I've been given this sentence and I need to shuffle the words of it:
char array[] = "today it is going to be a beautiful day.";
A correct output would be: "going it beautiful day is a be to today"
I've tried many things like turning it into a 2D array and shuffling the rows, but I can't get it to work.
Your instinct of creating a 2D array is solid. However in C that's more involved than you might expect:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <time.h>
int main()
{
char array[] = "today it is going to be a beautiful day.";
char out_array[sizeof(array)];
char words[sizeof(array)][46];
int word_count = 0;
int letter_count = 0;
int on_word = 0;
int count = 0;
int i = 0;
int j = 0;
srand(time(NULL));
// parse words into 2D array
for (i = 0; i < sizeof(array); i++) {
if (array[i] == ' ') {
if (on_word) {
words[word_count++][letter_count] = '\0';
letter_count = 0;
on_word = 0;
}
} else if (array[i] == '\0' || array[i] == '.') {
break;
} else {
on_word = 1;
words[word_count][letter_count++] = array[i];
}
}
words[word_count++][letter_count] = '\0';
// randomly swap around words
for (i = 0; i < word_count; i++) {
char temp[46];
int idx = rand() % word_count;
if (idx != i) {
strcpy(temp, words[idx]);
strcpy(words[idx], words[i]);
strcpy(words[i], temp);
}
}
// output words into out_array
for (i = 0; i < word_count; i++) {
for (j = 0; words[i][j] != '\0'; j++) {
out_array[count++] = words[i][j];
}
out_array[count++] = ' ';
}
out_array[count - 1] = '\0';
printf("%s", out_array);
return 0;
}
You need two basic algorithms to solve this problem.
Split the input string into a list of words.
Randomly sample your list of words until there are no more.
1. Split the input string into a list of words.
This is much simpler than you may think. You don’t need to actually copy any words, just find where each one begins in your input string.
today it is going to be a beautiful day.
^---- ^- ^- ^---- ^- ^- ^ ^-------- ^--
There are all kinds of ways you can store that information, but the two most useful would be either an array of integer indices or an array of pointers.
For your example sentence, the following would be a list of indices:
0, 6, 9, 12, 18, 21, 24, 26, 36
To do this, just create an array with a reasonable upper limit on words:
int words[100]; // I wanna use a list of index values
int nwords = 0;
char * words[100]; // I wanna use a list of pointers
int nwords = 0;
If you do it yourself either structure is just as easy.
If you use strtok life is much easier with a list of pointers.
All you need at this point is a loop over your input to find the words and populate your list. Remember, a words is any alphabetic or numeric value (and maybe hyphens, if you want to go that far). Everything else is not a word. If you #include <ctype.h> you get a very handy function for classifying a character is “word” or “not-word”:
if (isalnum( input[n] )) its_a_word_character;
else its_not_a_word_character_meaning_we_have_found_the_end_of_the_word;
Now that you have a list of words, you can:
2. Randomly sample your list of words until there are no more.
There are, again, a number of ways you could do this. Already suggested above is to randomly shuffle the list of words (array of indices or array of pointers), and then simply rebuild the sentence by taking the words in order.
→ Beware, Etian’s example is not a correct shuffle, though it would probably go unnoticed or ignored by everyone at your level of instruction as it will appear to work just fine. Google around “coding horror fisher yates” for more.
The other way would be to just select and remove a random word from your array until there are no words left.
The random sampling is not difficult, but it does require some precise thinking, making this the actually most difficult part of your project.
To start you first need to get a proper random number. There is a trick to this that people are generally not taught. Here you go:
int random( int N ) // Return an UNBIASED pseudorandom value in [0, N-1].
{
int max_value = (RAND_MAX / N) * N;
int result;
do result = rand(); while (result >= max_value);
return result % N;
}
And in main() the very first thing you should do is initialize the random number generator:
#include <stdlib.h>
#include <time.h>
int main()
{
srand( (unsigned)time( NULL ) );
Now you can sample / shuffle your array properly. You can google "Fisher-Yates Shuffle" (or follow the link in the comment below your question). Or you can just select the next word:
while (nwords)
{
int index = random( nwords );
// do something with word[index] here //
// Remove the word we just printed from our list of words
// • Do you see what trick we use to remove the word?
// • Do you also know why this does not affect our random selection?
words[index] = words[--nwords];
}
Hopefully you can see that both of these methods are essentially the same thing. Whichever you choose is up to you. I personally would use the latter because of the following consideration:
Output
You can create a new string and then print it, or you can just print each word directly. As the homework (as you presented it) does not require generation of a new string, I would just print the output directly. This makes life simpler in the sense that you do not have to mess with another string array.
As you print each word (or append it to a new string), remember how you separated them to begin with. If you use strtok you can just use something like:
printf( "%s", words[index] ); // print word directly to stdout
strcat( output, words[index] ); // append word to output string
If you found the beginnings of each word yourself, you will have to again loop until you find the end of the word:
// Print word, character by character, directly to stdout
for (int n = index; isalnum( words[index+n] ); n++)
{
putchar( words[index+n] );
}
// Append word, character by character, to output string
for (int n = index; isalnum( words[index+n] ); n++)
{
char * p = strchr( output, '\0' ); // (Find end of output[])
*p++ = words[index+n]; // (Add char)
*p = '\0'; // (Add null terminator)
}
All that’s left is to pay attention to spaces and periods in your output.
