How do you solve Find Beautiful Indices in the Given Array I on LeetCode?

Find Beautiful Indices in the Given Array I (LeetCode #3006) 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 indices to track occurrences can significantly reduce the number of checks needed.

What is the time complexity of Find Beautiful Indices in the Given Array I?

The optimal solution for Find Beautiful Indices in the Given Array I runs in O(n) time and uses O(n) space. The time complexity is O(n) because we make a single pass to collect indices of 'a' and 'b', and then check conditions in linear time.

What is the brute force solution for Find Beautiful Indices in the Given Array I?

The brute force approach for Find Beautiful Indices in the Given Array I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible index for string 'a' and then searches for string 'b' within a range defined by 'k'. This method is straightforward but can be inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Find Beautiful Indices in the Given Array I?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases, like when 'a' and 'b' are the same or when 'k' is very large. Discuss your thought process and approach with the interviewer before coding.

What are common mistakes when solving Find Beautiful Indices in the Given Array I?

Not checking the bounds when accessing substrings. Forgetting to handle duplicates in the result.

#3006

Find Beautiful Indices in the Given Array I

Medium

Two PointersStringBinary SearchRolling HashString MatchingHash FunctionHash MapArray

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 approach uses a single pass to find all occurrences of 'a' and 'b', storing their indices. This allows us to efficiently check the conditions for beauty without nested loops.

⚙️

Algorithm

4 steps

1Step 1: Create two lists to store indices of occurrences for strings 'a' and 'b'.
2Step 2: Iterate through the string 's' once to fill these lists.
3Step 3: For each index in the list of 'a', check if there exists an index in the list of 'b' such that the absolute difference is less than or equal to 'k'.
4Step 4: If such a 'b' index exists, add the 'a' index to the result list.

solution.py12 lines

1def beautiful_indices(s, a, b, k):
2    result = []
3    len_a, len_b = len(a), len(b)
4    a_indices = [i for i in range(len(s) - len_a + 1) if s[i:i + len_a] == a]
5    b_indices = [j for j in range(len(s) - len_b + 1) if s[j:j + len_b] == b]
6    b_set = set(b_indices)
7    for i in a_indices:
8        for j in b_indices:
9            if abs(j - i) <= k:
10                result.append(i)
11                break
12    return sorted(set(result))

ℹ

Complexity note: The time complexity is O(n) because we make a single pass to collect indices of 'a' and 'b', and then check conditions in linear time.

1Using indices to track occurrences can significantly reduce the number of checks needed.
2Sorting the result helps in returning the indices in the required order.

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