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.
// C++ program to demonstrate working of// is_permutation()#include <bits/stdc++.h>using namespace std; // Driver program to test aboveint 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
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.
// C++ program to count all permutation of// given text#include<bits/stdc++.h>using namespace std; // Function to count occurrences of anagrams of patint 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 mathced with current window if (is_permutation(text.begin()+i, text.begin()+i+p_len, pat.begin())) count++; return count;} // Driver codeint 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 contribute.geeksforgeeks.org or mail your article to contribute@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.
