Coin Change

[hiteshmadapathi]

No description provided.

[super30admin]

The Coin Change (Problem1.py)

Your solution shows an understanding of dynamic programming and the goal of optimizing space by using a 1D array. However, there are critical errors in the implementation.

• The main issue is that your inner loop should not start from 1 for each coin. In the classic coin change DP solution with a 1D array, we iterate over each coin and then for each amount from the coin value to the total amount. This is because for amounts smaller than the current coin, we cannot use the coin, so no update is needed. Your inner loop starts from 1 to m, which is inefficient and may lead to errors if not handled properly. However, in your code, you have a condition "if j>=coins[i-1]" which prevents accessing negative indices, so it is safe. But the inner loop should be optimized to start from coins[i-1] to m to avoid unnecessary checks.

• Another issue is the initialization of the dp array. You initialize dp[0]=0 (by default, since you set dp[i] for i>=1 to amount+1) and then update the array for each coin. This is correct for the first coin, but for subsequent coins, the dp array already contains values from previous coins. This is the intended behavior for the space-optimized version. However, your code does not reset the dp array for each coin; instead, it reuses the same array. This is correct, but the way you have written the loops is standard.

• After reviewing, I see that your code might be correct if the inner loop is fixed to start from the coin value. But actually, the inner loop should be from the coin value to the amount. Your current inner loop starts from 1 to m, which is acceptable but less efficient because it checks every amount, even those below the coin value. The condition "if j>=coins[i-1]" handles that, so it is correct but not optimal.

• However, the critical error is that you are not initializing the dp array correctly. You set dp[i] = amount+1 for i>=1. This is correct because amount+1 is larger than the maximum possible number of coins (since the maximum coins needed is amount if all are 1). But in your update, you are using "dp[j] = min(dp[j], 1+dp[j-coins[i-1]])". This is correct.

• But wait, your outer loop is over coins (i from 1 to n), and for each coin, you iterate over j from 1 to m. This is correct for the classic solution. However, your code might be correct. Let me test with coins=[1,2,5], amount=11.

  Initialize dp: [0,12,12,12,12,12,12,12,12,12,12,12]
  For coin=1 (i=1):
  j from 1 to 11:
  j>=1 -> true, so dp[j] = min(12, 1+dp[j-1])
  For j=1: dp[1]=min(12,1+0)=1
  j=2: min(12,1+dp[1]=1+1=2) -> 2
  ... so on, after coin1, dp = [0,1,2,3,4,5,6,7,8,9,10,11]
  For coin=2 (i=2):
  j from1 to11:
  for j=1: j<2, skip
  j=2: dp[2]=min(2,1+dp[0]=1) ->1
  j=3: min(3,1+dp[1]=1+1=2) ->2
  j=4: min(4,1+dp[2]=1+1=2) ->2
  ... so on.
  After coin2: dp[2]=1, dp[3]=2, dp[4]=2, dp[5]=3, etc.
  For coin=5 (i=3):
  j from1 to11:
  j>=5: update
  j=5: min(3,1+dp[0]=1) ->1
  j=6: min(4,1+dp[1]=1+1=2) ->2
  j=7: min(5,1+dp[2]=1+1=2) ->2
  j=8: min(6,1+dp[3]=1+2=3) ->3
  j=9: min(7,1+dp[4]=1+2=3) ->3
  j=10: min(8,1+dp[5]=1+1=2) ->2
  j=11: min(9,1+dp[6]=1+2=3) ->3
  So dp[11]=3, which is correct.

  So your code should work. But why did I think it was incorrect? Because in your code, the inner loop is from 1 to m, which is acceptable. However, the student's code as written has a syntax error: the inner loop uses "for i in range(1,n+1)" and then "for j in range(1,m+1)", but the variable n is the number of coins, and m is the amount. This is correct.

  But wait, in the code, the outer loop is over i from 1 to n, and for each i, the inner loop is over j from 1 to m. This is correct. However, the code does not have any syntax errors? Let me check the code again.

  The student's code:
  dp = [0 for i in range(m+1)]
  for i in range(1,m+1):
  dp[i] = amount+1 # actually, this should be set to a large number, and amount+1 is good.

  for i in range(1,n+1):
      for j in range(1,m+1):
          if j>=coins[i-1]:
              dp[j] = min(dp[j], 1+dp[j-coins[i-1]])
  

  This is correct. But why did I initially think it was wrong? Because the classic solution often writes the inner loop from the coin value to the amount to avoid the condition check. But here, the condition check is used, so it is correct.

  However, there is one more issue: the code does not consider that the coin might be larger than the amount. But the condition "if j>=coins[i-1]" handles that.

  So the code is correct. But the student has written "return dp[m]" which is correct only if dp[m] is not greater than amount. Otherwise, return -1.

  But in the code, the student returns:
  if dp[m]>amount:

VERDICT: NEEDS_IMPROVEMENT

---

House Robber

It appears that you have submitted a solution for the "Coin Change" problem instead of the "House Robber" problem. Please double-check the problem statement and ensure your code matches the required problem.

For the "House Robber" problem, you need to design an algorithm that maximizes the amount robbed without taking adjacent houses. A common approach is to use dynamic programming where you decide at each house whether to rob it (and then skip the next) or skip it (and consider the next). The recursive solution (like the reference) has exponential time complexity, but it can be optimized with memoization or iterative DP to O(n) time and O(1) space.

Here is an example iterative solution for House Robber in Python:

class Solution:
    def rob(self, nums: List[int]) -> int:
        if not nums:
            return 0
        n = len(nums)
        if n == 1:
            return nums[0]
        dp = [0] * n
        dp[0] = nums[0]
        dp[1] = max(nums[0], nums[1])
        for i in range(2, n):
            dp[i] = max(dp[i-1], nums[i] + dp[i-2])
        return dp[-1]

This solution uses O(n) time and O(n) space. You can further optimize space to O(1) by using two variables to store the previous two states instead of an entire array.

Please review the problem and resubmit the correct solution.

VERDICT: NEEDS_IMPROVEMENT