Given two strings s1 and s2write a function to return true if s2 contains the permutation of s1. In other words, one of the first string's permutations is the substring of the second string.
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The simplest method is to generate all the permutations of the short string and to check if the generated permutation is a substring of the longer string.
This function creates all the possible permutations of the short string s 1 s1 s 1. Then, it swaps the current element with every other element in the array, lying towards its right, so as to generate a new ordering of the array elements. After the swapping has been done, it makes another call to permute but this time with the index of the next element in the array.
While returning back, we reverse the swapping done in the current function call. Thus, when we reach the end of the array, a new ordering of the array's elements is generated. The following animation depicts the process of generating the permutations. Time complexity : O n! We match all the permutations of the short string s 1 s1 s 1of length s 1 s1 s 1with s 2 s2 s 2. Here, n n n refers to the length of s 1 s1 s 1. The depth of the recursion tree is n n n n n n refers to the length of the short string s 1 s1 s 1.
Every node of the recursion tree contains a string of max. The idea behind this approach is that one string will be a permutation of another string only if both of them contain the same characters the same number of times. In order to check this, we can sort the two strings and compare them. We sort the short string s 1 s1 s 1 and all the substrings of s 2 s2 s 2sort them and compare them with the sorted s 1 s1 s 1 string. If the two match completely, s 1 s1 s 1 's permutation is a substring of s 2 s2 s 2otherwise not.
As discussed above, one string will be a permutation of another string only if both of them contain the same charaters with the same frequency. We can consider every possible substring in the long string s 2 s2 s 2 of the same length as that of s 1 s1 s 1 and check the frequency of occurence of the characters appearing in the two.
If the frequencies of every letter match exactly, then only s 1 s1 s 1 's permutation can be a substring of s 2 s2 s 2. In order to implement this approach, instead of sorting and then comparing the elements for equality, we make use of a hashmap s 1 m a p s1map s 1 m a p which stores the frequency of occurence of all the characters in the short string s 1 s1 s 1.
We consider every possible substring of s 2 s2 s 2 of the same length as that of s 1 s1 s 1find its corresponding hashmap as well, namely s 2 m a p s2map s 2 m a p.
Thus, the substrings considered can be viewed as a window of length as that of s 1 s1 s 1 iterating over s 2 s2 s 2. If the two hashmaps obtained are identical for any such window, we can conclude that s 1 s1 s 1 's permutation is a substring of s 2 s2 s 2otherwise not. Space complexity : O 1 O 1 O 1.
Instead of making use of a special HashMap datastructure just to store the frequency of occurence of characters, we can use a simpler array data structure to store the frequencies. Given strings contains only lowercase alphabets 'a' to 'z'. So we need to take an array of size The rest of the process remains the same as the last approach. Instead of generating the hashmap afresh for every window considered in s 2 s2 s 2we can create the hashmap just once for the first window in s 2 s2 s 2.
Then, later on when we slide the window, we know that we add one preceding character and add a new succeeding character to the new window considered. Thus, we can update the hashmap by just updating the indices associated with those two characters only. Again, for every updated hashmap, we compare all the elements of the hashmap for equality to get the required result.
Constant space is used. The last approach can be optimized, if instead of comparing all the elements of the hashmaps for every updated s 2 m a p s2map s 2 m a p corresponding to every window of s 2 s2 s 2 considered, we keep a track of the number of elements which were already matching in the earlier hashmap and update just the count of matching elements when we shift the window towards the right.
To do so, we maintain a c o u n t count c o u n t variable, which stores the number of characters out of the 26 alphabetswhich have the same frequency of occurence in s 1 s1 s 1 and the current window in s 2 s2 s 2. When we slide the window, if the deduction of the last element and the addition of the new element leads to a new frequency match of any of the characters, we increment the c o u n t count c o u n t by 1.
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Change your preferences any time. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. I have a string. I want to generate all permutations from that string, by changing the order of characters in it. For example, say:.
Currently I am iterating on the list cast of the string, picking 2 letters randomly and transposing them to form a new string, and adding it to set cast of l. Based on the length of the string, I am calculating the number of permutations possible and continuing iterations till set size reaches the limit. There must be a better way to do this.
The itertools module has a useful method called permutations.
The documentation says:. If r is not specified or is None, then r defaults to the length of the iterable and all possible full-length permutations are generated. Permutations are emitted in lexicographic sort order. So, if the input iterable is sorted, the permutation tuples will be produced in sorted order. If you find yourself troubled by duplicates, try fitting your data into a structure with no duplicates like a set :.
Thanks to pst for pointing out that this is not what we'd traditionally think of as a type cast, but more of a call to the set constructor. Here is another way of doing the permutation of string with minimal code.
