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567. Permutation in String

MediumHash TableTwo PointersStringSliding Window
Leetcode Link

Problem Description

Given two strings s1 and s2, your task is to determine whether s2 contains a permutation of s1. In other words, you must check if any substring of s2 has the same characters as s1, in any order. The function should return true if at least one permutation of s1 is a substring of s2, otherwise, it should return false.

Intuition

The intuition behind the solution is to use a sliding window approach along with character counting. We want to slide a window of size equal to the length of s1 over s2 and check if the characters inside this window form a permutation of s1. The key idea is to avoid recomputing the frequency of characters in the window from scratch each time we slide the window; instead, we can update the count based on the character that is entering and the character that is leaving the window.

To implement this, we use a counter data structure to keep track of the difference between the number of occurrences of each character in the current window and the number of occurrences of each character in s1. Initially, the counter is set by decrementing for s1 characters and incrementing for the first window in s2. We can then iterate through s2, moving the window to the right by incrementing the count for the new character and decrementing for the character that's no longer in the window.

The difference count is the sum of the non-zero values in the counter. If at any point the difference count is zero, it means the current window is a permutation of s1, and we return true. If we reach the end of s2 without finding such a window, we return false.

Learn more about Two Pointers and Sliding Window patterns.

Solution Approach

The problem is solved efficiently by using the sliding window technique, coupled with a character counter that keeps track of the frequencies of characters within the window.

Here are the key steps of the algorithm:

  1. Initialize a Counter object that will track the frequency difference of characters between s1 and the current window of s2.
  2. Set up the initial count by decrementing for each character in s1 and incrementing for each character in the first window of s2.
  3. Calculate the initial difference count, which is the sum of non-zero counts in the Counter. This represents how many characters' frequencies do not match between s1 and the current window.
  4. Start traversing s2 with a window size of s1. For each step, do the following:
    • If the difference count is zero, return true.
    • Update the Counter by incrementing the count for the new character entering the window, and decrementing the count for the character leaving the window.
    • Adjust the difference count if the updated character counts change from zero to non-zero, or vice versa.
  5. If the loop completes without the difference count reaching zero, return false.

The implementation takes O(n + m) time where n is the length of s1 and m is the length of s2. The space complexity is O(1) since the counter size is limited to the number of possible characters, which is constant.

The key data structures and patterns used in this solution are:

  • Counter from the Python collections module to keep track of frequencies of characters.
  • Sliding window technique to efficiently inspect substrings of s2 without re-counting characters each time.
  • Two-pointers pattern to represent the current window's start and end within s2.

This approach effectively checks every possible window in s2 that could be a permutation of s1, doing so in a manner that only requires a constant amount of work for each move of the window.

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Example Walkthrough

Let's consider an example to illustrate the solution approach:

Suppose we have s1 = "abc" and s2 = "eidbacoo". We are tasked with determining if s2 contains a permutation of s1.

  1. First, we initialize the Counter object for s1 which would look like Counter({'a':1, 'b':1, 'c':1}) as each character in s1 occurs once.
  2. Next, we look at the first window of s2 with the same length as s1, which is eid. We initialize another Counter for this window, resulting in Counter({'e':1, 'i':1, 'd':1}).
  3. Now, we compute the initial difference count by comparing our two Counter objects. For characters e, i, and d the count increments as they appear in s2 but not s1. For characters a, b, and c, the counts decrement for their presence in s1 but absence in the initial window of s2. The sum of non-zero counts is 6, as we have three characters in s1 that are not in the window and three characters in the window that are not in s1.
  4. We start sliding the window in s2 to the right, one character at a time. The next window is idb. We increment the count for b (as it enters the window) and decrement the count for e (as it exits). Now the Counter updates, and we recalculate the difference count. Characters i and d still contribute to the difference count, but b does not anymore because it matches with s1.
  5. Continue sliding the window to the right to the window dba, updating the Counter by incrementing for a and decrementing for i. The counter is now matched for a and b, but not for d.
  6. Proceed to the window bac. Increment for c and decrement for d. Now the Counter should match s1 completely, which means the difference count will be 0.
  7. As the difference count is 0, it indicates that the bac window is a permutation of s1. Therefore, we return true.

By using the sliding window and the Counter, we moved through s2 efficiently, avoiding recalculating the frequency of characters from scratch. We found that s2 contains a permutation of s1, demonstrating the solution approach effectively.