Hopefully this should be enough to get you started.
Related
I'm trying to do a program that get number of names from the user, then it get the names from the user and save them in array in strings. After it, it sort the names in the array by abc and then print the names ordered. The program work good, but the problem is when I try to free the dynamic memory I defined.
Here is the code:
#include <stdio.h>
#include <string.h>
#define STR_LEN 51
void myFgets(char str[], int n);
void sortString(char** arr, int numberOfStrings);
int main(void)
{
int i = 0, numberOfFriends = 0, sizeOfMemory = 0;
char name[STR_LEN] = { 0 };
char** arrOfNames = (char*)malloc(sizeof(int) * sizeOfMemory);
printf("Enter number of friends: ");
scanf("%d", &numberOfFriends);
getchar();
for (i = 0; i < numberOfFriends; i++) // In this loop we save the names into the array.
{
printf("Enter name of friend %d: ", i + 1);
myFgets(name, STR_LEN); // Get the name from the user.
sizeOfMemory += 1;
arrOfNames = (char*)realloc(arrOfNames, sizeof(int) * sizeOfMemory); // Change the size of the memory to more place to pointer from the last time.
arrOfNames[i] = (char*)malloc(sizeof(char) * strlen(name) + 1); // Set dynamic size to the name.
*(arrOfNames[i]) = '\0'; // We remove the string in the currnet name.
strncat(arrOfNames[i], name, strlen(name) + 1); // Then, we save the name of the user into the string.
}
sortString(arrOfNames, numberOfFriends); // We use this function to sort the array.
for (i = 0; i < numberOfFriends; i++)
{
printf("Friend %d: %s\n", i + 1, arrOfNames[i]);
}
for (i = 0; i < numberOfFriends; i++)
{
free(arrOfNames[i]);
}
free(arrOfNames);
getchar();
return 0;
}
/*
Function will perform the fgets command and also remove the newline
that might be at the end of the string - a known issue with fgets.
input: the buffer to read into, the number of chars to read
*/
void myFgets(char str[], int n)
{
fgets(str, n, stdin);
str[strcspn(str, "\n")] = 0;
}
/*In this function we get array of strings and sort the array by abc.
Input: The array and the long.
Output: None*/
void sortString(char** arr, int numberOfStrings)
{
int i = 0, x = 0;
char tmp[STR_LEN] = { 0 };
for (i = 0; i < numberOfStrings; i++) // In this loop we run on all the indexes of the array. From the first string to the last.
{
for (x = i + 1; x < numberOfStrings; x++) // In this loop we run on the next indexes and check if is there smaller string than the currnet.
{
if (strcmp(arr[i], arr[x]) > 0) // If the original string is bigger than the currnet string.
{
strncat(tmp, arr[i], strlen(arr[i])); // Save the original string to temp string.
// Switch between the orginal to the smaller string.
arr[i][0] = '\0';
strncat(arr[i], arr[x], strlen(arr[x]));
arr[x][0] = '\0';
strncat(arr[x], tmp, strlen(tmp));
tmp[0] = '\0';
}
}
}
}
After the print of the names, when I want to free the names and the array, in the first try to free, I get an error of: "HEAP CORRUPTION DETECTED: after normal block(#87)". By the way, I get this error only when I enter 4 or more players. If I enter 3 or less players, the program work properly.
Why does that happen and what I should do to fix it?
First of all remove the unnecessary (and partly wrong) casts of the return value of malloc and realloc. In other words: replace (char*)malloc(... with malloc(..., and the same for realloc.
Then there is a big problem here: realloc(arrOfNames, sizeof(int) * sizeOfMemory) : you want to allocate an array of pointers not an array of int and the size of a pointer may or may not be the same as the size of an int. You need sizeof(char**) or rather the less error prone sizeof(*arrOfNames) here.
Furthermore this in too convoluted (but not actually wrong):
*(arrOfNames[i]) = '\0';
strncat(arrOfNames[i], name, strlen(name) + 1);
instead you can simply use this:
strcpy(arrOfNames[i], name);
Same thing in the sort function.
Keep your code simple.
But actually there are more problems in your sort function. You naively swap the contents of the strings (which by the way is inefficient), but the real problem is that if you copy a longer string, say "Walter" into a shorter one, say "Joe", you'll write beyond the end of the allocated memory for "Joe".
Instead of swapping the content of the strings just swap the pointers.
I suggest you take a pencil and a piece of paper and draw the pointers and the memory they point to.
I am a beginner in C with some experience in python and java. I want to solve a problem with C. The problem goes like this:
Take an input as a sentence with words separated by blank spaces only (assume lower case only), re-write the sentence with the following rules:
1) If a word occurs the first time, keep it the same.
