CareerCup

Yes. 200 in total.

What if we do an inorder traversal and find the largest length subarray which is sorted.

Time Complexity : O(n)
Space Complexity : O(n)

public void transformArray(int in[]) {
Stack<Integer> stack = new Stack<Integer>();
int out[] = new int[in.length];
stack.push(0);
for (int i = 1; i < in.length; i++) {
if (in[stack.peek()] <= in[i])
stack.push(i);
else {
while (in[stack.peek()] > in[i]) {
out[stack.pop()] = in[i];
}
stack.push(i);
}
}

for (int i = 0; i < out.length; i++) {
System.out.print(out[i] + " ");
}
}

public int count_on_bits(int n)
{
   int count=0;
   while(n!=0) 
   {
     n = n & n-1 ;
     count++;
   }
   return count;
}

mynode *temp= NULL; // declare a temporary variable named count.

The function mynode *reverse_recurse_n(mynode *start, int n, int count) takes the pointer of the start node of a linked list and a number n as the block size.

count is used to keep the count of node for internal logic while reversing the linked list.

# include <stdio.h>
# include <stdlib.h>
  
   typedef struct node
   {
    int data;
    struct node *next;
   }mynode;


   void printlist ( struct node *p )
   {
        printf("\nThe data values in the list are\n");
         while (p!= NULL)
         {
            printf("%d\t",p-> data);
            p = p-> next;
         }
   }

   mynode *reverse_recurse_n(mynode *start, int n, int count)
   {
      if(start->next != NULL)
      {   
         mynode *head=reverse_recurse_n(start->next, n, count+1);              
                   
         if(count % n !=0)
         {
            start->next->next=start; 
            start->next = NULL;
         }else{
            temp = head;   
            head = start ;  
         }
         if(count % n == 1)
         {  
            start->next = temp;             
         }   
                                             
         return(head); 
                  
      }else{                    
         return start;
      }
      
   }

   int main()
   {
     struct node *start = NULL ;
     start = insert ( start, 1 ); start = insert ( start, 2 );
     start = insert ( start, 3 ); start = insert ( start, 4 );
     start = insert ( start, 5 ); start = insert ( start, 6 );
     start = insert ( start, 7 ); start = insert ( start, 8 );
     start = insert ( start, 9 ); start = insert ( start, 10 );
     start = insert ( start, 11 );  
     printf("The created list is\n");
     printlist (start);
     start = reverse_recurse_n(start,5,1);
     printlist (start);
     getch();
     return 0;
   }