How do you solve Partition Equal Subset Sum on LeetCode?

Partition Equal Subset Sum (LeetCode #416) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * target) time complexity and O(target) space complexity. Key insight: The total sum must be even for a partition to be possible.

What is the time complexity of Partition Equal Subset Sum?

The optimal solution for Partition Equal Subset Sum runs in O(n * target) time and uses O(target) space. This complexity comes from iterating through the array and the target sum, leading to a more manageable polynomial time complexity.

What is the brute force solution for Partition Equal Subset Sum?

The brute force approach for Partition Equal Subset Sum is Brute Force, with O(2^n) time complexity and O(n) space complexity. The brute-force approach tries all possible combinations of subsets to see if any of them sum to half of the total sum of the array. This is like trying every possible way to split a pizza into two equal halves. The optimal Optimal Solution approach improves this to O(n * target).

What algorithm or data structure is used to solve Partition Equal Subset Sum?

Partition Equal Subset Sum uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Subset Sum Problem.

What are the interview tips for Partition Equal Subset Sum?

Always consider edge cases, such as arrays with a single element or all elements being the same. Explain your thought process clearly while coding, as it helps the interviewer understand your approach. If you get stuck, try to break the problem down into smaller parts or consider simpler examples.

What are common mistakes when solving Partition Equal Subset Sum?

Forgetting to check if the total sum is odd before proceeding. Not properly updating the DP array in reverse order, which can lead to incorrect results.

#416

Partition Equal Subset Sum

Medium

ArrayDynamic ProgrammingDynamic ProgrammingSubset Sum Problem

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(2^n)	O(n * target)
Space	O(n)	O(target)

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Intuition

Time O(n * target)Space O(target)

The optimal approach uses dynamic programming to build a table that tracks whether a subset sum can be achieved for each possible sum up to half of the total. It's like filling a backpack with items to reach a specific weight without exceeding it.

⚙️

Algorithm

4 steps

1Step 1: Calculate the total sum of the array. If it's odd, return false.
2Step 2: Create a boolean DP array of size (target + 1) initialized to false, with DP[0] = true (0 sum is always achievable).
3Step 3: Iterate through each number in the array and update the DP array from back to front to avoid overwriting results from the current iteration.
4Step 4: Return the value of DP[target] which indicates if the target sum can be achieved.

solution.py11 lines

1def canPartition(nums):
2    total = sum(nums)
3    if total % 2 != 0:
4        return False
5    target = total // 2
6    dp = [False] * (target + 1)
7    dp[0] = True
8    for num in nums:
9        for i in range(target, num - 1, -1):
10            dp[i] = dp[i] or dp[i - num]
11    return dp[target]

ℹ

Complexity note: This complexity comes from iterating through the array and the target sum, leading to a more manageable polynomial time complexity.

1The total sum must be even for a partition to be possible.
2Dynamic programming can significantly reduce the time complexity from exponential to polynomial.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.