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Anagram Pattern Search
Anagrams are basically all permutations of a given string or pattern. This pattern searching algorithm is slightly different. In this case, not only the exact pattern is searched, it searches all possible arrangements of the given pattern in the text.
To solve this problem, we will divide the whole texts into several windows of length same as patterns. Then count on each character of the pattern is found and stored in an array. For each window, we also try to find the count array, then check whether they are matching or not.
The time Complexity of Anagram Pattern Search Algorithm is O(n).
Input and Output
Input: The main String “AABAACBABBCABAABBA”. The pattern “AABC”. Output: Anagram found at position: 2 Anagram found at position: 3 Anagram found at position: 4 Anagram found at position: 10
Algorithm
anagramSearch(text, pattern)
Input − The main string and the pattern
Output − All locations where pattern and it’s all anagrams are found.
Begin define patternFreq array and stringFreq array patLne := length of pattern stringLen := length of the text set all entries of patternFreq array to 0 for all characters present in pattern, do increase the frequency. done for i := 0 to i<= stringLen – patLen, do set all entries of stringFreq to 0 for all characters of each window, do increase the frequency done if the stringFreq and patternFreq are same, then display the value of i, as anagram found at that location done End
Example
#include<iostream>
#include<cstring>
#define LETTER 26
using namespace std;
bool arrayCompare(int *array1, int *array2, int n) {
for(int i = 0; i<n; i++) {
if(array1[i] != array2[i])
return false; //if there is one mismatch stop working
}
return true; //arrays are identical
}
void setArray(int *array, int n, int value) {
for(int i = 0; i<n; i++)
array[i] = value; //put value for all places in the array
}
void anagramSearch(string mainString, string patt, int *array, int *index) {
int strFreq[LETTER], pattFreq[LETTER];
int patLen = patt.size();
int stringLen = mainString.size();
setArray(pattFreq, LETTER, 0); //initialize all frequency to 0
for(int i = 0; i<patLen; i++) {
int patIndex = patt[i] - 'A'; //subtract ASCII of A
pattFreq[patIndex]++; //increase frequency
}
for(int i = 0; i<=(stringLen - patLen); i++) { //the range where window will move
setArray(strFreq, LETTER, 0); //initialize all frequency to 0 for main string
for(int j = i; j<(i+patLen); j++){ //update frequency for each window.
int strIndex = mainString[j] - 'A';
strFreq[strIndex]++; //increase frequency
}
if(arrayCompare(strFreq, pattFreq, LETTER)) { //when both arrays are identical
(*index)++;
array[*index] = i; //anagram found at ith position
}
}
}
int main() {
string mainStrng = "AABAACBABBCABAABBA";
string pattern = "AABC";
int matchLocation[mainStrng.size()];
int index = -1;
anagramSearch(mainStrng, pattern, matchLocation, &index);
for(int i = 0; i<=index; i++) {
cout << "Anagram found at position: " << matchLocation[i] << endl;
}
}Output
Anagram found at position: 2 Anagram found at position: 3 Anagram found at position: 4 Anagram found at position: 10
Updated on: 15-Jun-2020
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