In O(n^2)
eg.
#include
#include
using namespace std;
int a[]={9,8,10,15,6,22,23,4,111,112};
int p[10]={-1,-1,-1,-1,-1,-1,-1,-1,-1,-1};
int m[10];
int main()
{
int max, pos, i,j;
for(i=9; i>=0; i--)
{
for(j=i+1; j<10; j++)
{
if(a[i]http://groups.google.com/
According to me we can apply the equatin method :
Let the missing number be y.
Let the duplicated no. be x.
1. then find the product of the 1 to 100 and the sum 1 to 100 :A
Product would be of the form...1 *2*3x*..y*100 : B
Sum : 1+2+3.+x+y.+100
2. find the prduct
@Rahul : I know that u are using table of base b2 in base b1 and then
dividing the number using the table ...but the real problem is to code
it
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You can refer to this sample code i had tried to sort in a single
pass:
#include
#include
using namespace std;
int main()
{
int a = 0, b=1, c=9;
int s[10]= { 2,1,0,1,1,0,2,1,0,1};
while(ab)
c--;
if(!(a>=b) && !(b>=c))
{
if(s[b]==0)
to traverse the tree in a spiral manner...
1. Insert the root into stack.
2. then in a loop check not both stack and queue are empty
3. {
while(stack not empty)
{
node =pop(stack);
coutleft);
insertintoqueue(node
SPIRAL Traversal
a
/\
b c
/\ /\
d e f g
Output : abcgfed
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With reference to the first question...I would say that both the
statements were given seperately.
The answer to the question is : Hello Hello...as i have tried it out.
The reason for the output as i think is that the class functions are
allocated the memory even if no object of the class has been
With reference to the 2nd question i would lyk to add that the array
size is not 3 but any numberwhat i meant to say was that the array
nly contains 0,1,2 and no other number...and with reference to the 1st
question it should be noted that there is no explicitly defined
constructors involved wi
1. Suppose a class A is given as :
class A
{
public :
void print()
{
cout<<"Hello";
}
}
int main()
{
A * a;
A* b = NULL;
a->print();
b->print();
I would again say that you don't have to use any intermediate base
while converting.
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I had proposed an algorithm of repeatedly subtracting 1 from the given
number and subsequently adding 1 to the new number initialised to 0,
till the given number becomes 0. However as soon as the digit reaches
the limit , the digit becomes 0 and you add 1 to the next digit. I was
not able to code i
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