![]() ![]() Java Solution 1 Based on Permutation, we can add a set to track if an element is duplicate and no need to swap. For example, 1,1,2 have the following unique permutations: 1,1,2, 1,2,1, and 2,1,1. See the image above for clarification.ĭo this for all the cases and it will generate all possible permutations of the given array. Given a collection of numbers that might contain duplicates, return all possible unique permutations. Fixing the second position automatically fixes the third position. In the first column of second-level 1 is fixed at the first position, in the second column 2 is fixed at the first position and in the third column 3 is fixed at the first position.Īfter fixing an element at the first position, fix an element at the second position, consider the case in the second level and the first column, that is,, 1 is fixed at the first position, so we have 2 choices for the second position that is either 2 or 3. The image below the second level represents this situation. Thus, we need to rearrange the numbers to the right of ai1 including. To understand this better, let’s assume an element is a and another element is b, both the elements are. ![]() We then divide by n to get the required value to return. We make use of the equation nums i nums i + (n (nums nums in)) to store the new values in the nums array. Now, no rearrangements to the right of ai1 can create a larger permutation since that subarray consists of numbers in descending order. This solves the problem of adding extra space to our solution. Explanation for Leetcode problem Permutationsįix an element in the first position, we have three choices 1, or 2, or 3. First, we observe that for any given sequence that is in descending order, no next larger permutation is possible.
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