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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

  • 
    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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