What is the brute force solution for Lexicographically Smallest String After Applying Operations?

The brute force approach for Lexicographically Smallest String After Applying Operations is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible strings by applying the operations in various sequences. Since the operations can be applied multiple times, we explore every combination to find the lexicographically smallest string. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Lexicographically Smallest String After Applying Operations?

Clearly explain your thought process and how you plan to generate possible strings. Be mindful of edge cases, such as when a or b are 0. Practice generating combinations of operations to build intuition.

What are common mistakes when solving Lexicographically Smallest String After Applying Operations?

Not considering the cyclic nature of digits when applying the add operation. Overlooking the need to check all rotations of the string.

#1625

Lexicographically Smallest String After Applying Operations

Medium

StringDepth-First SearchBreadth-First SearchEnumerationDepth-First SearchBreadth-First SearchEnumeration

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 the fact that we can generate all possible strings by focusing on the even and odd indices separately. By applying the add operation strategically and using rotations, we can minimize the number of generated strings and efficiently find the smallest one.

⚙️

Algorithm

3 steps

1Step 1: Generate all possible values for the odd indices by applying the add operation up to 10 times (since digits cycle from 0 to 9).
2Step 2: For each generated odd index configuration, rotate the string and check all configurations.
3Step 3: Keep track of the smallest string encountered.

solution.py18 lines

1# Full working Python code
2from itertools import product
3
4def lexicographically_smallest_string(s, a, b):
5    n = len(s)
6    odd_indices = [int(s[i]) for i in range(1, n, 2)]
7    even_indices = [s[i] for i in range(0, n, 2)]
8    smallest = s
9    for add_ops in product(range(10), repeat=len(odd_indices)):
10        new_odd = [(odd + a * op) % 10 for odd, op in zip(odd_indices, add_ops)]
11        new_string = ''.join(even_indices[i] + str(new_odd[i]) for i in range(len(new_odd)))
12        for rotation in range(n):
13            rotated = new_string[rotation:] + new_string[:rotation]
14            smallest = min(smallest, rotated)
15    return smallest
16
17# Example usage
18print(lexicographically_smallest_string('5525', 9, 2))

ℹ

Complexity note: The time complexity is O(n) because we efficiently generate possible strings by focusing on odd indices and then rotating them, leading to a linear growth in operations based on the string length.

1The operations can be applied in any order, leading to a combinatorial explosion of possibilities.
2Focusing on odd and even indices separately helps reduce the complexity.

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