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438. Find All Anagrams in a String

Description

Given two strings s and p, return an array of all the start indices of p's anagrams in s. You may return the answer in any order.

An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

 

Example 1:

Input: s = "cbaebabacd", p = "abc"
Output: [0,6]
Explanation:
The substring with start index = 0 is "cba", which is an anagram of "abc".
The substring with start index = 6 is "bac", which is an anagram of "abc".

Example 2:

Input: s = "abab", p = "ab"
Output: [0,1,2]
Explanation:
The substring with start index = 0 is "ab", which is an anagram of "ab".
The substring with start index = 1 is "ba", which is an anagram of "ab".
The substring with start index = 2 is "ab", which is an anagram of "ab".

 

Constraints:

  • 1 <= s.length, p.length <= 3 * 104
  • s and p consist of lowercase English letters.

Solutions

  • class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        int m = s.length(), n = p.length();
        List<Integer> ans = new ArrayList<>();
        if (m < n) {
            return ans;
        }
        int[] cnt1 = new int[26];
        for (int i = 0; i < n; ++i) {
            ++cnt1[p.charAt(i) - 'a'];
        }
        int[] cnt2 = new int[26];
        for (int i = 0; i < n - 1; ++i) {
            ++cnt2[s.charAt(i) - 'a'];
        }
        for (int i = n - 1; i < m; ++i) {
            ++cnt2[s.charAt(i) - 'a'];
            if (Arrays.equals(cnt1, cnt2)) {
                ans.add(i - n + 1);
            }
            --cnt2[s.charAt(i - n + 1) - 'a'];
        }
        return ans;
    }
}

  • class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        int m = s.size(), n = p.size();
        vector<int> ans;
        if (m < n) {
            return ans;
        }
        vector<int> cnt1(26);
        for (char& c : p) {
            ++cnt1[c - 'a'];
        }
        vector<int> cnt2(26);
        for (int i = 0; i < n - 1; ++i) {
            ++cnt2[s[i] - 'a'];
        }
        for (int i = n - 1; i < m; ++i) {
            ++cnt2[s[i] - 'a'];
            if (cnt1 == cnt2) {
                ans.push_back(i - n + 1);
            }
            --cnt2[s[i - n + 1] - 'a'];
        }
        return ans;
    }
};

  • class Solution:
    def findAnagrams(self, s: str, p: str) -> List[int]:
        m, n = len(s), len(p)
        ans = []
        if m < n:
            return ans
        cnt1 = Counter(p)
        cnt2 = Counter(s[: n - 1])
        for i in range(n - 1, m):
            cnt2[s[i]] += 1
            if cnt1 == cnt2:
                ans.append(i - n + 1)
            cnt2[s[i - n + 1]] -= 1
        return ans


  • func findAnagrams(s string, p string) (ans []int) {
	m, n := len(s), len(p)
	if m < n {
		return
	}
	cnt1 := [26]int{}
	cnt2 := [26]int{}
	for _, c := range p {
		cnt1[c-'a']++
	}
	for _, c := range s[:n-1] {
		cnt2[c-'a']++
	}
	for i := n - 1; i < m; i++ {
		cnt2[s[i]-'a']++
		if cnt1 == cnt2 {
			ans = append(ans, i-n+1)
		}
		cnt2[s[i-n+1]-'a']--
	}
	return
}

  • function findAnagrams(s: string, p: string): number[] {
    const m = s.length;
    const n = p.length;
    const ans: number[] = [];
    if (m < n) {
        return ans;
    }
    const cnt1: number[] = new Array(26).fill(0);
    const cnt2: number[] = new Array(26).fill(0);
    const idx = (c: string) => c.charCodeAt(0) - 'a'.charCodeAt(0);
    for (const c of p) {
        ++cnt1[idx(c)];
    }
    for (const c of s.slice(0, n - 1)) {
        ++cnt2[idx(c)];
    }
    for (let i = n - 1; i < m; ++i) {
        ++cnt2[idx(s[i])];
        if (cnt1.toString() === cnt2.toString()) {
            ans.push(i - n + 1);
        }
        --cnt2[idx(s[i - n + 1])];
    }
    return ans;
}


  • public class Solution {
    public IList<int> FindAnagrams(string s, string p) {
        int m = s.Length, n = p.Length;
        IList<int> ans = new List<int>();
        if (m < n) {
            return ans;
        }
        int[] cnt1 = new int[26];
        int[] cnt2 = new int[26];
        for (int i = 0; i < n; ++i) {
            ++cnt1[p[i] - 'a'];
        }
        for (int i = 0, j = 0; i < m; ++i) {
            int k = s[i] - 'a';
            ++cnt2[k];
            while (cnt2[k] > cnt1[k]) {
                --cnt2[s[j++] - 'a'];
            }
            if (i - j + 1 == n) {
                ans.Add(j);
            }
        }
        return ans;
    }
}

  • impl Solution {
    pub fn find_anagrams(s: String, p: String) -> Vec<i32> {
        let (s, p) = (s.as_bytes(), p.as_bytes());
        let (m, n) = (s.len(), p.len());
        let mut ans = vec![];
        if m < n {
            return ans;
        }

        let mut cnt = [0; 26];
        for i in 0..n {
            cnt[(p[i] - b'a') as usize] += 1;
            cnt[(s[i] - b'a') as usize] -= 1;
        }
        for i in n..m {
            if cnt.iter().all(|&v| v == 0) {
                ans.push((i - n) as i32);
            }
            cnt[(s[i] - b'a') as usize] -= 1;
            cnt[(s[i - n] - b'a') as usize] += 1;
        }
        if cnt.iter().all(|&v| v == 0) {
            ans.push((m - n) as i32);
        }
        ans
    }
}

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