What is the time complexity of Partition Array Into Two Arrays to Minimize Sum Difference?

The optimal solution for Partition Array Into Two Arrays to Minimize Sum Difference runs in O(n * target) time and uses O(target) space. The time complexity is O(n * target) where n is the number of elements and target is half of the total sum. The space complexity is O(target) for the DP array.

What is the brute force solution for Partition Array Into Two Arrays to Minimize Sum Difference?

The brute force approach for Partition Array Into Two Arrays to Minimize Sum Difference is Brute Force, with O(2^n * n) time complexity and O(1) space complexity. The brute force approach involves generating all possible partitions of the array into two groups and calculating the sum difference for each partition. This is straightforward but inefficient due to the exponential number of combinations. The optimal Optimal Solution approach improves this to O(n * target).

What are the interview tips for Partition Array Into Two Arrays to Minimize Sum Difference?

Clearly explain your thought process and how you arrived at each step. Consider edge cases, such as all elements being the same or having negative values. Practice dynamic programming problems to get comfortable with state transitions.

What are common mistakes when solving Partition Array Into Two Arrays to Minimize Sum Difference?

Not considering negative numbers and their impact on sums. Overlooking the need to only consider combinations of exactly n elements.

#2035

Partition Array Into Two Arrays to Minimize Sum Difference

Hard

ArrayTwo PointersBinary SearchDynamic ProgrammingBit ManipulationSortingOrdered SetBitmaskDynamic ProgrammingSubset Sum Problem

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(2^n * n)	O(n * target)
Space	O(1)	O(target)

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Intuition

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

The optimal solution uses dynamic programming to explore possible sums that can be achieved with half of the elements. It focuses on finding the closest sum to half of the total sum, which minimizes the difference between the two partitions.

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Algorithm

4 steps

1Step 1: Calculate the total sum of the array and determine the target sum as total_sum / 2.
2Step 2: Use a dynamic programming array to track achievable sums with n elements from the array.
3Step 3: Iterate through the nums array and update the DP array for possible sums.
4Step 4: Find the maximum achievable sum that is less than or equal to the target sum, and calculate the minimum difference.

solution.py12 lines

1def minAbsDifference(nums):
2    n = len(nums) // 2
3    total_sum = sum(nums)
4    target = total_sum // 2
5    dp = [False] * (target + 1)
6    dp[0] = True
7    for num in nums:
8        for j in range(target, num - 1, -1):
9            dp[j] = dp[j] or dp[j - num]
10    for i in range(target, -1, -1):
11        if dp[i]:
12            return abs(total_sum - 2 * i)

ℹ

Complexity note: The time complexity is O(n * target) where n is the number of elements and target is half of the total sum. The space complexity is O(target) for the DP array.

1The problem can be viewed as a variation of the subset-sum problem.
2Finding a partition that minimizes the difference between sums is akin to balancing weights on a scale.

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