# 390. Elimination Game

## Description

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 = [6]


Example 2:

Input: n = 1
Output: 1


Constraints:

• 1 <= n <= 109

## Solutions

• 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 {
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;
}
};

• 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


• 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
}