We basically create a loop and then we keep swapping two characters at a time, Inside the loop we'll have the recursion. Notice,we only print when indexers reaches the length of our string. Example: ABC i for our starting point and our recursion param j for our loop. Stack Overflow users have already posted some strong solutions but I wanted to show yet another solution.
This one I find to be more intuitive. Here is another approach different from what Adriano and illerucis posted.Given two strings s1 and s2write a function to return true if s2 contains the permutation of s1. In other words, one of the first string's permutations is the substring of the second string.
File Here! Discuss Permutation in String.AMAZON - GROUPED ANAGRAMS (LeetCode)
The length of both given strings is in range [1, 10,]. Related Topics. Two Pointers Sliding Window. Similar Questions. Minimum Window Substring Hard. Find All Anagrams in a String Medium. Show Hint 1. Show Hint 2. How will you check whether one string is a permutation of another string? Show Hint 3. One way is to sort the string and then compare. But, Is there a better way?
Show Hint 4. If one string is a permutation of another string then they must one common metric. What is that? Show Hint 5. Both strings must have same character frequencies, if one is permutation of another.
Which data structure should be used to store frequencies? Show Hint 6. Sign in to view your submissions. Sign in. Console Contribute. Run Code Submit. All Problems. Need more space? Try using full screen mode! Got it!Print all the permutations of the input string using STL functions. Read more Permutations STL.
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If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment. Tags: Expert. January 29, January 26, December 8, One suggestion to this post: We do not need second swap and when left and size are equal then just print the string.
Permutations of Strings
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Subscribe No Spams!! Difficulty Level Beginner Intermediate Expert.Given a string strthe task is to print all the permutations of str. A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. Approach: Write a recursive function that prints every permutation of the given string. Terminating condition will be when the passed string is empty.
Approach: Write a recursive function that print distinct permutations. If the character has not been used then the recursive call will take place. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.
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All permutations of a string using iteration
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Output: geek geke gkee egek egke eegk eekg ekge ekeg kgee kege keeg. Recommended Posts: Print all the permutations of a string without repetition using Collections in Java Java Program to print distinct permutations of a string Print all palindrome permutations of a string Print all distinct permutations of a given string with duplicates Write a program to print all permutations of a given string Print all the palindromic permutations of given string in alphabetic order Distinct permutations of a string containing duplicates using HashSet in Java Print the two possible permutations from a given sequence Find Kth largest string from the permutations of the string with two characters Print all permutations with repetition of characters Iterative approach to print all permutations of an Array Print all permutations in sorted lexicographic order Print distinct sorted permutations with duplicates allowed in input Permutations of a given string using STL All permutations of a string using iteration.
Check out this Author's contributed articles. Load Comments.Please Login in order to post a comment. Agree on the difficulty. I was able to come up with the algorithm on my own, but it wasn't easy. This was the majority of my line of thinking:.
An iterative solution, I would imagine would be difficult. However, this is a somewhat straightforward recursive problem and it actually scales fairly well since the number of recursive calls is the length of the array at worst unlike something like the Fibonacci numbers where you get exponential growth in recursive calls.
So think of it this way:. The trick is working your way backwards. So let's look at 3 4. The part in parentheses is guaranteed to be in reverse order because you only need to look at it if you couldn't find another permutation therefore search in reverse order for something bigger than 3 that's obviously just 4 in this case. Swap them out, then reverse the part in the parentheses nothing to reverse when you only have 3.
So this gives:. Since you successfully permute 3 4the permutation 1 2 4 3 changes nothing, then therefore 1 2 4 3 changes nothing.
Try to permute 4 3. You can't becaus 4! That failed so now you're trying to permute: 2 4 3. Since this was successful, nothing else changes the 1 at the beginning :. Next will obviously be 1 3 4 2 in the same way we went from 1 2 3 4 to 1 2 4 3. And then the next two will be in a similar fashion to above 1 4 2 3, then 1 4 3 2. Next again, we'll try to permute 4 3 2 but will fail 4 isn't less than 2 or 3.
This is an important part that I missed initially. You need to keep that order. You can't reverse to get 4 2 3then realize 4 isn't less than 2 or 3 because then you'll be working with 1 4 2 3 which is incorrect. Which gives the first few:. And then it goes for a while from there there would be 24 permutations in this case.
The last one will be 4 3 2 1. Going through it, 2 12! This would return a false and would signal that there are no more permutations. Totally agree. I spent many hours before looking for the pseudo code.
Although it was fun to code. Abhishek, that's essentially what I did as well. But since you know it's in reverse order, your method is more efficient.
Luckily the arrays aren't big. I found a weird thing, I undersatand conditionals short-circuting and the logic of this code, but can't pinpoint the why the SegFault is happening. Hope that makes sense! Thank you for the great explanation!