Partition Equal Subset Sum

Clarify the problem:

• The problem requires us to determine whether it is possible to partition a given array into two subsets with equal sums.
• We need to return a boolean value indicating whether such a partition is possible.

Analyze the problem:

• Input: An array of positive integers.
• Output: A boolean value indicating whether the array can be partitioned into two subsets with equal sums.
• Constraints:
  • The length of the array is between 1 and 200.
  • The values in the array are between 1 and 100.

Design an algorithm:

• We can solve this problem using a dynamic programming approach.
• We need to determine whether there is a subset of the given array that sums up to half of the total sum of the array.
• If such a subset exists, it means we can partition the array into two subsets with equal sums.
• We can create a 1D boolean array dp of size equal to half of the total sum of the array.
• We iterate over the elements of the array and update the dp array.
• At each iteration, we update the dp array by marking dp[j] as True if dp[j - num] is True, where num is the current element.
• We start with dp[0] as True since an empty subset can have a sum of 0.
• Finally, we return dp[half], where half is half of the total sum of the array.

Explain your approach:

• We will implement a dynamic programming solution to determine whether the array can be partitioned into two subsets with equal sums.
• We initialize a boolean array dp of size equal to half of the total sum of the array.
• We set dp[0] as True since an empty subset can have a sum of 0.
• We iterate over the elements of the array.
• For each element num, we iterate from half down to num and update the dp array.
• We mark dp[j] as True if dp[j - num] is True.
• Finally, we return dp[half].

Write clean and readable code:

python

def canPartition(nums):
    total_sum = sum(nums)
    if total_sum % 2 != 0:
        return False

    half = total_sum // 2
    dp = [False] * (half + 1)
    dp[0] = True

    for num in nums:
        for j in range(half, num - 1, -1):
            dp[j] = dp[j] or dp[j - num]

    return dp[half]

Test your code:

• We can test the code with multiple test cases, including edge cases and corner cases, to ensure its correctness.
• For example:

  python

• # Example 1
  nums = [1, 5, 11, 5]
  print(canPartition(nums))
  # Output: True
  
  # Example 2
  nums = [1, 2, 3, 5]
  print(canPartition(nums))
  # Output: False
  
  # Example 3
  nums = [1, 2, 3, 4, 5, 6, 7]
  print(canPartition(nums))
  # Output: True
  

Optimize if necessary:

• The provided solution already has an optimal time complexity of O(n * sum), where n is the length of the array and sum is the total sum of the array.
• The space complexity is O(sum) since we use a 1D boolean array of size equal to half of the total sum of the array.

Handle error cases:

• If the total sum of the array is odd (total_sum % 2 != 0), it is not possible to partition the array into two subsets with equal sums. In such cases, we can return False.

Discuss complexity analysis:

• The time complexity of the solution is O(n * sum), where n is the length of the array and sum is the total sum of the array.
• This is because we iterate over the elements of the array and perform calculations for each element in the range from half to num.
• The space complexity is O(sum) since we use a 1D boolean array of size equal to half of the total sum of the array.
• The solution is efficient and optimal considering the requirements of the problem.