Answer Posted / vijay bharti

#include<stdio.h>
#include<conio.h>

void main()
{
int num,i,count=0;
clrscr();
for(num=1;num<=300;num++)
{
for(i=2;i<=num/2;i++)
{
if(num%i!=0)
count=count+1;
}

}

printf("number of prime numbers between 1 to 300 are : %d",count);
getch();
}

You have given 2 array. You need to find whether they will create the same BST or not. For example: Array1:10 5 20 15 30 Array2:10 20 15 30 5 Result: True Array1:10 5 20 15 30 Array2:10 15 20 30 5 Result: False One Approach is Pretty Clear by creating BST O(nlogn) then checking two tree for identical O(N) overall O(nlogn) ..we need there exist O(N) Time & O(1) Space also without extra space .Algorithm ?? DevoCoder guest Posted 3 months ago # #define true 1 #define false 0 int check(int a1[],int a2[],int n1,int n2) { int i; //n1 size of array a1[] and n2 size of a2[] if(n1!=n2) return false; //n1 and n2 must be same for(i=0;ia1[i+1]) && (a2[i]>a2[i+1]) ) ) return false; } return true;//assumed that each array doesn't contain duplicate elements in themshelves }

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