What is the time complexity of Splitting a String Into Descending Consecutive Values?

The optimal solution for Splitting a String Into Descending Consecutive Values runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only traverse the string once, and the space complexity is O(n) due to the recursion stack.

What is the brute force solution for Splitting a String Into Descending Consecutive Values?

The brute force approach for Splitting a String Into Descending Consecutive Values is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible ways to split the string into substrings and checking each combination to see if they meet the criteria of being in descending order with a difference of 1. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Splitting a String Into Descending Consecutive Values?

Splitting a String Into Descending Consecutive Values uses the following concepts: String, Backtracking, Enumeration. The recommended patterns to study are: Backtracking, Recursion.

What are the interview tips for Splitting a String Into Descending Consecutive Values?

Think about edge cases, such as strings with leading zeros. Clearly explain your thought process while coding to show your understanding. Practice writing recursive functions, as they are often used in backtracking problems.

What are common mistakes when solving Splitting a String Into Descending Consecutive Values?

Not handling leading zeros correctly, which can lead to incorrect numerical values. Forgetting to check if the previous number is valid before making a recursive call.

#1849

Splitting a String Into Descending Consecutive Values

Medium

StringBacktrackingEnumerationBacktrackingRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses backtracking with early stopping to avoid unnecessary checks. By keeping track of the last number used, we can efficiently determine if the next number can be part of a valid split.

⚙️

Algorithm

3 steps

1Step 1: Use a recursive function to explore possible splits of the string.
2Step 2: For each split, convert the substring to a number and check if it meets the descending order and difference criteria.
3Step 3: If a valid split is found, return true; otherwise, backtrack.

solution.py11 lines

1def canSplit(s):
2    def backtrack(start, prev):
3        if start == len(s):
4            return prev is not None
5        for end in range(start + 1, len(s) + 1):
6            num = int(s[start:end])
7            if (prev is None or (prev - num == 1 and num < prev)):
8                if backtrack(end, num):
9                    return True
10        return False
11    return backtrack(0, None)

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string once, and the space complexity is O(n) due to the recursion stack.

1Check for leading zeros when converting substrings to integers.
2Ensure the last number is always smaller than the previous one.

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