So i was doing some coding exercice on Talentbuddy (for those who know), and i cant get why i cant finish this one.
The exercice is removing a substring from a string, given as input the string, the position P where beginning to remove characters and N the number of characters needed to be remove.
Here is what i've done :
#include <stdio.h>
#include <unistd.h>
void remove_substring(char *s, int p, int n)
{
int idx;
idx = -1;
while (s[++idx] != '\0')
write(1, &s[idx == p - 1 ? idx + n : idx], 1);
}
When the input is "abcdefghi", P = 9 and N = 1, the result given is "abcdefgh" exactly the same as the one i get with my function. But TalentBuddy keep saying me that my output is wrong and i dont thing he (talentbuddy) is wrong.
Maybe there is a blank space or something between the "h" and the '\0'.
But i cant figure it cause when i add another write(1, "END", 3) at the end it appears like "abcdefghEND".
If the question is exclusively for strings( NULL Terminated )
Why can't this be as simple as this, unless it is a homework.
void removesubstr( const char *string, const char *substring )
{
char *p = strstr(string, substring);
if(p)
{
strcpy(p,p+strlen(substring));
}
}
Your problem is that you write something for every original index, even if it should be suppressed. What you write looks like abcdefgh, but it is abcdefgh<nul>, where the terminal doesn't render the <nul>.
You are mixing two methods here. Either filter out the removed substring:
void remove_substring(char *s, int p, int n)
{
int i = 0;
p--; /* convert to C-style index */
while (s[i] != '\0') {
if (i < p || i >= p + n) putchar(s[i]);
i++;
}
}
or skip the substring by jumping over it:
void remove_substring(char *s, int p, int n)
{
int i = 0;
int l = strlen(s);
while (i < l) {
if (i + 1 == p) {
i += n;
} else {
putchar(s[i++]);
}
}
}
You're trying to do a bit of both.
(I've avoided the awkward combination of prefix increment and starting at minus 1. And I've used putchar instead of unistd's write. And the termination by length is so you don't inadvertently jump beyond the terminating <nul>.)
Related
I want to code for the question :
define the function getCharacter which accepts a string S and an integer K as input. the function must return K-th character when the reverse of the string S is repeated infinite number of times.
I try this Many times..But I couldn't get the output..I'm beginner to C..My code is here..
#include<stdio.h>
#include<stdlib.h>
char getCharacter(char *s,int k)
{
char rev[100];
int b,e,c=0;
while(s[c]!='\0')
{
c++;
}
e=c-1;
for(b=0;b<c;b++)
{
rev[b]=s[e];
e--;
}
rev[b]='\0';
char res;
res=rev[k+1];
return res;
}
int main()
{
char s[100],f;
int k;
gets(s);
scanf("%d",&k);
f=getCharacter(s,k);
printf("%c",f);
return 0;
}
Required Output :
INPUT :
skillrack
13
OUTPUT:
r
Ok, since this appears to be a homework/practice question, there's going to be no code but here are some pointers based on your question:
1st requirement is to reverse the string followed by infinite repetitions. Now, since computers are limited resources of memory/storage, storing an infinite string is out of the question, so one needs to improvise.
Note that the reverse of a string abcd can be printed to the console by just printing the string beginning from the end starting with d in this case and then moving the index backward, so that's d, c, b and a. Unless one doesn't need to store the reversed string, that's a good enough method to just obtain the reverse.
The next thing to note is that if one places the string dcba and joins another copy of this reversed string, it becomes dcbadcba. Now if one needs to access the 0th character of this reversed string, it's d but so is the 4th character, which is d too. Figure out why that is. (Hint: look at the original string and find how 4 is related to it). So, once this is established, one can figure out that the infinite reverse of the string actually doesn't need to be copied (or joined) from the original string at all but can be just obtained by manipulating the index. This should give enough pointers to solve the question.
Also, since you'll require it, this is a link to modular arithmetic.
Thanks for all of your Reply..
I done my code successfully..
here is my code..