Solution Implementation

1from collections import Counter
2
3class Solution:
4    def check_inclusion(self, pattern: str, text: str) -> bool:
5        # Calculate the length of both the pattern and text
6        pattern_length, text_length = len(pattern), len(text)
7      
8        # If the pattern is longer than the text, the inclusion is not possible
9        if pattern_length > text_length:
10            return False
11      
12        # Initialize a counter for the characters in both strings
13        char_counter = Counter()
14      
15        # Decrease the count for pattern characters and increase for the first window in text
16        for pattern_char, text_char in zip(pattern, text[:pattern_length]):
17            char_counter[pattern_char] -= 1
18            char_counter[text_char] += 1
19      
20        # Calculate the number of characters that are different
21        diff_count = sum(x != 0 for x in char_counter.values())
22      
23        # If no characters are different, we found an inclusion
24        if diff_count == 0:
25            return True
26      
27        # Slide the window over text, one character at a time
28        for i in range(pattern_length, text_length):
29            # Get the character that will be removed from the window and the one that will be added
30            char_out = text[i - pattern_length]
31            char_in = text[i]
32          
33            # Update diff_count if the incoming character impacts the balance
34            if char_counter[char_in] == 0:
35                diff_count += 1
36            char_counter[char_in] += 1
37            if char_counter[char_in] == 0:
38                diff_count -= 1
39          
40            # Update diff_count if the outgoing character impacts the balance
41            if char_counter[char_out] == 0:
42                diff_count += 1
43            char_counter[char_out] -= 1
44            if char_counter[char_out] == 0:
45                diff_count -= 1
46          
47            # If no characters are different, we have found an inclusion
48            if diff_count == 0:
49                return True
50      
51        # If inclusion has not been found by the end of the text, return False
52        return False
53
1class Solution {
2    public boolean checkInclusion(String s1, String s2) {
3        int length1 = s1.length();
4        int length2 = s2.length();
5      
6        // If the first string is longer than the second string,
7        // it's not possible for s1 to be a permutation of s2.
8        if (length1 > length2) {
9            return false;
10        }
11      
12        // Array to hold the difference in character counts between s1 and s2.
13        int[] charCountDelta = new int[26];
14      
15        // Populate the array with initial counts
16        for (int i = 0; i < length1; ++i) {
17            charCountDelta[s1.charAt(i) - 'a']--;
18            charCountDelta[s2.charAt(i) - 'a']++;
19        }
20      
21        // Counts the number of characters with non-zero delta counts.
22        int nonZeroCount = 0;
23        for (int count : charCountDelta) {
24            if (count != 0) {
25                nonZeroCount++;
26            }
27        }
28      
29        // If all deltas are zero, s1 is a permutation of the first part of s2.
30        if (nonZeroCount == 0) {
31            return true;
32        }
33      
34        // Slide the window of length1 through s2
35        for (int i = length1; i < length2; ++i) {
36            int charLeft = s2.charAt(i - length1) - 'a'; // Character going out of the window
37            int charRight = s2.charAt(i) - 'a'; // Character coming into the window
38          
39            // Update counts for the exiting character
40            if (charCountDelta[charRight] == 0) {
41                nonZeroCount++;
42            }
43            charCountDelta[charRight]++;
44            if (charCountDelta[charRight] == 0) {
45                nonZeroCount--;
46            }
47          
48            // Update counts for the entering character
49            if (charCountDelta[charLeft] == 0) {
50                nonZeroCount++;
51            }
52            charCountDelta[charLeft]--;
53            if (charCountDelta[charLeft] == 0) {
54                nonZeroCount--;
55            }
56          
57            // If all deltas are zero, s1's permutation is found in s2.
58            if (nonZeroCount == 0) {
59                return true;
60            }
61        }
62      
63        // If we reach here, no permutation of s1 is found in s2.
64        return false;
65    }
66}
67
1class Solution {
2public:
3    // This function checks if s1's permutation is a substring of s2
4    bool checkInclusion(string s1, string s2) {
5        int len1 = s1.size(), len2 = s2.size();
6      
7        // If length of s1 is greater than s2, permutation is not possible
8        if (len1 > len2) {
9            return false;
10        }
11
12        // Vector to store character counts
13        vector<int> charCount(26, 0);
14      
15        // Initialize the character count vector with the first len1 characters
16        for (int i = 0; i < len1; ++i) {
17            --charCount[s1[i] - 'a']; // Decrement for characters in s1
