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. Consider the following program
main()
{
int a[5]={1,3,6,7,0};
int *b;
b=&a[2];
}
The value of b[-1] is
(A) 1 (B) 3 (C) -6 (D) none
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Answer Posted / bipin chandra sai.s
ans is none,bcoz b has been assigned address &[2],but it
has been asked that ans for b[-1],so the location -1 is not
there,we have locations from 0,1,2,3..,so none is the ans
| Is This Answer Correct ? | 4 Yes | 2 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
