Java

import java.util.Arrays;

/*
Given two strings s1 and s2, write a function to return true if s2 contains the permutation of s1.
In other words, one of the first string's permutations is the substring of the second string.

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
Note:
The input strings only contain lower case letters.
The length of both given strings is in range [1, 10,000].

 */

public class Permutation_in_String {

	public static void main(String[] args) {
		Permutation_in_String out = new Permutation_in_String();
		Solution s = out.new Solution();

		System.out.println(s.checkInclusion("ab", "eidbaooo"));
		System.out.println(s.checkInclusion("ab", "eidboaoo"));
	}

	public class Solution {
		public boolean checkInclusion(String s1, String s2) {
			if (s1.length() > s2.length()) {
				return false;
			}

			int window = s1.length();

			// N * N
			for(int i = 0; i + window - 1 < s2.length(); i++) {
				if (isPerm(s1, s2.substring(i, i + window))) {
					return true;
				}
			}

			return false;
		}

		// better permutation check: o(N)
		public boolean isPerm(String s1, String s2) {

			if(s1.length() != s2.length()) {
				return false;
			}

			int[] count = new int[256];

			// each char of s1: +1
			// each char of s2: -1
			// final check: permutation if all 0s array
			for(int i = 0; i < s1.length(); i++) {
				char s1char = s1.charAt(i);
				count[s1char] = count[s1char] + 1;

				char s2char = s2.charAt(i);
				count[s2char] = count[s2char] - 1;
			}

			for(int each: count) {
				if (each != 0) {
					return false;
				}
			}

			return true;
		}
	}

	// Time Limit Exceeded
	public class Solution2 {
		public boolean checkInclusion(String s1, String s2) {
			if (s1.length() > s2.length()) {
				return false;
			}

			int window = s1.length();

			// N * NlogN
			for(int i = 0; i + window - 1 < s2.length(); i++) {
				if (isPerm(s1, s2.substring(i, i + window))) {
					return true;
				}
			}

			return false;
		}

		// NlogN
		public boolean isPerm(String s1, String s2) {

			if(s1.length() != s2.length()) {
				return false;
			}

			char[] c1 = s1.toCharArray();
			char[] c2 = s2.toCharArray();

			Arrays.sort(c1); // NlogN
			Arrays.sort(c2);

			for(int i = 0; i < c1.length; i++) {
				if(c1[i] != c2[i]) {
					return false;
				}
			}
			return true;
		}
	}
}

// OJ: https://leetcode.com/problems/permutation-in-string/
// Time: O(B)
// Space: O(1)
class Solution {
public:
    bool checkInclusion(string a, string b) {
        if (a.size() > b.size()) return false;
        int N = a.size(), cnts[26] = {};
        for (char c : a) cnts[c - 'a']++;
        for (int i = 0; i < b.size(); ++i) {
            cnts[b[i] - 'a']--;
            if (i - N >= 0) cnts[b[i - N] - 'a']++;
            bool match = true;
            for (int j = 0; j < 26 && match; ++j) {
                if (cnts[j]) match = false;
            }
            if (match) return true;
        }
        return false;
    }
};

class Solution(object):
  def checkInclusion(self, s1, s2):
    """
    :type s1: str
    :type s2: str
    :rtype: bool
    """
    d = {}
    n = len(s1)
    for c in s1:
      d[c] = d.get(c, 0) + 1
    window = {}
    for i, c in enumerate(s2):
      window[c] = window.get(c, 0) + 1
      if i >= len(s1):
        window[s2[i - n]] = window.get(s2[i - n], 0) - 1
        if window[s2[i - n]] == 0:
          del window[s2[i - n]]
      if window == d:
        return True
    return False

Java

class Solution {
    public boolean checkInclusion(String s1, String s2) {
        if (s1 == null || s2 == null || s1.length() > s2.length())
            return false;
        if (s2.indexOf(s1) >= 0)
            return true;
        int[] counts1 = new int[26];
        int[] counts2 = new int[26];
        int length1 = s1.length(), length2 = s2.length();
        for (int i = 0; i < length1; i++) {
            char c1 = s1.charAt(i);
            counts1[c1 - 'a']++;
            char c2 = s2.charAt(i);
            counts2[c2 - 'a']++;
        }
        if (checkSubstring(counts1, counts2))
            return true;
        for (int i = length1; i < length2; i++) {
            char prevC = s2.charAt(i - length1), curC = s2.charAt(i);
            counts2[prevC - 'a']--;
            counts2[curC - 'a']++;
            if (checkSubstring(counts1, counts2))
                return true;
        }
        return false;
    }

    public boolean checkSubstring(int[] counts1, int[] counts2) {
        for (int i = 0; i < 26; i++) {
            if (counts1[i] != counts2[i])
                return false;
        }
        return true;
    }
}

// OJ: https://leetcode.com/problems/permutation-in-string/
// Time: O(B)
// Space: O(1)
class Solution {
public:
    bool checkInclusion(string a, string b) {
        if (a.size() > b.size()) return false;
        int N = a.size(), cnts[26] = {};
        for (char c : a) cnts[c - 'a']++;
        for (int i = 0; i < b.size(); ++i) {
            cnts[b[i] - 'a']--;
            if (i - N >= 0) cnts[b[i - N] - 'a']++;
            bool match = true;
            for (int j = 0; j < 26 && match; ++j) {
                if (cnts[j]) match = false;
            }
            if (match) return true;
        }
        return false;
    }
};

class Solution(object):
  def checkInclusion(self, s1, s2):
    """
    :type s1: str
    :type s2: str
    :rtype: bool
    """
    d = {}
    n = len(s1)
    for c in s1:
      d[c] = d.get(c, 0) + 1
    window = {}
    for i, c in enumerate(s2):
      window[c] = window.get(c, 0) + 1
      if i >= len(s1):
        window[s2[i - n]] = window.get(s2[i - n], 0) - 1
        if window[s2[i - n]] == 0:
          del window[s2[i - n]]
      if window == d:
        return True
    return False

567 - Permutation in String

All Problems

All Solutions