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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 = [6]

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

• Java
• Python
• C++
• Go

• 
  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
      head = 1
      isFromLeft = True
      step = 1
      while count > 1:
        if isFromLeft or count % 2 == 1:
          head = head + step
        count /= 2
        step *= 2
        isFromLeft = not isFromLeft
      return head
  
  

• 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
  }
  

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