# Question

Formatted question description: https://leetcode.ca/all/390.html

You have a list arr of all integers in the range [1, n] sorted in a strictly increasing order. Apply the following algorithm on arr:

• 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 rightmost number and every other number from the remaining numbers.
• Keep repeating the steps again, alternating left to right and right to left, until a single number remains.

Given the integer n, return the last number that remains in arr.

Example 1:

Input: n = 9
Output: 6
Explanation:
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9]
arr = [2, 4, 6, 8]
arr = [2, 6]
arr = 


Example 2:

Input: n = 1
Output: 1


Constraints:

• 1 <= n <= 109

# Algorithm

It is very simple to do it with recursion. Use a bool type variable left2right. True means to traverse from left to right, and false means to traverse from right to left.

When n is 1, it returns 1 no matter from left to right or from right to left.

If n is greater than 1,

• If it is from left to right, return twice the traversal of n/2 from right to left;
• If it is from right to left, it will be a little more troublesome. It is definitely necessary to call the recursive function on n/2, but it should be divided into parity.
• If n is an odd number, return twice the value of n/2 traversal from left to right;
• If n is an even number, double the traversal value of n/2 from left to right, and subtract 1

# Code

• 
public class Elimination_Game {

public static void main(String[] args) {
Elimination_Game out = new Elimination_Game();
Solution s = out.new Solution();

System.out.println(s.lastRemaining(9));
}

class Solution {
public int lastRemaining(int n) {
return dfs(n, true);
}

int dfs(int n, boolean left2right) {
if (n == 1) return 1;
if (left2right) {
return 2 * dfs(n / 2, false);
} else {
return 2 * dfs(n / 2, true) - 1 + n % 2;
}
}
}

class Solution2 {
public int lastRemaining(int n) {
return n == 1 ? 1 : 2 * (1 + n / 2 - lastRemaining(n / 2));
}

}
}

############

class Solution {
public int lastRemaining(int n) {
int a1 = 1, an = n, step = 1;
for (int i = 0, cnt = n; cnt > 1; cnt >>= 1, step <<= 1, ++i) {
if (i % 2 == 1) {
an -= step;
if (cnt % 2 == 1) {
a1 += step;
}
} else {
a1 += step;
if (cnt % 2 == 1) {
an -= step;
}
}
}
return a1;
}
}

• class Solution:
def lastRemaining(self, n: int) -> int:
a1, an = 1, n
i, step, cnt = 0, 1, n
while cnt > 1:
if i % 2:
an -= step
if cnt % 2:
a1 += step
else:
a1 += step
if cnt % 2:
an -= step
cnt >>= 1
step <<= 1
i += 1
return a1

############

class Solution(object):
def lastRemaining(self, n):
"""
:type n: int
:rtype: int
"""
count = n
isFromLeft = True
step = 1
while count > 1:
if isFromLeft or count % 2 == 1:
count /= 2
step *= 2
isFromLeft = not isFromLeft


• class Solution {
public:
int lastRemaining(int n) {
int a1 = 1, an = n, step = 1;
for (int i = 0, cnt = n; cnt > 1; cnt >>= 1, step <<= 1, ++i) {
if (i % 2) {
an -= step;
if (cnt % 2) a1 += step;
} else {
a1 += step;
if (cnt % 2) an -= step;
}
}
return a1;
}
};

• func lastRemaining(n int) int {
a1, an, step := 1, n, 1
for i, cnt := 0, n; cnt > 1; cnt, step, i = cnt>>1, step<<1, i+1 {
if i%2 == 1 {
an -= step
if cnt%2 == 1 {
a1 += step
}
} else {
a1 += step
if cnt%2 == 1 {
an -= step
}
}
}
return a1
}