In other words, one of the first string's permutations is the substring of the second string.Įxplanation: s2 contains one permutation of s1 ("ba"). I hope this helps! Let me know if you have any questions.Given two strings s1 and s2, write a function to return true if s2 contains the permutation of s1. This problem asks us to find all possible permutations of a. That’s it! We have successfully implemented a solution to check if s2 is a permutation of s1 using the sliding window technique. Welcome to another exciting problem-solving article Today’s challenge revolves around LeetCode problem 46, titled Permutations. s1 = "ab"Īll the test cases pass, which means our solution is correct. Example 1: Input: ABC Output: ABC ACB BAC BCA CAB CBA Explanation: Given string ABC has permutations in 6 forms as ABC, ACB, BAC, BCA, CAB and CBA. Now we can test our solution with some test cases. Here is the implementation: from collections import Counterĭef checkInclusion(s1: str, s2: str) -> bool: Given two strings s1 and s2, write a function to return true if s2 contains the. If the counter becomes 0 at the end of the loop, it means that the window list contains a permutation of s1, so we return True.įinally, we return False if we have checked all the windows and none of them is a permutation of s1. Learn about the Permutation In String problem on LeetCode solutions. Otherwise, we break the loop and continue with the next iteration. Then, we can iterate over the characters in the window list and check if each character is in the s1_set. We can create a set s1_set from the characters in s1 and a counter counter initialized to the length of s1. Next, we can use a set and a counter to check if the characters in the window list are a permutation of s1. This way, the window list always stores the characters in the current window that we are checking. If the length of the window list is greater than the length of s1, we remove the first character from the list. :( Sorry, it is possible that the version of your browser is too low to load the code-editor, please try to update browser to revert to using code-editor.update. In each iteration, we append the current character to the window list. Then, we can use a for loop to iterate over the characters in s2. This list will store the characters in the current window that we are checking. We can start by initializing a variable window to an empty list. Now let’s implement the solution in Python.įirst, we need to define the function check_permutation(s1: str, s2: str) -> bool that takes in two strings s1 and s2 and returns a boolean indicating if s2 is a permutation of s1. If we have checked all the windows and none of them is a permutation of s1, we return False. Otherwise, we slide the window by one character to the right and repeat the process until we reach the end of s2. We can slide a window of length len(s1) over the string s2 and check if the characters in the window are a permutation of s1. One approach to solve this problem is to use sliding window technique. C++ C Java Recursion Backtracking Depth-First Search String Bit Manipulation Breadth-First Search Iterator Bitmask Ordered Set Combinatorics Probability and Statistics Queue Array Dynamic Programming Stack Math Binary Search Tree Memoization Binary Tree. If such an arrangement is not possible, it must rearrange it as the lowest possible order (i.e., sorted in ascending order). Given an array nums of distinct integers, return all the possible permutations.
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