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
char getCharacter(char *s,int k)
{
char rev[100],rev1[1000];
int b,e,c=0,len,n=100;
len=strlen(s);
while(s[c]!='\0')
{
c++;
}
e=c-1;
for(b=0;b<c;b++)
{
rev[b]=s[e];
e--;
}
rev[b]='\0';
char *result;
result = malloc(sizeof(char) * len + 1);
while (n--)
{
strcat(result,rev);
}
char res;
res=result[k-1];
return res;
}
int main()
{
char s[100],f;
int k;
gets(s);
scanf("%d",&k);
f=getCharacter(s,k);
printf("%c",f);
return 0;
}
char getCharacter(char *s,int k)
{
int len = strlen(s);
if (k % len == 0) return s[0];
if (k < len) return s[len-k];
return (s[len - 1 - (k % len) + 1]);
}
Input: s = "A man, a plan, a canal: Panama"
Output: true
Explanation: "amanaplanacanalpanama" is a palindrome.
bool isPalindrome(char * s){
if(strlen(s) == 0) return true;
int m = 0;
for(int i = 0; i < strlen(s); i++)
if(isalnum(s[i])) s[m++] = tolower(s[i]);
int i = 0;
while(i<m)
if(s[i++] != s[--m]) return false;
return true;
}
My code's running time is 173ms. My instructor suggested me to use two pointers to improve the performance and memory usage, but I have no idea where to start.
Just position the two pointers like this
char* first = someString;
char* end = someString + strlen(s) - 1;
Now for it to be a palindrome what first and end point to must be the same
e.g. char someString[] = "1331";
So you in the first iteration *first == *last i.e. '1'
Now move the pointers towards each other until there is nothing left to compare or when they differ
++first, --end;
now *first and *last point to '3'
and so on, check if they are pointing to the same or have passed each other it is a palindrome.
Something like this
#include <stdio.h>
#include <string.h>
int palindrome(char* str)
{
char* start = str;
char* end = str + strlen(str) - 1;
for (; start < end; ++start, --end )
{
if (*start != *end)
{
return 0;
}
}
return 1;
}
int main()
{
printf("palindrome: %d\n", palindrome("1331"));
printf("palindrome: %d\n", palindrome("132331"));
printf("palindrome: %d\n", palindrome("74547"));
return 0;
}
You should add error checks, there are no error checks in the function.
My code's running time is 173ms. My instructor suggested me to use two pointers to improve the performance and memory usage, but I have no idea where to start.
It's already running in O(n) so you cannot reduce the time complexity (except for the iterative call to strlen, see below), although there are some room for improving performance.
Your function does not declare any arrays, and only use a few variables and the memory usage does not depend at all on input size. The memory usage is already O(1) and very low, so it's not a real concern.
But if you want to do it with pointers, here is one:
bool isPalindrome(char * s){
char *end = s + strlen(s);
char *a = s;
char *b = end-1;
while(true) {
// Skip characters that's not alphanumeric
while( a != end && !isalnum(*a) ) a++;
while( b != s && !isalnum(*b) ) b--;
// We're done when we have passed the middle
if(b < a) break;
// Perform the check
if(tolower(*a) != tolower(*b)) return false;
// Step to next character
a++;
b--;
}
return true;
}
When it comes to performance, your code has two issues, none of which gets solved by pointers. First one is that you're calling strlen for each iteration. The second is that you don't need to loop through the whole array, because that's checking it twice.
for(int i = 0; i < strlen(s); i++)
should be
size_t len = strlen(s);
for(size_t i = 0; i < len/2; i++)
Another remark I have on your code is that it changes the input string. That's not necessary. If I have a function that is called isPalindrome I'd expect it to ONLY check if the string is a palindrome or not. IMO, the signature should be bool isPalindrome(const char * s)
For an assignment in class, we have been instructed to write a program which takes a string and a delimiter and then takes "words" and stores them in a new array of strings. i.e., the input ("my name is", " ") would return an array with elements "my" "name" "is".
