![]() ![]() This means that the right-most digit (which carries the maximum value) is also numerically the most superior. We know that the digits next to the Smaller Candidate (now replaced by the Larger Candidate) are arranged in a descending order. Now, that we know the Smaller Candidate and the Larger Candidate, we swap them. Thus, if we go right-to-left starting with the right-most digit, we get an ascending order, and the first digit that is larger than the Smaller Candidate will also be the smallest digit that is greater than the Smaller Candidate. In other words, while sweeping right-to-left, each succeeding element was either equal to or greater than the preceding digit. This is because, each of those digits failed the condition nums < nums. Why the first digit? Each digit to right of the Smaller Candidate will be arranged in descending order (if we go left-to-right from the Smaller Candidate). So, we do another right-to-left sweep, and find out the first digit that is greater than the Smaller Candidate. But there may be another more suitable candidate on the right-hand side of the Smaller Candidate. Now, how do we choose the Larger Candidate? We know that the candidate immediately to the right of the Smaller Candidate is a potential candidate, because it is greater than the Smaller Candidate. The first pair that satisfies this, gives us the Smaller Candidate: namely, the nums candidate. At each index, we check if the digit to its left is smaller than itself, i.e., nums < nums. This is to ensure that we stick to the smallest-integer-greater-than rule.Īrmed with this knowledge, we do a right-to-left sweep. To achieve this, we need to know that there is at least one candidate to the right of the Smaller Candidate that is greater than Smaller Candidate.Īlso, the Smaller Candidate should be the right-most digit, as long as we can guarantee that there is at least one digit to its right that is greater than itself. Thus, the Larger Candidate needs to be selected from the digits on the left of the Smaller Candidate. But because digits carry greater value as we go towards left, if we choose the Larger Candidate from the left of the Smaller Candidate, we will result with a smaller integer. The Larger Candidate can either be selected from its right-hand side or its left hand side. Now, lets assume, we somehow find the Smaller Candidate. contains the exact same digits as the given integer (we can only replace the relative positions of the existing digits).is smaller than every other integer that is greater than the given integer.Thus, increasing 2 by 1 will create a much greater result, than increasing 4 with 1.Īlso note that, to find the ‘next’ integer, we need to find an integer, that: Thus, one digit will be smaller (lets call it Smaller Candidate) and another will be greater ( Larger Candidate).īefore proceeding further, we should note that the values carried by digits (in an integer) decrease from left-to-right. The pair of digits being replaced should not be equal to each other because otherwise we will end up with the original integer itself. ![]() Suitable Pair = Larger Candidate + Smaller Candidate In order to find the ‘next’ integer, we need to swap the positions of a specific pair of digits. To simplify the explanation, I will use the example of integers, in the following part of this post. So, when we want to find the next permutation in Lexicographical order, it is similar to finding the next non-negative integer on the number scale (achieved solely by rearranging the existing digits). Not just dictionaries, lexicographical order is used in non-negative number systems as well. The word containing the letter which is alphabetically superior (at that particular index) gets a higher order.įor example, between the words, peek and peak, the word peak will be placed first, because a > e in the alphabetical order. We stop at the index, where the letters do not match. By going left-to-right, we compare the corresponding letters of each index. Words in a dictionary are ordered by the comparing alphabetical order of the letters, starting with the first (or left-most) place. Lexicographical order is a generalisation of the ‘aphpabetical order’ we see in dicitionaries. If there is no such permutation possible (i.e., it is the greatest lexicographical permutation), the array should be rearranged to its “lowest possible order”, i.e., ascending order. Given an array of integers, nums, rearrange the elements of the array to produce the “lexicographically next greater permutation of numbers”. In this post, I discuss the solution to the LeetCode Problem, titled Next Permutation. Finding the Lexicographically Next Permutation
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