![]() ![]() Original Assignee Apple Inc International Business Machines Corp Motorola Inc Priority date (The priority date is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation or warranty as to the accuracy of the list.) ![]() Google has not performed a legal analysis and makes no representation as to the accuracy of the status listed.) Expired - Lifetime Application number US09/062,100 Inventor Hunter Ledbetter Scales, III Keith Everett Diefendorff Brett Olsson Pradeep Kumar Dubey Ronald Ray Hochsprung Current Assignee (The listed assignees may be inaccurate. Google Patents Data processing system and method of permutation with replication within a vector register fileĭownload PDF Info Publication number US5996057A US5996057A US09/062,100 US6210098A US5996057A US 5996057 A US5996057 A US 5996057A US 6210098 A US6210098 A US 6210098A US 5996057 A US5996057 A US 5996057A Authority US United States Prior art keywords vector input register elements data Prior art date Legal status (The legal status is an assumption and is not a legal conclusion. Google Patents US5996057A - Data processing system and method of permutation with replication within a vector register file US5996057A - Data processing system and method of permutation with replication within a vector register file If you also wish to share your knowledge with the takeUforward fam, please check out this article. Special thanks to Dewanshi Paul for contributing to this article on takeUforward. Step 7: Reversing from i1+1 index to last of the array. Step 6: Swapping values present in i1 and i2 indices. Step 5: i2 indexed element is greater than i1 indexed element. Step 4: i2 will be another index to find just greater than i1 indexed elements in the array. Step 3: Since 1 is less than 2, we achieved our start of the increasing sequence. Step 2: Since 3 is not less than 2, we decrease i1 by 1. Step 1: First find an increasing sequence. Space Complexity: Since no extra storage is required. This sums up to 3*O(N) which is approximately O(N). Time Complexity: For the first iteration backward, the second interaction backward and reversal at the end takes O(N) for each, where N is the number of elements in the input array. Step 4: Reverse array from index+1 where the index is found at step 1 till the end of the array.ĭef nextPermutation(self, nums: List) -> None: ![]() Step 3: Swap values present in indices found in the above two steps. Find an index that has a value greater than the previously found index. Hence, we will reverse the input array to get the minimum or starting permutation. This means the given input array is the largest lexicographical permutation. Step 2: If the index value received from step 1 is less than 0. Step 1: Linearly traverse array from backward such that ith index value of the array is less than (i+1)th index value. We achieve these from all remaining steps of our approach. To make rank as few as possible but greater than input array, just perverse array from breakpoint achieved from the first step of the approach. Then, we need to get just a larger value than the point where the peak occurs. This is what we achieve from our first step of the approach. Hence, our approach aims to get a peak from where the increasing sequence starts. Thus, we can see every sequence has increasing order. If such an arrangement is not possible, it must rearrange it as the lowest possible order (i.e., sorted in ascending order).Įxample 1 : Input format: Arr =. Problem Statement: Given an array Arr of integers, rearrange the numbers of the given array into the lexicographically next greater permutation of numbers. ![]()
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