What is the time complexity of Minimum Sum of Four Digit Number After Splitting Digits?

The optimal solution for Minimum Sum of Four Digit Number After Splitting Digits 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, while the space complexity is O(n) for storing the digits. This is efficient given the fixed size of n (4).

What is the brute force solution for Minimum Sum of Four Digit Number After Splitting Digits?

The brute force approach for Minimum Sum of Four Digit Number After Splitting Digits is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible pairs of two-digit numbers from the four digits of the input number and calculating their sums. This method is straightforward but inefficient as it explores all combinations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Sum of Four Digit Number After Splitting Digits?

Minimum Sum of Four Digit Number After Splitting Digits uses the following concepts: Math, Greedy, Sorting. The recommended patterns to study are: Sorting, Greedy Algorithms.

What are the interview tips for Minimum Sum of Four Digit Number After Splitting Digits?

Always think about edge cases, such as leading zeros. Communicate your thought process clearly when explaining your approach. Practice simplifying the problem to its core components before diving into code.

What are common mistakes when solving Minimum Sum of Four Digit Number After Splitting Digits?

Not considering leading zeros when forming new numbers. Overcomplicating the problem by trying to generate all combinations instead of focusing on the smallest digits.

#2160

Minimum Sum of Four Digit Number After Splitting Digits

Easy

MathGreedySortingSortingGreedy Algorithms

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Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

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

By sorting the digits and strategically pairing the smallest digits, we can construct the two smallest two-digit numbers efficiently. This minimizes the sum without needing to check every combination.

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Algorithm

4 steps

1Step 1: Extract the digits from the number and store them in an array.
2Step 2: Sort the array of digits.
3Step 3: Form two two-digit numbers using the smallest digits: new1 from the first and third smallest, new2 from the second and fourth smallest.
4Step 4: Return the sum of new1 and new2.

solution.py3 lines

1def minimumSum(num):
2    digits = sorted([int(d) for d in str(num)])
3    return (digits[0] * 10 + digits[2]) + (digits[1] * 10 + digits[3])

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) for storing the digits. This is efficient given the fixed size of n (4).

1Sorting the digits allows for optimal pairing to minimize the sum.
2Using the two smallest digits in the tens place of the two numbers minimizes their overall contribution.

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