Roughly, what I've attempted is to:
Use a separate helper called number_of_delimeters() to determine the size of the array of strings
Iterate through the initial array to find the number of elements in a given string which would be placed in the array
Allocate storage within my array for each string
Store the elements within the allocated memory
Include directives:
#include <stdlib.h>
#include <stdio.h>
This is the separate helper:
int number_of_delimiters (char* s, int d)
{
int numdelim = 0;
for (int i = 0; s[i] != '\0'; i++)
{
if (s[i] == d)
{
numdelim++;
}
}
return numdelim;
}
`This is the function itself:
char** split_at (char* s, char d)
{
int numdelim = number_of_delimiters(s, d);
int a = 0;
int b = 0;
char** final = (char**)malloc((numdelim+1) * sizeof(char*));
for (int i = 0; i <= numdelim; i++)
{
int sizeofj = 0;
while (s[a] != d)
{
sizeofj++;
a++;
}
final[i] = (char*)malloc(sizeofj);
a++;
int j = 0;
while (j < sizeofj)
{
final[i][j] = s[b];
j++;
b++;
}
b++;
final[i][j+1] = '\0';
}
return final;
}
To print:
void print_string_array(char* a[], unsigned int alen)
{
printf("{");
for (int i = 0; i < alen; i++)
{
if (i == alen - 1)
{
printf("%s", a[i]);
}
else
{
printf("%s ", a[i]);
}
}
printf("}");
}
int main(int argc, char *argv[])
{
print_string_array(split_at("Hi, my name is none.", ' '), 5);
return 0;
}
This currently returns {Hi, my name is none.}
After doing some research, I realized that the purpose of this function is either similar or identical to strtok. However, looking at the source code for this proved to be little help because it included concepts we have not yet used in class.
I know the question is vague, and the code rough to read, but what can you point to as immediately problematic with this approach to the problem?
The program has several problems.
while (s[a] != d) is wrong, there is no delimiter after the last word in the string.
final[i][j+1] = '\0'; is wrong, j+1 is one position too much.
The returned array is unusable, unless you know beforehand how many elements are there.
Just for explanation:
strtok will modify the array you pass in! After
char test[] = "a b c ";
for(char* t = test; strtok(t, " "); t = NULL);
test content will be:
{ 'a', 0, 'b', 0, 'c', 0, 0 }
You get subsequently these pointers to your test array: test + 0, test + 2, test + 4, NULL.
strtok remembers the pointer you pass to it internally (most likely, you saw a static variable in your source code...) so you can (and must) pass NULL the next time you call it (as long as you want to operate on the same source string).
You, in contrast, apparently want to copy the data. Fine, one can do so. But here we get a problem:
char** final = //...
return final;
void print_string_array(char* a[], unsigned int alen)
You just return the array, but you are losing length information!
How do you want to pass the length to your print function then?
char** tokens = split_at(...);
print_string_array(tokens, sizeof(tokens));
will fail, because sizeof(tokens) will always return the size of a pointer on your local system (most likely 8, possibly 4 on older hardware)!
My personal recommendation: create a null terminated array of c strings:
char** final = (char**)malloc((numdelim + 2) * sizeof(char*));
// ^ (!)
// ...
final[numdelim + 1] = NULL;
Then your print function could look like this:
void print_string_array(char* a[]) // no len parameter any more!
{
printf("{");
if(*a)
{
printf("%s", *a); // printing first element without space
for (++a; *a; ++a) // *a: checking, if current pointer is not NULL
{
printf(" %s", *a); // next elements with spaces
}
}
printf("}");
}
No problems with length any more. Actually, this is exactly the same principle C strings use themselves (the terminating null character, remember?).
Additionally, here is a problem in your own code:
while (j < sizeofj)
{
final[i][j] = s[b];
j++; // j will always point behind your string!
b++;
}
b++;
// thus, you need:
final[i][j] = '\0'; // no +1 !
For completeness (this was discovered by n.m. already, see the other answer): If there is no trailing delimiter in your source string,
while (s[a] != d)
will read beyond your input string (which is undefined behaviour and could result in your program crashing). You need to check for the terminating null character, too:
while(s[a] && s[a] != d)
Finally: how do you want to handle subsequent delimiters? Currently, you will insert empty strings into your array? Print out your strings as follows (with two delimiting symbols - I used * and + like birth and death...):
printf("*%s+", *a);
and you will see. Is this intended?
