Answer / sandeep ambekar

//
// MAIN -- Write a program for deleting duplicate elements in an array
// Example: # a.out string substr
// I/P example: ./a.out Bangalore a
//
// Author -- Sandeep Ambekar

#include <stdio.h>
#define TRUE 1
#define FALSE 0

void
purge_char_from_string (char *str, char *c)
{
int i, k, flag, count; // i/j for index, flag to
keep track of
// 'c' and count for iteration..
char *buf = str; // save the original string
start:
str = buf; // we could enter the loop
again.
i = 0;
k = 1;
count = 0;
flag = FALSE; // INIT variables...
printf ("Input String [%s] [%c] \n", str, *c);

while (str[i] != '\0')
{
printf ("\t (%c) <==> (%c)\n", str[i], str[k]);
if (str[i] == c[0])
{
str[i] = str[k];
flag = TRUE;
count++;
}
else if (flag == TRUE)
{
str[i] = str[k]; // later have a while loop
to find a char
// to which is !c and replace them.
}
i++;
k++;
}
printf ("[%s]\n", buf);
if ((count - 1) >= 1)
goto start;
} // end of
purge_char_from_string ....

//
// MAIN -- Program to Purge a given character from a string.
//

int
main (int argc, char *argv[])
{
if (argc < 3)
{
printf ("Input <String> <char> \n");
return 1;
}
printf (" ## Input string %s : Char [%s]\n", argv[1],
argv[2]);
purge_char_from_string (argv[1], argv[2]);

printf (" Trimmed String ---> %s\n", argv[1]);
return 0;
}

Answer / bhanuprakash a

/* here is the actual logic */
for(i=1; i<n; i++)
{
for(j=i+1;j<=n;j++)
{
if(a[i] = a[j])
{
for(k=j;k<n;k++)
{
a[k]=a[k+1]; //deletion
n=n-1;
}
}
}
}

Answer / vrushali

Any program without goto.

#include<stdio.h> void main() { int a [5]; for (i=0; i<=4; i++) printf(“%d” ,a[i]); }

8 Answers  

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which operator is known as dummy operator in c?

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what does keyword ‘extern’ mean in a function declaration?

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A function 'q' that accepts a pointer to a character as argument and returns a pointer to an array of integer can be declared as: A)int (*q(char*)) [] B)int *q(char*) [] C)int(*q)(char*) [] D)None of the Above

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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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Differentiate between full, complete & perfect binary trees.

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