What is the time complexity of Replace Question Marks in String to Minimize Its Value?

The optimal solution for Replace Question Marks in String to Minimize Its Value runs in O(n) time and uses O(n) space. The complexity is O(n) because we only traverse the string a couple of times: once for counting and once for replacing, making it linear in terms of input size.

What is the brute force solution for Replace Question Marks in String to Minimize Its Value?

The brute force approach for Replace Question Marks in String to Minimize Its Value is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will try every possible combination of replacing '?' with letters from 'a' to 'z'. This is straightforward but inefficient as we will check all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Replace Question Marks in String to Minimize Its Value?

Replace Question Marks in String to Minimize Its Value uses the following concepts: Hash Table, String, Greedy, Sorting, Heap (Priority Queue), Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Replace Question Marks in String to Minimize Its Value?

Always think about the constraints and how they affect your approach. Discuss your thought process clearly; explain why you're choosing a particular method. Practice identifying patterns in problems, as they can often lead to more efficient solutions.

What are common mistakes when solving Replace Question Marks in String to Minimize Its Value?

Not considering the lexicographical order when multiple strings have the same cost. Failing to efficiently count character frequencies, leading to unnecessary complexity.

#3081

Replace Question Marks in String to Minimize Its Value

Medium

Hash TableStringGreedySortingHeap (Priority Queue)CountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

Instead of generating all combinations, we can replace '?' with the least frequent letters in the string to minimize the cost. This approach is efficient and ensures we get the lexicographically smallest string.

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Algorithm

3 steps

1Step 1: Count the frequency of each character in the string, ignoring '?' characters.
2Step 2: For each '?', replace it with the character that has the lowest frequency, ensuring that we maintain the lexicographical order.
3Step 3: Construct the final string with the replacements.

solution.py13 lines

1# Full working Python code
2from collections import Counter
3
4def min_value_string(s):
5    count = Counter(c for c in s if c != '?')
6    result = list(s)
7    for i in range(len(result)):
8        if result[i] == '?':
9            # Find the character with the minimum frequency
10            min_char = min((chr(c) for c in range(ord('a'), ord('z') + 1)), key=lambda x: count[x])
11            result[i] = min_char
12            count[min_char] += 1
13    return ''.join(result)

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Complexity note: The complexity is O(n) because we only traverse the string a couple of times: once for counting and once for replacing, making it linear in terms of input size.

1The cost function depends on the frequency of characters, not their order.
2Replacing '?' with the least frequent characters minimizes the overall cost.

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