18            ++charCount[s2[i] - 'a']; // Increment for characters in the first window of s2
19        }
20
21        // Calculate the difference count
22        int diffCount = 0;
23        for (int count : charCount) {
24            if (count != 0) {
25                ++diffCount;
26            }
27        }
28      
29        // If diffCount is zero, a permutation exists in the first window
30        if (diffCount == 0) {
31            return true;
32        }
33
34        // Slide the window over s2 and update the counts and diffCount
35        for (int i = len1; i < len2; ++i) {
36            int index1 = s2[i - len1] - 'a'; // Index for the old character in the window
37            int index2 = s2[i] - 'a'; // Index for the new character in the window
38
39            // Before updating charCount for the new character
40            if (charCount[index2] == 0) {
41                ++diffCount;
42            }
43            ++charCount[index2]; // Include the new character in the window
44
45            // After updating charCount for the new character
46            if (charCount[index2] == 0) {
47                --diffCount;
48            }
49
50            // Before updating charCount for the old character
51            if (charCount[index1] == 0) {
52                ++diffCount;
53            }
54            --charCount[index1]; // Remove the old character as we move the window
55
56            // After updating charCount for the old character
57            if (charCount[index1] == 0) {
58                --diffCount;
59            }
60
61            // If the diffCount is zero after the updates, a permutation is found
62            if (diffCount == 0) {
63                return true;
64            }
65        }
66
67        // No permutation was found
68        return false;
69    }
70};
71
1function checkInclusion(s1: string, s2: string): boolean {
2    // If s1 is longer than s2, it's impossible for s1 to be a permutation of s2.
3    if (s1.length > s2.length) {
4        return false;
5    }
6
7    // Helper function to convert characters into zero-based indices
8    function charToIndex(char: string): number {
9        return char.charCodeAt(0) - 'a'.charCodeAt(0);
10    }
11
12    // Helper function to check if both character frequency arrays match
13    function doArraysMatch(freqArray1: number[], freqArray2: number[]): boolean {
14        for (let i = 0; i < 26; i++) {
15            if (freqArray1[i] !== freqArray2[i]) {
16                return false;
17            }
18        }
19        return true;
20    }
21
22    const s1Length = s1.length;
23    const s2Length = s2.length;
24    // Arrays to store the frequency of each letter in s1 and the current window in s2
25    const freqArray1 = new Array(26).fill(0);
26    const freqArray2 = new Array(26).fill(0);
27
28    // Filling freqArray1 with frequencies of letters in s1
29    for (const char of s1) {
30        const index = charToIndex(char);
31        freqArray1[index]++;
32    }
33
34    // Filling freqArray2 with frequencies of the first window of s2 with size equal to s1 length
35    for (let i = 0; i < s1Length; i++) {
36        const index = charToIndex(s2[i]);
37        freqArray2[index]++;
38    }
39
40    // Sliding window to check each substring in s2
41    for (let left = 0, right = s1Length; right < s2Length; left++, right++) {
42        // Check if the current window is a permutation of s1
43        if (doArraysMatch(freqArray1, freqArray2)) {
44            return true;
45        }
46
47        // Slide the window forward: remove the left character and add the right character
48        const leftIndex = charToIndex(s2[left]);
49        const rightIndex = charToIndex(s2[right]);
50        freqArray2[leftIndex]--;
51        freqArray2[rightIndex]++;
52    }
53
54    // Check the last window after the loop
55    return doArraysMatch(freqArray1, freqArray2);
56}
57

Time and Space Complexity

Time Complexity

The time complexity of the provided code is O(n + m), where n is the length of s1 and m is the length of s2. Here's why:

  1. zip(s1, s2) takes O(n) time to iterate through the elements of the shorter string, which is s1 in this case as we return False immediately if s1 is longer than s2.
  2. The sum(x != 0 for x in cnt.values()) takes O(1) time since there can be at most 26 characters (assuming lowercase English letters), so the number of different characters is constant.
  3. The main loop runs from n to m, which executes m - n + 1 times (inclusive of n). Each iteration of the loop has a constant number of operations that do not depend on the size of n or m. Therefore, this part also takes O(m) time.
  4. Combining these parts, we get a total time complexity of O(n + m).

Space Complexity

The space complexity of the code is O(1) because the cnt counter will contain at most 26 key-value pairs (if we are considering the English alphabet). The number of keys in cnt does not grow with the size of the input strings s1 and s2, thus it is a constant space overhead.

Learn more about how to find time and space complexity quickly using problem constraints.

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