finding position.txt

Problem Description

Given an integer A which denotes the number of people standing in the queue.

A selection process follows a rule where people standing on even positions are selected. Of the selected people a queue is formed and again out of these only people on even position are selected.

This continues until we are left with one person. Find and return the position of that person in the original queue.

int Solution::solve(int A) {
    /*--------------Algorithm----------
    we know that we have to choose even positions.so it means that even postion elements gets divided each time.
    so example:
    
    1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
                                      
                                index:1 2 3 4 5  6  7 8 
    when we choose even pos elements: 2 4 6 8 10 12 14 16 
    
    
                                index:   1 2 3  4       look at the pos each time each element is divided by 2.
                
class Solution:
# 	__________algorithm_________
    # As we all knopw that when A%26==0 means that value will be z so for that we have to just decrease vaue by 1;
	def convertToTitle(self, A):
	    l=[]
	   # A-=1
	    while A>0:
	        A-=1
	    
	        l.append(chr((A%26)+ord('A')))
	        A//=26
	    l.reverse()
	    return ''.join(l)
                elements:4 8 12 16
                                
                                index:   1 2      
                                elements:8 16
    
    */
    return (1<<int(log2(A)));
}