What is the time complexity of Partition String Into Substrings With Values at Most K?

The optimal solution for Partition String Into Substrings With Values at Most K runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only make a single pass through the string, and the space complexity is O(1) since we are using a fixed amount of extra space.

What is the brute force solution for Partition String Into Substrings With Values at Most K?

The brute force approach for Partition String Into Substrings With Values at Most K is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible substrings of the input string and checking if their integer values are less than or equal to k. This method is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Partition String Into Substrings With Values at Most K?

Partition String Into Substrings With Values at Most K uses the following concepts: String, Dynamic Programming, Greedy. The recommended patterns to study are: Greedy, Two Pointers.

What are the interview tips for Partition String Into Substrings With Values at Most K?

Clearly explain your thought process and approach before coding. Consider edge cases, such as single digits and maximum values for k. Optimize your solution step by step, explaining why each optimization is necessary.

What are common mistakes when solving Partition String Into Substrings With Values at Most K?

Not checking if a single digit exceeds k and returning -1 immediately. Failing to reset the current substring value correctly when it exceeds k.

#2522

Partition String Into Substrings With Values at Most K

Medium

StringDynamic ProgrammingGreedyGreedyTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal solution uses a greedy approach to minimize the number of substrings. It builds substrings from the left and checks their values against k, ensuring that we only create valid partitions.

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Algorithm

6 steps

1Step 1: Initialize a counter for substrings and a variable to store the current substring value.
2Step 2: Iterate through each character in the string.
3Step 3: For each character, update the current substring value by multiplying the existing value by 10 and adding the integer value of the current character.
4Step 4: If the current substring value exceeds k, increment the substring counter, reset the current value to the current character's value.
5Step 5: If any character itself is greater than k, return -1.
6Step 6: After the loop, account for the last substring and return the counter.

solution.py18 lines

1# Full working Python code
2
3def minPartitions(s, k):
4    count = 0
5    current_value = 0
6    for char in s:
7        digit = int(char)
8        if digit > k:
9            return -1
10        current_value = current_value * 10 + digit
11        if current_value > k:
12            count += 1
13            current_value = digit
14    return count + (1 if current_value > 0 else 0)
15
16# Example usage
17print(minPartitions('165462', 60))  # Output: 4
18

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Complexity note: The time complexity is O(n) because we only make a single pass through the string, and the space complexity is O(1) since we are using a fixed amount of extra space.

1If any single digit is greater than k, a valid partition is impossible.
2The greedy approach ensures we minimize the number of substrings by always trying to extend the current substring until it exceeds k.

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