Edit 2: The variant with pointer arithmetic (only):
char** split_at (char* s, char d)
{
int numdelim = 0;
char* t = s; // need a copy
while(*t)
{
numdelim += *t == d;
++t;
}
char** final = (char**)malloc((numdelim + 2) * sizeof(char*));
char** f = final; // pointer to current position within final
t = s; // re-assign t, using s as start pointer for new strings
while(*t) // see above
{
if(*t == d) // delimiter found!
{
// can subtract pointers --
// as long as they point to the same array!!!
char* n = (char*)malloc(t - s + 1); // +1: terminating null
*f++ = n; // store in position pointer and increment it
while(s != t) // copy the string from start to current t
*n++ = *s++;
*n = 0; // terminate the new string
}
++t; // next character...
}
*f = NULL; // and finally terminate the string array
return final;
}
While I've now been shown a more elegant solution, I've found and rectified the issues in my code:
char** split_at (char* s, char d)
{
int numdelim = 0;
int x;
for (x = 0; s[x] != '\0'; x++)
{
if (s[x] == d)
{
numdelim++;
}
}
int a = 0;
int b = 0;
char** final = (char**)malloc((numdelim+1) * sizeof(char*));
for (int i = 0; i <= numdelim; i++)
{
int sizeofj = 0;
while ((s[a] != d) && (a < x))
{
sizeofj++;
a++;
}
final[i] = (char*)malloc(sizeofj);
a++;
int j = 0;
while (j < sizeofj)
{
final[i][j] = s[b];
j++;
b++;
}
final[i][j] = '\0';
b++;
}
return final;
}
I consolidated what I previously had as a helper function, and modified some points where I incorrectly incremented .
trying to write function that returns 1 if every letter in “word” appears in “s”.
for example:

containsLetters1("this_is_a_long_string","gas") returns 1
containsLetters1("this_is_a_longstring","gaz") returns 0
containsLetters1("hello","p") returns 0
Can't understand why its not right:
#include <stdio.h>
#include <string.h>
#define MAX_STRING 100
int containsLetters1(char *s, char *word)
{
int j,i, flag;
long len;
len=strlen(word);
for (i=0; i<=len; i++) {
flag=0;
for (j=0; j<MAX_STRING; j++) {
if (word==s) {
flag=1;
word++;
s++;
break;
}
s++;
}
if (flag==0) {
break;
}
}
return flag;
}
int main() {
char string1[MAX_STRING] , string2[MAX_STRING] ;
printf("Enter 2 strings for containsLetters1\n");
scanf ("%s %s", string1, string2);
printf("Return value from containsLetters1 is: %d\n",containsLetters1(string1,string2));
return 0;
Try these:
for (i=0; i < len; i++)... (use < instead of <=, since otherwise you would take one additional character);
if (word==s) should be if (*word==*s) (you compare characters stored at the pointed locations, not pointers);
Pointer s advances, but it should get back to the start of the word s, after reaching its end, i.e. s -= len after the for (j=...);
s++ after word++ is not needed, you advance the pointer by the same amount, whether or not you found a match;
flag should be initialized with 1 when declared.
Ah, that should be if(*word == *s) you need to use the indirection operator. Also as hackss said, the flag = 0; must be outside the first for() loop.
Unrelated but probably replace scanf with fgets or use scanf with length specifier For example
scanf("%99s",string1)
Things I can see wrong at first glance:
Your loop goes over MAX_STRING, it only needs to go over the length of s.
Your iteration should cover only the length of the string, but indexes start at 0 and not 1. for (i=0; i<=len; i++) is not correct.
You should also compare the contents of the pointer and not the pointers themselves. if(*word == *s)
The pointer advance logic is incorrect. Maybe treating the pointer as an array could simplify your logic.
Another unrelated point: A different algorithm is to hash the characters of string1 to a map, then check each character of the string2 and see if it is present in the map. If all characters are present then return 1 and when you encounter the first one that is not present then return 0. If you are only limited to using ASCII characters a hashing function is very easy. The longer your ASCII strings are the better the performance of the second approach.
Here is a one-liner solution, in keeping with Henry Spencer's Commandment 7 for C Programmers.
#include <string.h>
/*
* Does l contain every character that appears in r?
*
* Note degenerate cases: true if r is an empty string, even if l is empty.
