Complete DP-1

[PrakarshKamal]

No description provided.

[super30admin]

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.