Answer / sakthipriya

#include<stdio.h>
#include<conio.h>
void main()
{
int n=4, c, k, space, count = 1;
space = n;
for ( c = 1 ; c <= n ; c++)
{
for( k = 1 ; k < space ; k++)
printf(" ");
for ( k = 1 ; k <= c ; k++)
{
printf("*");
if ( c > 1 && count < c)
{
printf(" ");
count++;
}
}
printf("\n");
space--;
count = 1;
}
getch();
}

Answer / prasanna

int i,j;
for(i=o;i<4;i++)
{
for(j=0;j<=i;j++)
{
printf("*");
}
}

What is difference between far and near pointers?

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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 }

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why TCS selected more student in the software field from all institution.

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