*/
int contains(const char *l, const char *r)
{
return strspn(r, l) == strlen(r);
}
However, the problem statement is not about characters, but about letters. To solve the problem as literally given in the question, we must remove non-letters from the right string. For instance if r is the word error-prone, and l does not contain a hyphen, then the function returns 0, even if l contains every letter in r.
If we are allowed to modify the string r in place, then what we can do is replace every non-letter in the string with one of the letters that it does contain. (If it contains no letters, then we can just turn it into an empty string.)
void nuke_non_letters(char *r)
{
static const char *alpha =
"abcdefghijklmnopqrstuvwxyz"
"ABCDEFGHIJKLMNOPQRSTUVWXYZ";
while (*r) {
size_t letter_span = strspn(r, alpha);
size_t non_letter_span = strcspn(r + letter_span, alpha);
char replace = (letter_span != 0) ? *r : 0;
memset(r + letter_span, replace, non_letter_span);
r += letter_span + non_letter_span;
}
}
This also brings up another flaw: letters can be upper and lower case. If the right string is A, and the left one contains only a lower-case a, then we have failure.
One way to fix it is to filter the characters of both strings through tolower or toupper.
A third problem is that a letter is more than just the 26 letters of the English alphabet. A modern program should work with wide characters and recognize all Unicode letters as such so that it works in any language.
By the time we deal with all that, we may well surpass the length of some of the other answers.
Extending the idea in Rajiv's answer, you might build the character map incrementally, as in containsLetters2() below.
The containsLetters1() function is a simple brute force implementation using the standard string functions. If there are N characters in the string (haystack) and M in the word (needle), it has a worst-case performance of O(N*M) when the characters of the word being looked for only appear at the very end of the searched string. The strchr(needle, needle[i]) >= &needle[i] test is an optimization if there are likely to be repeated characters in the needle; if there won't be any repeats, it is a pessimization (but it can be removed and the code still works fine).
The containsLetters2() function searches through the string (haystack) at most once and searches through the word (needle) at most once, for a worst case performance of O(N+M).
#include <assert.h>
#include <stdio.h>
#include <string.h>
static int containsLetters1(char const *haystack, char const *needle)
{
for (int i = 0; needle[i] != '\0'; i++)
{
if (strchr(needle, needle[i]) >= &needle[i] &&
strchr(haystack, needle[i]) == 0)
return 0;
}
return 1;
}
static int containsLetters2(char const *haystack, char const *needle)
{
char map[256] = { 0 };
size_t j = 0;
for (int i = 0; needle[i] != '\0'; i++)
{
unsigned char c_needle = needle[i];
if (map[c_needle] == 0)
{
/* We don't know whether needle[i] is in the haystack yet */
unsigned char c_stack;
do
{
c_stack = haystack[j++];
if (c_stack == 0)
return 0;
map[c_stack] = 1;
} while (c_stack != c_needle);
}
}
return 1;
}
int main(void)
{
assert(containsLetters1("this_is_a_long_string","gagahats") == 1);
assert(containsLetters1("this_is_a_longstring","gaz") == 0);
assert(containsLetters1("hello","p") == 0);
assert(containsLetters2("this_is_a_long_string","gagahats") == 1);
assert(containsLetters2("this_is_a_longstring","gaz") == 0);
assert(containsLetters2("hello","p") == 0);
}
Since you can see the entire scope of the testing, this is not anything like thoroughly tested, but I believe it should work fine, regardless of how many repeats there are in the needle.
I can't write a workable code for a function that deletes N characters from the string S, starting from position P. How you guys would you write such a function?
void remove_substring(char *s, int p, int n) {
int i;
if(n == 0) {
printf("%s", s);
}
for (i = 0; i < p - 1; i++) {
printf("%c", s[i]);
}
for (i = strlen(s) - n; i < strlen(s); i++) {
printf("%c", s[i]);
}
}
Example:
s: "abcdefghi"
p: 4
n: 3
output:
abcghi
But for a case like n = 0 and p = 1 it's not working!
Thanks a lot!
A few people have shown you how to do this, but most of their solutions are highly condensed, use standard library functions or simply don't explain what's going on. Here's a version that includes not only some very basic error checking but some explanation of what's happening:
void remove_substr(char *s, size_t p, size_t n)
{
// p is 1-indexed for some reason... adjust it.
p--;
// ensure that we're not being asked to access
// memory past the current end of the string.
