![]() Next, we swap the "next largest digit" to the front with the iter_swap() statement and then since we know that digit was the next largest, we know that the digits to the right are still in descending order, so to put it in ascending order, we just have to reverse() it. The replacement must be in place and use only constant extra memory. If such an arrangement is not possible, it must rearrange it as the lowest possible order (i.e., sorted in ascending order). Well, since the things to the right are in descending order, to find the "next largest digit" we just have to iterate from the end, which we see in the first 3 lines of code. Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. We see that when everything to the right of a digit is in descending order, we find the next largest digit and put it in front and then put the remaining digits in ascending order. The ascending order if statement is essentially finding the leftmost place where "everything to the right is in descending order". " which supports our previous observation that we only need to do something to a digit "when everything to the right is in descending order". We now understand the if (i = begin) part so all we need to understand is the if (*i < *j) part.Īlso note: "Then if the elements are in ascending order. Otherwise, we continue and we see that j and i are essentially decremented. Otherwise, if the whole thing is in descending order, ( if (i = begin)) then this is the last permutation. Then, if the elements are in ascending order, ( if (*i < *j)) do something. Let's now go back to the code: while (true)įrom the first 2 lines in the loop, j is an element and i is the element before it. We only change the position of a "digit" when everything to the right is in descending order because if it isn't in descending order then there are still more permutations to go (ie we can "increase" the permutation by a smaller amount). So when do we finally "use" the 1? When there are only no more permutations of the last 3 digits.Īnd when are there no more permutations of the last 3 digits? When the last 3 digits are in descending order.Īha! This is key to understanding the algorithm. In the example above we see that 1 stays as the first number for a long time as there are many reorderings of the last 3 "digits" which "increase" the permutation by a smaller amount. in between and although 10 is larger than 3, there are missing numbers which can be gotten by increasing 3 by a smaller amount. For example when counting we don't count 1, 2, 3, 10. When we order numbers we want to "increase them by the smallest amount". Viewing the problem in this way we want to order the permutations/numbers in "ascending" order. We can view the elements as digits and the permutations as numbers. How do we go from one permutation to the next? Firstly, let's look at things a little differently. What is going on in the body of the main loop? At entry of the main loop i is pointing to the last element (not one past end) and the list is at least 2 elements long. My questions are: How does it work? What is the meaning of i, j and k? What value do they hold at the different parts of execution? What is a sketch of a proof of its correctness?Ĭlearly before entering the main loop it just checks the trivial 0 or 1 element list cases. While (::next_permutation(v.begin(), v.end())) ![]() ![]() not sure if you root for your hometown team, but the wizards are gonna be terrible this NBA season.I was curious how std:next_permutation was implemented so I extracted the the gnu libstdc++ 4.7 version and sanitized the identifiers and formatting to produce the following demo. ![]() To be exact, we will swap ‘3’ with one of the numbers in that sublist, namely with the next higher number. And hey, Microsoft thinks I'm employable.Įven though In almost two decades older than you, your videos helped me realize everything is possible even at an 'advanced' age. This means that there is no next permutation for the sublist 9, 4, 2, 1, so we are forced to do something with the ‘3’. They made a huge difference for me, if nothing else, it boosted my confidence. Somehow they gave me a verbal offer today, and even though for family reasons I can't move to Seattle, I just never thought that I could even get an offer before your videos. I failed the first one, and miraculously passed the second while using the free wifi from WholeFoods to take the call.Īnyway, I went on-site in September and got tripped up on level order traversal on the last interview of the day because I didn't finish all the videos you guys have. They gave me two chances to do a technical for screen for two different roles. I recently found your videos and soon later applied to Microsoft on a whim even though I didn't feel prepared. ![]()
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