567. Permutation in String

Permutation in String

Difficulty: Medium

Topics: Hash Table, Two Pointers, String, Sliding Window

Given two strings s1 and s2, return true if s2 contains a permutation[^1] of s1, or false otherwise.

In other words, return true if one of s1‘s permutations is the substring of s2.

Example 1:

• Input: s1 = “ab”, s2 = “eidbaooo”
• Output: true
• Explanation: s2 contains one permutation of s1 (“ba”).

Example 2:

• Input: s1 = “ab”, s2 = “eidboaoo”
• Output: false

Constraints:

• <code>1 <= s1.length, s2.length <= 10⁴
• s1 and s2 consist of lowercase English letters.

Hint:

1. Obviously, brute force will result in TLE. Think of something else.
2. How will you check whether one string is a permutation of another string?
3. One way is to sort the string and then compare. But, Is there a better way?
4. If one string is a permutation of another string then they must have one common metric. What is that?
5. Both strings must have same character frequencies, if one is permutation of another. Which data structure should be used to store frequencies?
6. What about hash table? An array of size 26?

[^1]: Permutation A permutation is a rearrangement of all the characters of a string.

Solution:

We can use the sliding window technique with a frequency counter, rather than trying every possible substring permutation (which would be too slow for large inputs).

Key Idea:

1. If a permutation of s1 is a substring of s2, both strings must have the same character frequencies for the letters present in s1.
2. We can use a sliding window of length s1 over s2 to check if the substring within the window is a permutation of s1.

Steps:

• Create a frequency counter for s1 (i.e., count the occurrences of each character).
• Slide a window over s2 of size equal to s1 and check if the frequency of characters in the window matches the frequency of characters in s1.
• If the frequencies match, it means that the substring in the window is a permutation of s1.

Algorithm:

1. Create a frequency array (count1) for s1.
2. Use another frequency array (count2) for the current window in s2.
3. Slide the window over s2 and update the frequency array for the window as you move.
4. If the frequency arrays match at any point, return true.
5. If you slide through all of s2 without finding a match, return false.

Let’s implement this solution in PHP: 567. Permutation in String

<?php
/**
 * @param String $s1
 * @param String $s2
 * @return Boolean
 */
function checkInclusion($s1, $s2) {
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Example usage
$s1 = "ab";
$s2 = "eidbaooo";
echo checkInclusion($s1, $s2) ? "true" : "false"; // Output: true
?>

Explanation:

1. Frequency Arrays:
   • We use two arrays (count1 for s1 and count2 for the sliding window in s2) of size 26, corresponding to the 26 lowercase letters.
   • Each array keeps track of how many times each character appears.
2. Sliding Window:
   • We first initialize the frequency of the first window (the first len1 characters of s2) and compare it to count1.
   • Then, we slide the window one character at a time. For each step, we update the frequency array by:
     • Adding the character that is entering the window.
     • Removing the character that is leaving the window.
   • After each slide, we compare the two frequency arrays.
3. Efficiency:
   • Instead of recalculating the frequency for every new window, we efficiently update the frequency array by adjusting just two characters.
   • This gives an overall time complexity of O(n), where n is the length of s2.

Time and Space Complexity:

• Time Complexity: O(n) where n is the length of s2. We iterate over s2 once, updating the frequency arrays in constant time.
• Space Complexity: O(1), as we are only using fixed-size arrays (size 26) to store the frequency counts of characters.

This approach ensures that the solution is efficient even for larger inputs.

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