// Note that if p is already past the end of
// string then p + n will, necessarily, also be
// past the end of the string so this one check
// is sufficient.
if(p + n >= strlen(s))
return;
// Offset n to account for the data we will be
// skipping.
n += p;
// We copy one character at a time until we
// find the end-of-string character
while(s[n] != 0)
s[p++] = s[n++];
// And make sure our string is properly terminated.
s[p] = 0;
}
One caveat to watch out for: please don't call this function like this:
remove_substr("abcdefghi", 4, 3);
Or like this:
char *s = "abcdefghi";
remove_substr(s, 4, 3);
Doing so will result in undefined behavior, as string literals are read-only and modifying them is not allowed by the standard.
Strictly speaking, you didn't implement a removal of a substring: your code prints the original string with a range of characters removed.
Another thing to note is that according to your example, the index p is one-based, not zero-based like it is in C. Otherwise the output for "abcdefghi", 4, 3 would have been "abcdhi", not "abcghi".
With this in mind, let's make some changes. First, your math is a little off: the last loop should look like this:
for (i = p+n-1; i < strlen(s); i++) {
printf("%c", s[i]);
}
Demo on ideone.
If you would like to use C's zero-based indexing scheme, change your loops as follows:
for (i = 0; i < p; i++) {
printf("%c", s[i]);
}
for (i = p+n; i < strlen(s); i++) {
printf("%c", s[i]);
}
In addition, you should return from the if at the top, or add an else:
if(n == 0) {
printf("%s", s);
return;
}
or
if(n == 0) {
printf("%s", s);
} else {
// The rest of your code here
...
}
or remove the if altogether: it's only an optimization, your code is going to work fine without it, too.
Currently, you code would print the original string twice when n is 0.
If you would like to make your code remove the substring and return a result, you need to allocate the result, and replace printing with copying, like this:
char *remove_substring(char *s, int p, int n) {
// You need to do some checking before calling malloc
if (n == 0) return s;
size_t len = strlen(s);
if (n < 0 || p < 0 || p+n > len) return NULL;
size_t rlen = len-n+1;
char *res = malloc(rlen);
if (res == NULL) return NULL;
char *pt = res;
// Now let's use the two familiar loops,
// except printf("%c"...) will be replaced with *p++ = ...
for (int i = 0; i < p; i++) {
*pt++ = s[i];
}
for (int i = p+n; i < strlen(s); i++) {
*pt++ = s[i];
}
*pt='\0';
return res;
}
Note that this new version of your code returns dynamically allocated memory, which needs to be freed after use.
Here is a demo of this modified version on ideone.
Try copying the first part of the string, then the second
char result[10];
const char input[] = "abcdefg";
int n = 3;
int p = 4;
strncpy(result, input, p);
strncpy(result+p, input+p+n, length(input)-p-n);
printf("%s", result);
If you are looking to do this without the use of functions like strcpy or strncpy (which I see you said in a comment) then use a similar approach to how strcpy (or at least one possible variant) works under the hood:
void strnewcpy(char *dest, char *origin, int n, int p) {
while(p-- && *dest++ = *origin++)
;
origin += n;
while(*dest++ = *origin++)
;
}
metacode:
allocate a buffer for the destination
decalre a pointer s to your source string
advance the pointer "p-1" positions in your source string and copy them on the fly to destination
advance "n" positions
copy rest to destination
What did you try? Doesn't strcpy(s+p, s+p+n) work?
Edit: Fixed to not rely on undefined behaviour in strcpy:
void remove_substring(char *s, int p, int n)
{
p--; // 1 indexed - why?
memmove(s+p, s+p+n, strlen(s) - n);
}
If your heart's really set on it, you can also replace the memmove call with a loop:
char *dst = s + p;
char *src = s + p + n;
for (int i = 0; i < strlen(s) - n; i++)
*dst++ = *src++;
And if you do that, you can strip out the strlen call, too:
while ((*dst++ = *src++) != '\0);
But I'm not sure I recommend compressing it that much.