![]() Space Complexity Analysisįor some of the code, a space of O ( n 1 ) ∗ t O(n1)*t O ( n 1 ) ∗ t can be used for storing the sorted array. ![]() So, the total time complexity taken for the sorting approach will be: O ( n 1 l o g n 1 ) + ( n 2 − n 1 ) ∗ n 2 l o g n 2 O(n_1logn_1)+(n_2-n_1)*n_2logn_2 O ( n 1 l o g n 1 ) + ( n 2 − n 1 ) ∗ n 2 l o g n 2 . And to sort the substrings of string s2, for ( n 2 − n 1 ) (n_2-n_1) ( n 2 − n 1 ) times, the total time taken for will be ( n 2 − n 1 ) ∗ n 2 l o g n 2 (n_2-n_1)*n_2logn_2 ( n 2 − n 1 ) ∗ n 2 l o g n 2 . Now, to sort the first string the time taken will be O ( n 1 l o g n 1 ) O(n_1logn_1) O ( n 1 l o g n 1 ). ![]() Suppose, the length of our string s1 is n 1 n_1 n 1 and the string s2 is n 2 n_2 n 2 . Let us analyze the time complexity for this code: If both of them completely match, then we can say that the string s1's permutation is a substring of s2, and we can return True,Īs we must be already familiar, that the sorting function takes a worst-case time complexity of O ( N l o g N ) O(NlogN) O ( N l o g N ), our code also has somewhat similar time complexity. We may compare the two strings after sorting them to verify whether they contain the same number of characters or not after sorting.Īfter that, we find all the substrings of the string s2, of length s1, sort them and compare them with the string s1. Only if s o r t e d ( x ) = s o r t e d ( y ) sorted(x) = sorted(y) s o r t e d ( x ) = s o r t e d ( y ), then we can say that the string x is a permutation of the string y. This method is majorly based on the principle that two strings can only be permuted if they both contain the same characters the same number of times. Now let us look at some conditions before implying the method: In this approach, we will be using the sorting technique to find out our results. Let us now look at our second approach to solving the Permutation of String problem. Approach #2 of Permutation of String Using sorting : Also, in our recursion tree, every node will contain a string having a maximum length of n n n. Please note that n n n refers to the length of the shorter string. The reason for this is that the depth of the recursion tree will be n n n. The Space complexity for the above approach will be O ( n 2 ) O(n^2) O ( n 2 ) in the worst case. Also, at every step we compare the permutation of the substring S1 with the string S2. And, to generate the permutations of a string of length N N N, we take a time of O ( N ! ) O(N!) O ( N ! ). The major reason for this exponential time complexity is that, first of all, we try to match all the permutations of our shorter string S1, which is having a length of N N N, with our larger string S2. And, M M M denotes the length of the larger string. The N N N in the time complexity depicts the length of the shorter string. The worst-case time complexity of the above brute force approach is O ( N ! ∗ M ) O(N!*M) O ( N ! ∗ M ). Let us look into the code of it, and then we will learn more approaches to solving this of string permutation. The procedure for creating the permutations is shown in the following image.īut doing so will result in the TLE (Time Limit Exceed) in most of the coding platforms. Permute accepts as one of the inputs the index of the current element current indexcurrent index, as one of its parameters.Īfter that, to create a new ordering of the array items, it then swaps the current element with every other member in the array that is located to its right.įollowing the swapping, it calls permute(string 1, string 2, current index) once again, but this time using the index of the subsequent element in the array.Īs a result, a new ordering of the array's items is created when reaches the end of the array. To create all possible combinations of the shorter string, we utilise the function permute(string 1, string 2, current index) to create every feasible pairing.Ībove method generates every combination of the shorter string s1 that is possible. The most straightforward approach is to create all possible combinations of the shorter string given and then determine whether each combination produced is a substring of the longer string or not. Let us begin with the Brute Force approach for getting to our solution for the Permutation of String problem. s 1 s1 s 1 and s 2 s2 s 2 consists of lowercase English letters.Īpproach #1 of Permutation of String : Brute Force Approach.Following will be the constraints attached with them: In the input, you will be give two strings s1 and s2. Constraintsīelow given are the constraints for the above code: Now, since no permutation of our string s1 = "xt" is present in our string s2, we simply return a False. In the above example, we are given two strings - s1 = "xt" and s2 = "axv".
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