> For the complete documentation index, see [llms.txt](https://maksimdan.gitbook.io/interview-practice-problems/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://maksimdan.gitbook.io/interview-practice-problems/leetcode_sessions/438-find-all-anagrams-in-a-string.md).

# 438 Find All Anagrams in a String

Given a string s and a **non-empty** string **p**, find all the start indices of **p**'s anagrams in**s**.

Strings consists of lowercase English letters only and the length of both strings**s**and**p**will not be larger than 20,100.

The order of output does not matter.

**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".
```

**The Idea**

* This is probably the most creative solution I've come up with for a leetcode problem. I also think it has the potential to beat top submissions given a few optimizations, and perhaps no account on the preprocessing step of performing a uniform distribution.
* My first thought was simply iterate through the string, and sum together the integers. e.g.

```
in = apple
sub = ppl

int key = p + p + l;

a + p + p = key ? NO
p + p + l = key ? YES
p + l + e = ket ? NO
```

* This works because of the associative property of addition.
* But the main problem with this is that multiple permutation sums can produce the same key identifier. Consider:

```
in = 'af'
sub = 'be'

key = b + e

a + f = key? YES

Why this happens:
*    * 
abcdef
 *  *

a is smaller than b by 1, but f is greater than e by 1, so their sums are the same.
```

* So to avoid this, why won't develop our own random mapping for the alphabet? The greater our bounds for a random number, the less likely it is that the sum of a word will equal the sum of another word, unless every single character in the word hashes the exact location. We can produce use a uniform distribution to that every number without the bounds of our range has an equal probability of becoming selected. Finally, we perform the same algorithm as above.

```cpp
unordered_map<char, int> generate_char_map()
{
    unordered_map<char, int> alpha_map;
    alpha_map.reserve(26);

    random_device rd;
    mt19937 eng(rd());
    uniform_int_distribution<> u_dis(0, INT_MAX);

    string alphabet = "abcdefghijklmnopqrstuvwxyz";
    for (char c : alphabet) 
        alpha_map.insert({ c, u_dis(eng) });

    return alpha_map;
}

vector<int> findAnagrams(string &in,  string &sub) 
{
    vector<int> anagram_i;
    if (sub.length() > in.length() || in.empty() || sub.empty())
        return anagram_i;

    unordered_map<char, int> alpha_map = generate_char_map();

    long long sub_key = 0;
    for (int i = 0; i < sub.length(); i++)
        sub_key += alpha_map[sub[i]];


    long long temp_key = 0;

    // base case: first sub.length() chars
    for (int i = 0; i < sub.length(); i++) 
        temp_key += alpha_map[in[i]];

    int back_ptr = 0;
    if (temp_key == sub_key)
        anagram_i.push_back(back_ptr);

    // every other case
    for (int i = sub.length(); i < in.length(); i++) {
        // remove old
        temp_key -= alpha_map[in[back_ptr++]];

        // add new
        temp_key += alpha_map[in[i]];
        if (temp_key == sub_key)
            anagram_i.push_back(back_ptr);
    }

    return anagram_i;
}

int main()
{
    string one3 = "af", two3 = "be";
    vector<int> test3 = findAnagrams(one3, two3);
    print(test3);

    string one = "cbaebabacd", two = "abc";
    vector<int> test = findAnagrams(one, two);
    print(test);

    string one1 = "abab", two1 = "ab";
    vector<int> test1 = findAnagrams(one1, two1);
    print(test1);

    string one2 = "", two2 = "a";
    vector<int> test2 = findAnagrams(one2, two2);
    print(test2);
}
```

$$O(n^2)$$ solution with constant space. (too slow)

```cpp
vector<int> findAnagrams( string &in,  string &sub) 
{
    sort(sub.begin(), sub.end());

    string subtemp;
    vector<int> anagram_i;
    const int go_to = in.length() - sub.length();

    for (int i = 0; i <= go_to; i++) {
        subtemp = in.substr(i, sub.length());
        sort(subtemp.begin(), subtemp.end());
        if (subtemp == sub) 
            anagram_i.push_back(i);
    }

    return anagram_i;
}

int main()
{
    string one2 = "", two2 = "a";
    vector<int> test2 = findAnagrams(one2, two2);
    print(test2);

    string one = "cbaebabacd", two = "abc";
    vector<int> test = findAnagrams(one, two);
    print(test);

    string one1 = "abab", two1 = "ab";
    vector<int> test1 = findAnagrams(one1, two1);
    print(test1);
}
```


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