How do you solve Split With Minimum Sum on LeetCode?

Split With Minimum Sum (LeetCode #2578) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting the digits helps in minimizing the sum.

What is the time complexity of Split With Minimum Sum?

The optimal solution for Split With Minimum Sum runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the sorting step, and the space complexity is O(n) because we store the digits in a list.

What is the brute force solution for Split With Minimum Sum?

The brute force approach for Split With Minimum Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible pairs of splits of the digits of the number and calculating their sums. This method is straightforward but inefficient, as it checks every possible combination. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Split With Minimum Sum?

Split With Minimum Sum uses the following concepts: Math, Greedy, Sorting. The recommended patterns to study are: Sorting, Greedy.

What are the interview tips for Split With Minimum Sum?

Explain your thought process clearly while coding. Consider edge cases, such as the smallest and largest inputs. Practice similar problems to recognize patterns in digit manipulation.

What are common mistakes when solving Split With Minimum Sum?

Not sorting the digits before distribution. Ignoring leading zeros when forming numbers.

#2578

Split With Minimum Sum

Easy

MathGreedySortingSortingGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

The optimal solution leverages sorting the digits and then distributing them alternately between two numbers. This ensures that the smaller digits contribute to the overall sum in a balanced manner, minimizing the final sum.

⚙️

Algorithm

4 steps

1Step 1: Convert the number into a list of its digits.
2Step 2: Sort the digits in non-decreasing order.
3Step 3: Distribute the digits between num1 and num2 alternately.
4Step 4: Convert num1 and num2 back to integers and return their sum.

solution.py12 lines

1# Full working Python code
2
3def min_sum_split(num):
4    digits = sorted(str(num))
5    num1, num2 = '', ''
6    for i in range(len(digits)):
7        if i % 2 == 0:
8            num1 += digits[i]
9        else:
10            num2 += digits[i]
11    return int(num1) + int(num2)
12

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, and the space complexity is O(n) because we store the digits in a list.

1Sorting the digits helps in minimizing the sum.
2Distributing digits alternately balances the contribution to both numbers.

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