How do you solve Construct Smallest Number From DI String on LeetCode?

Construct Smallest Number From DI String (LeetCode #2375) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a stack allows us to efficiently manage the order of digits based on the pattern.

What is the brute force solution for Construct Smallest Number From DI String?

The brute force approach for Construct Smallest Number From DI String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible permutations of the digits '1' to '9' and checking which ones satisfy the given pattern of 'I' and 'D'. This method guarantees that we will find the smallest valid number, but it is inefficient due to the large number of permutations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Construct Smallest Number From DI String?

Construct Smallest Number From DI String uses the following concepts: String, Backtracking, Stack, Greedy. The recommended patterns to study are: Stack, Greedy.

What are the interview tips for Construct Smallest Number From DI String?

Always clarify the constraints and edge cases with the interviewer. Explain your thought process clearly while coding to show your understanding. Practice writing clean and efficient code to minimize time spent debugging.

What are common mistakes when solving Construct Smallest Number From DI String?

Students often forget to handle the last character of the pattern correctly. Not considering edge cases where the pattern is all 'I's or all 'D's.

#2375

Construct Smallest Number From DI String

Medium

StringBacktrackingStackGreedyStackGreedy

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 leverages a greedy approach by constructing the result string based on the pattern directly. We can use a stack to manage the digits and ensure that we maintain the required order without generating all permutations.

⚙️

Algorithm

4 steps

1Step 1: Initialize an empty stack and a result string.
2Step 2: Iterate through the pattern, pushing digits onto the stack.
3Step 3: When encountering 'I', pop from the stack to the result to maintain the increasing order.
4Step 4: At the end, pop any remaining digits in the stack to the result.

solution.py10 lines

1def smallestNumber(pattern):
2    stack = []
3    result = ''
4    n = len(pattern)
5    for i in range(n + 1):
6        stack.append(str(i + 1))
7        if i == n or pattern[i] == 'I':
8            while stack:
9                result += stack.pop()
10    return result

ℹ

Complexity note: The time complexity is O(n) because we iterate through the pattern and push/pop from the stack, which takes constant time. The space complexity is O(n) due to the stack storing up to n elements.

1Using a stack allows us to efficiently manage the order of digits based on the pattern.
2Greedy approaches can often yield optimal solutions without the need for exhaustive search.

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