You are given a 0-indexed string s, a string a, a string b, and an integer k.
An index i is beautiful if:
0 <= i <= s.length - a.lengths[i..(i + a.length - 1)] == aj such that:
0 <= j <= s.length - b.lengths[j..(j + b.length - 1)] == b|j - i| <= kReturn the array that contains beautiful indices in sorted order from smallest to largest.
Example 1:
Input: s = "isawsquirrelnearmysquirrelhouseohmy", a = "my", b = "squirrel", k = 15 Output: [16,33] Explanation: There are 2 beautiful indices: [16,33]. - The index 16 is beautiful as s[16..17] == "my" and there exists an index 4 with s[4..11] == "squirrel" and |16 - 4| <= 15. - The index 33 is beautiful as s[33..34] == "my" and there exists an index 18 with s[18..25] == "squirrel" and |33 - 18| <= 15. Thus we return [16,33] as the result.
Example 2:
Input: s = "abcd", a = "a", b = "a", k = 4 Output: [0] Explanation: There is 1 beautiful index: [0]. - The index 0 is beautiful as s[0..0] == "a" and there exists an index 0 with s[0..0] == "a" and |0 - 0| <= 4. Thus we return [0] as the result.
Constraints:
1 <= k <= s.length <= 1051 <= a.length, b.length <= 10s, a, and b contain only lowercase English letters.This approach involves directly comparing each element with every other element. This method solves the problem by exploring all possible combinations without utilizing any advanced optimization techniques.
The C code iterates over every pair of elements in the array via nested loops, and checks if their sum matches the target. Upon finding a match, it prints the pair.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n^2)
Space Complexity: O(1)
This approach leverages a hash map to store the elements and checks for the complement required to reach the target as it iterates through the array. This significantly reduces the number of checks needed compared to the brute force method.
The C code creates a simple hash table using linked lists to store previous numbers, allowing for efficient look-up of complements during iteration, thus reducing overall complexity.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n)
Space Complexity: O(n)
| Approach | Complexity |
|---|---|
| Brute Force Approach | Time Complexity: O(n^2) |
| Hash Map Approach | Time Complexity: O(n) |
Knuth–Morris–Pratt KMP - Find the Index of the First Occurrence in a String - Leetcode 28 - Python • NeetCode • 152,214 views views
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