints.
• The code is clean and well-commented.

Areas for Improvement:

• While the 2D DP solution is correct, you can optimize the space to O(amount) by using a 1D array. This is because when processing the i-th coin, you only need the state from the previous coin and the current state for smaller amounts. Here's how you can do it:
  Initialize dp[0] = 0 and for j from 1 to amount, dp[j] = a large value.
  Then for each coin, iterate j from coin value to amount:
  dp[j] = Math.min(dp[j], 1 + dp[j - coin]);
• Avoid including solutions for multiple problems in the same file. It's better to have separate files for each problem to maintain clarity.
• In the DP solution, you used Integer.MAX_VALUE-1 to prevent integer overflow when adding 1. This is good, but you should note that if the amount is very large and the coins are small, the value might still be large. However, with the constraints given, it is safe.

Example of optimized code (1D DP):

public int coinChange(int[] coins, int amount) {
    int[] dp = new int[amount+1];
    Arrays.fill(dp, Integer.MAX_VALUE-1);
    dp[0] = 0;
    for (int coin : coins) {
        for (int j = coin; j <= amount; j++) {
            dp[j] = Math.min(dp[j], 1 + dp[j - coin]);
        }
    }
    return dp[amount] == Integer.MAX_VALUE-1 ? -1 : dp[amount];
}

Note: This optimized version is more efficient in space and has the same time complexity.

Strengths:

• The student has provided both a recursive solution (commented out) and an iterative DP solution, showing a good understanding of the problem and the transition from naive to optimized approach.
• The DP solution for Coin Change is correctly implemented with proper initialization and recurrence.
• The code is clean and well-commented.

Areas for Improvement:

• For the Coin Change problem, the space complexity can be optimized from O(n * amount) to O(amount) by using a 1D DP array. This is because when building the DP table for the next coin, we only need the previous row (or the current state). Here's how it can be done:

  int[] dp = new int[amount+1];
  Arrays.fill(dp, Integer.MAX_VALUE-1);
  dp[0] = 0;
  for (int coin : coins) {
      for (int j = coin; j <= amount; j++) {
          dp[j] = Math.min(dp[j], 1 + dp[j - coin]);
      }
  }
  

• The use of Integer.MAX_VALUE-1 is to avoid overflow when adding 1. However, it would be better to use a large number that is definitely bigger than the maximum possible coins (which is at most 10000) instead of Integer.MAX_VALUE-1 to avoid any unintended issues. For example, we can set the initial value to amount+1 (since the maximum coins cannot exceed amount) which is safe.
• The solution currently initializes the first row (for i=0) to Integer.MAX_VALUE-1 for all amounts greater than 0. This is correct, but note that in the 1D optimization, we can initialize the dp array with amount+1 (which is a value that is impossible to achieve but safe) and then set dp[0]=0.
• The student should also consider edge cases, such as when amount is 0 (which returns 0). The current solution handles it correctly because dp[0][0]=0 and for i=0 and j>0, it is set to a large value.

Overall, the solution for Coin Change is correct and efficient, but can be optimized in space.