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is_permutation() in C++ and its application for anagram search

  • Difficulty Level : Medium
  • Last Updated : 16 Nov, 2021

is_permutations() is used to check if two containers like string and vector are permutation of each other. It accepts three parameters, the first two parameters are begin and end positions of first object and third parameter begin position of second object.
 

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CPP




// C++ program to demonstrate working of
// is_permutation()
#include <bits/stdc++.h>
using namespace std;
 
// Driver program to test above
int main()
{
    vector<int> v1{1, 2, 3, 4};
    vector<int> v2{2, 3, 1, 4};
 
    // v1 and v2 are permutation of each other
    if (is_permutation(v1.begin(), v1.end(), v2.begin()))
        cout << "True\n";
    else
        cout << "False\n";
 
    // v1 and v3 are NOT permutation of each other
    vector<int> v3{5, 3, 1, 4};
    if (is_permutation(v1.begin(), v1.end(), v3.begin()))
        cout << "True\n";
    else
        cout << "False\n";
 
    return 0;
}

Output : 

True
False

 
Application : 
Given a pattern and a text, find all occurrences of pattern and its anagrams in text.
Examples: 
 



Input : text ="forxxorfxdofr"  
        pat = "for"
Output :  3
There are three anagrams of "for"
int text.

Input : word = "aabaabaa" 
        text = "aaba"
Output : 4

 

We have discussed a (n) solution her. But in this post it is done using is_permutation(). Although the complexity is higher than previously discussed method, but the purpose is to explain application of is_permutation().
Let size of pattern to be searched be pat_len. The idea is to traverse given text and for every window of size pat_len, check if it is a permutation of given pattern or not.
 

CPP




// C++ program to count all permutation of
// given text
#include<bits/stdc++.h>
using namespace std;
 
// Function to count occurrences of anagrams of pat
int countAnagrams(string text, string pat)
{
    int t_len = text.length();
    int p_len = pat.length();
 
    // Start traversing the text
    int count = 0; // Initialize result
    for (int i=0; i<=t_len-p_len; i++)
 
        // Check if substring text[i..i+p_len]
        // is a permutation of pat[].
        // Three parameters are :
        // 1) Beginning position of current window in text
        // 2) End position of current window in text
        // 3) Pattern to be matched with current window
        if (is_permutation(text.begin()+i,
                           text.begin()+i+p_len,
                           pat.begin()))
            count++;
 
    return count;
}
 
// Driver code
int main()
{
    string str = "forxxorfxdofr";
    string pat = "for";
    cout << countAnagrams(str, pat) << endl;
    return 0;
}

Output: 
 

3

This article is contributed by Sahil Chhabra. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 




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