Given a text txt[0..n-1] and a pattern pat[0..m-1], write a function search(char pat[], char txt[]) that prints all occurrences of pat[] and its permutations (or anagrams) in txt[]. You may assume that n > m.

Expected time complexity is O(n)

**Examples:**

1) Input: txt[] = "BACDGABCDA" pat[] = "ABCD" Output: Found at Index 0 Found at Index 5 Found at Index 6 2) Input: txt[] = "AAABABAA" pat[] = "AABA" Output: Found at Index 0 Found at Index 1 Found at Index 4

## We strongly recommend that you click here and practice it, before moving on to the solution.

A simple idea is to modify Rabin Karp Algorithm. For example we can keep the hash value as sum of ASCII values of all characters under modulo of a big prime number. For every character of text, we can add the current character to hash value and subtract the first character of previous window. This solution looks good, but like standard Rabin Karp, the worst case time complexity of this solution is O(mn). The worst case occurs when all hash values match and we one by one match all characters.

We can achieve O(n) time complexity under the assumption that alphabet size is fixed which is typically true as we have maximum 256 possible characters in ASCII. The idea is to use two count arrays:

1) The first count array store frequencies of characters in pattern.

2) The second count array stores frequencies of characters in current window of text.

The important thing to note is, time complexity to compare two count arrays is O(1) as the number of elements in them are fixed (independent of pattern and text sizes). Following are steps of this algorithm.

1) Store counts of frequencies of pattern in first count array *countP[]*. Also store counts of frequencies of characters in first window of text in array *countTW[]*.

2) Now run a loop from i = M to N-1. Do following in loop.

…..a) If the two count arrays are identical, we found an occurrence.

…..b) Increment count of current character of text in countTW[]

…..c) Decrement count of first character in previous window in countWT[]

3) The last window is not checked by above loop, so explicitly check it.

## Python3

`# Python program to search all ` `# anagrams of a pattern in a text ` ` ` `MAX` `=` `256` ` ` `# This function returns true ` `# if contents of arr1[] and arr2[] ` `# are same, otherwise false. ` `def` `compare(arr1, arr2): ` ` ` `for` `i ` `in` `range` `(` `MAX` `): ` ` ` `if` `arr1[i] !` `=` `arr2[i]: ` ` ` `return` `False` ` ` `return` `True` ` ` `# This function search for all ` `# permutations of pat[] in txt[] ` `def` `search(pat, txt): ` ` ` ` ` `M ` `=` `len` `(pat) ` ` ` `N ` `=` `len` `(txt) ` ` ` ` ` `# countP[]: Store count of ` ` ` `# all characters of pattern ` ` ` `# countTW[]: Store count of ` ` ` `# current window of text ` ` ` `countP ` `=` `[` `0` `]` `*` `MAX` ` ` ` ` `countTW ` `=` `[` `0` `]` `*` `MAX` ` ` ` ` `for` `i ` `in` `range` `(M): ` ` ` `(countP[` `ord` `(pat[i]) ]) ` `+` `=` `1` ` ` `(countTW[` `ord` `(txt[i]) ]) ` `+` `=` `1` ` ` ` ` `# Traverse through remaining ` ` ` `# characters of pattern ` ` ` `for` `i ` `in` `range` `(M, N): ` ` ` ` ` `# Compare counts of current ` ` ` `# window of text with ` ` ` `# counts of pattern[] ` ` ` `if` `compare(countP, countTW): ` ` ` `print` `(` `"Found at Index"` `, (i` `-` `M)) ` ` ` ` ` `# Add current character to current window ` ` ` `(countTW[ ` `ord` `(txt[i]) ]) ` `+` `=` `1` ` ` ` ` `# Remove the first character of previous window ` ` ` `(countTW[ ` `ord` `(txt[i` `-` `M]) ]) ` `-` `=` `1` ` ` ` ` `# Check for the last window in text ` ` ` `if` `compare(countP, countTW): ` ` ` `print` `(` `"Found at Index"` `, N` `-` `M) ` ` ` `# Driver program to test above function ` `txt ` `=` `"BACDGABCDA"` `pat ` `=` `"ABCD"` `search(pat, txt) ` ` ` `# This code is contributed ` `# by Upendra Singh Bartwal ` |

*chevron_right*

*filter_none*

**Output:**

('Found at Index', 0) ('Found at Index', 5) ('Found at Index', 6)

Please refer complete article on Anagram Substring Search (Or Search for all permutations) for more details!