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

Find Beautiful Indices in the Given Array II (LeetCode #3008) 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 efficient string matching algorithms like KMP can drastically reduce the time complexity.

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

The optimal solution for Find Beautiful Indices in the Given Array II runs in O(n) time and uses O(n) space. The time complexity is O(n) because we efficiently find all occurrences of 'a' and 'b' using KMP, and then we only traverse the lists of indices once.

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

The brute force approach for Find Beautiful Indices in the Given Array II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible index in the string to see if it matches the substring 'a'. For each valid index of 'a', it then checks all possible indices for 'b' to see if they are within the allowed distance 'k'. The optimal Optimal Solution approach improves this to O(n).

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

Always discuss your thought process and approach before jumping into coding. Consider edge cases and constraints early on to avoid surprises later. Practice implementing string matching algorithms like KMP or Rabin-Karp as they are commonly used in string problems.

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

Students often forget to handle edge cases where indices might go out of bounds. Not using efficient string searching techniques can lead to timeouts in larger inputs.

#3008

Find Beautiful Indices in the Given Array II

Hard

Two PointersStringBinary SearchRolling HashString MatchingHash FunctionString MatchingTwo Pointers

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)

Using efficient string searching techniques like the KMP algorithm allows us to find all occurrences of 'a' and 'b' in linear time. This reduces the need for nested loops and significantly speeds up the process.

⚙️

Algorithm

3 steps

1Step 1: Use KMP or a similar algorithm to find all starting indices of substring 'a' in 's'.
2Step 2: Use KMP or a similar algorithm to find all starting indices of substring 'b' in 's'.
3Step 3: Use a two-pointer technique to check for each index of 'a' if there exists an index of 'b' such that the absolute difference is less than or equal to 'k'.

solution.py44 lines

1def kmp_search(s, pattern):
2    m, n = len(pattern), len(s)
3    lps = [0] * m
4    j = 0
5    i = 1
6    while i < m:
7        if pattern[i] == pattern[j]:
8            j += 1
9            lps[i] = j
10            i += 1
11        else:
12            if j != 0:
13                j = lps[j - 1]
14            else:
15                lps[i] = 0
16                i += 1
17    indices = []
18    i = 0
19    j = 0
20    while i < n:
21        if pattern[j] == s[i]:
22            i += 1
23            j += 1
24        if j == m:
25            indices.append(i - j)
26            j = lps[j - 1]
27        elif i < n and pattern[j] != s[i]:
28            if j != 0:
29                j = lps[j - 1]
30            else:
31                i += 1
32    return indices
33
34def beautiful_indices(s, a, b, k):
35    indices_a = kmp_search(s, a)
36    indices_b = kmp_search(s, b)
37    beautiful = []
38    j = 0
39    for i in indices_a:
40        while j < len(indices_b) and indices_b[j] < i - k:
41            j += 1
42        if j < len(indices_b) and abs(indices_b[j] - i) <= k:
43            beautiful.append(i)
44    return sorted(set(beautiful))

ℹ

Complexity note: The time complexity is O(n) because we efficiently find all occurrences of 'a' and 'b' using KMP, and then we only traverse the lists of indices once.

1Using efficient string matching algorithms like KMP can drastically reduce the time complexity.
2The two-pointer technique helps in efficiently checking the distance condition without unnecessary comparisons.

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