CareerCup

superrrrrr... thanks dude...

zeros in higher order we can leave.Just consider the bits that are required to represent the numbers.

public static int search(int a[], int l, int u, int x) { 
		while (l <= u) { 
			int m = (l + u) / 2; 
			if (x == a[m]) {
				return m; 
			} else if (a[l] <= a[m]) {
				if (x > a[m]) {
					l = m+1; 
				} else if (x >=a [l]) {
					u = m-1; 
				} else {
					l = m+1;
				}
			} 
			else if (x < a[m]) u = m-1; 
			else if (x <= a[u]) l = m+1; 
			else u = m - 1; 
		} 
		return -1; 
	}

Developer your code is working perfectly.Can u explain the logic behind c=c&c-1?

it will fail if array contains duplicates

oh sorry guys above algo is failing if array contains duplicates...

int left=0;
		int right=a.length-1;
		int mid;
		while(left<=right)
		{
		  mid=left+(right-left)/2;
		  if(a[mid]==mid)
		  {
			  return mid;
		  }
		  else if(a[mid]>mid)
			  right=mid-1;
		  else
			  left=mid+1;
		  
		}
         return -1;

This is working perfectly guys....

import java.util.Arrays;


public class FindNeartoZero {
	
	public static void main(String[] args) {
		int a[]={-10,-20,-30,-10,-50,-1,1};
		Arrays.sort(a);
		int i;
		for(i=0;i<a.length;i++)
		{
			if(a[i]>0)
			{
				if(i==0)
				{
					System.out.println(a[i]);
				     System.exit(0);
				}
				int k=a[i-1];
				System.out.println(Math.abs(k)<Math.abs(a[i])?k:a[i]);
				System.exit(0);
			}
		}
		if(i==a.length)
			System.out.println(a[--i]);
	}
}

Hi Sparky,
Even i lost there but i find out another solution for finding where the loop is starting.

1)First find the meeting point using slow and fast pointers
2)Find out the number of elements present in the loop(say n nodes)
3)Find out the nth node from last(Starting from meeting point) that nth node become the start of the loop.

public static LinkedListNode FindBeginning(LinkedListNode head) {
		LinkedListNode n1 = head;
		LinkedListNode n2 = head; 
		
		// Find meeting point
		while (n2.next != null) { 
			n1 = n1.next; 
			n2 = n2.next.next; 
			if (n1 == n2) { 
				break; 
			}

}