- 39.4%

https://leetcode.com/problems/elimination-game/?tab=Description

There is a list of sorted integers from 1 to n. Starting from left to right, remove the first number and every other number afterward until you reach the end of the list.

Repeat the previous step again, but this time from right to left, remove the right most number and every other number from the remaining numbers.

We keep repeating the steps again, alternating left to right and right to left, until a single number remains.

Find the last number that remains starting with a list of length n.

1 | Example: |

#### java

https://discuss.leetcode.com/topic/59293/java-easiest-solution-o-logn-with-explanation

JAVA: Easiest solution O(logN) with explanation

1 | public int lastRemaining(int n) { |

My idea is to update and record head in each turn. when the total number becomes 1, head is the only number left.

When will head be updated?

- if we move from left
- if we move from right and the total remaining number % 2 == 1. like 2 4 6 8 10, we move from 10, we will take out 10, 6 and 2, head is deleted and move to 4. like 2 4 6 8 10 12, we move from 12, we will take out 12, 8, 4, head is still remaining 2

then we find a rule to update our head.

example:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Let us start with head = 1, left = true, step = 1 (times 2 each turn), remaining = n(24)

we first move from left, we definitely need to move head to next position. (head = head + step)

So after first loop we will have:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 - > 2 4 6 8 10 12 14 16 18 20 22 24

head = 2, left = false, step = 1 * 2 = 2, remaining = remaining / 2 = 12

second loop, we move from right, in what situation we need to move head?

only if the remaining % 2 == 1, in this case we have 12 % 2 == 0, we don’t touch head.

so after this second loop we will have:

2 4 6 8 10 12 14 16 18 20 22 24 - > 2 6 10 14 18 22

head = 2, left = true, step = 2 * 2 = 4, remaining = remaining / 2 = 6

third loop, we move from left, move head to next position

after third loop we will have:

2 6 10 14 18 22 - > 6 14 22

head = 6, left = false, step = 4 * 2 = 8, remaining = remaining / 2 = 3

fourth loop, we move from right, NOTICE HERE:

we have remaining(3) % 2 == 1, so we know we need to move head to next position

after this loop, we will have

6 14 22 - > 14

head = 14, left = true, step = 8 * 2 = 16, remaining = remaining / 2 = 1

while loop end, return head

https://discuss.leetcode.com/topic/61875/one-line-java-solution-based-on-josephus-problem

one line java solution based on Josephus Problem

This problem is similar to Josephus problem when k=2, the recursive version is easy after referring to the josephus problem on wiki.

it is highly recommend to refer to Josephus problem first, because i am chinese, my english is poor, my explanation may not be good, but the wiki explanation is very good.

1 | public int lastRemaining(int n) { |

1 | recursive version |

1 | public int lastRemaining(int n) { |

non-recursive version:

1 | public int lastRemaining(int n) { |

#### c

https://discuss.leetcode.com/topic/58042/c-1-line-solution-with-explanation

C 1 line solution with explanation

After first elimination, all the rest numbers are even numbers.

Divide by 2, we get a continuous new sequence from 1 to n / 2.

For this sequence we start from right to left as the first elimination.

Then the original result should be two times the mirroring result of lastRemaining(n / 2).

1 | int lastRemaining(int n) { |

Great answer. In fact, we can prove that “ML(1…n) + MR(1…n) = 1 + n” holds for any n >= 2, where ML(1…n) means removing elements from left to right first and MR(1…n) means removing elements from right to left first.