Reverse the bit order in a single macro.
eg. i/p = 10010101 --> o/p = 10101001
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Answer / balaji ganesh
#include<stdio.h>
#define f(a) strrev(a)
main()
{
char c[20];
scanf("%s",c,printf("enter bit string;"));
printf("%s",f(c));
}
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / vishnu
with out using strrev
=======================
#include<stdio.h>
#include<conio.h>
int i ;
void binary(int retval[],int num)
{
int k;
while(num >1)
{
k = num;
num =num/2;
retval[i] = k%2;
i++;
}
retval[i] =1;
}
int main()
{
int num;
int bin[20];
scanf("%d",&num);
binary(bin,num);
for(num=i;num >=0;num--)
printf("%d",bin[num]);
printf("\n");
for(num =0;num<=i;num++)
printf("%d",bin[num]);
getch();
}
| Is This Answer Correct ? | 0 Yes | 0 No |
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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;i<n1-1;i++) { if( !( (a1[i]>a1[i+1]) && (a2[i]>a2[i+1]) ) ) return false; } return true;//assumed that each array doesn't contain duplicate elements in themshelves }