2) If the word occurs twice, replace the second occurrence with the word being copied twice (e.g. two --> twotwo).
3) If the word occurs three times or more, delete all the occurrences after the second one.
Print the output as a sentence. The maximum lengths of the input sentence and each individual word is 500 chars and 50 chars.
Exemplary input: jingle bells jingle bells jingle all the way
Exemplary output: jingle bells jinglejingle bellsbells all the way
The approach I take is:
1) Read the input, separate each word and put them into an array of char pointers.
2) Use nested for loop to go through the array. For each word after the first word:
A - If there is no word before it that is equal to it, nothing happens.
B - If there is already one word before it that is equal to it, change the word as its "doubled form".
C - If there is already a "doubled form" of itself that exists before it, delete the word (set the element to NULL.
3) Print the modified array.
I am fairly confident about the correctness of this approach. However, when I actually write the code:
'''
int main()
{
char input[500];
char *output[500];
// Gets the input
printf("Enter a string: ");
gets(input);
// Gets the first token, put it in the array
char *token = strtok(input, " ");
output[0] = token;
// Keeps getting tokens and filling the array, untill no blank space is found
int i = 1;
while (token != NULL) {
token = strtok(NULL, " ");
output[i] = token;
i++;
}
// Processes the array, starting from the second element
int j, k;
char *doubled;
for (j = 1; j < 500; j++) {
strcpy(doubled, output[j]);
strcat(doubled, doubled); // Create the "doubled form"
for (k = 0; k < j; k++) {
if (strcmp(output[k], output[j]) == 0) { // Situation B
output[j] = doubled;
}
if (strcmp(output[k], doubled) == 0) { // Situation C
output[j] = ' ';
}
}
}
// Convert the array to a string
char *result = output[0]; // Initialize a string with the first element in the array
int l;
char *blank_space = " "; // The blank spaces that need to be addded into the sentence
for (l = 1; l < 500; l++) {
if (output[l] != '\0'){ // If there is a word that exists at the given index, add it
strcat(result, blank_space);
strcat(result, output[l]);
}
else { // If reaches the end of the sentence
break;
}
}
// Prints out the result string
printf("%s", result);
return 0;
}
'''
I did a bunch of tests on each individual block. There are several issues:
1) When processing the array, strcmp, strcat, and strcpy in the loop seem to give Segmentation fault error reports.
2) When printing the array, the words did not show the order that they are supposed to do.
I am now frustrated because it seems that the issues are all coming from some internal structural defects of my code and they are very much related to the memory mechanism of C which I am not really too familiar with. How should I fix this?
One issues jumps out at me. This code is wrong:
char *doubled;
for (j = 1; j < 500; j++) {
strcpy(doubled, output[j]);
strcat(doubled, doubled); // Create the "doubled form"
doubled doesn't point to any actual memory. So trying to copy data to where it points is undefined behavior and will almost certainly cause a SIGSEGV - and it will corrupt memory if it doesn't cause a SIGSEGV.
That needs to be fixed - you can't copy a string with strcpy() or strcat() to an pointer that doesn't point to actual memory.
This would be better, but still not ideal as no checking is done to ensure there's no buffer overflow:
char doubled[ 2000 ];
for (j = 1; j < 500; j++) {
strcpy(doubled, output[j]);
strcat(doubled, doubled); // Create the "doubled form"
This is also a problem with doubled defined like that:
if (strcmp(output[k], output[j]) == 0) { // Situation B
output[j] = doubled;
}
That just points output[j] at doubled. The next loop iteration will overwrite doubled, and the data that output[j] still points to to will get changed.
This would address that problem:
if (strcmp(output[k], output[j]) == 0) { // Situation B
output[j] = strdup( doubled );
}
strdup() is a POSIX function that, unsurprising, duplicates a string. That string will need to be free()'d later, though, as strdup() is the same as:
char *strdup( const char *input )
{
char *duplicate = malloc( 1 + strlen( input ) );
strcpy( duplicate, input );
return( duplicate );
}
As pointed out, strcat(doubled, doubled); is also a problem. One possible solution:
memmove(doubled + strlen( doubled ), doubled, 1 + strlen( doubled ) );
That copies the contents of the doubled string to the memory starting at the original '\0' terminator. Note that since the original '\0' terminator is part of the string, you can't use strcpy( doubled + strlen( doubled ), doubled );. Nor can you use memcpy(), for the same reason.
Your code invokes undefined behavior in several places by writing to memory that is not yet owned, and is likely the cause of the segmentation fault you are seeing. For example, in this segment:
char *result = output[0]; // Initialize a string with the first element in the array
int l;
char *blank_space = " "; // The blank spaces that need to be addded into the sentence
for (l = 1; l < 500; l++) {
if (output[l] != '\0'){ // If there is a word that exists at the given index, add it
strcat(result, blank_space);
strcat(result, output[l]);
By itself...
char *result = output[0]; //creates a pointer, but provides no memory.
...is not sufficient to receive content such as
strcat(result, blank_space);
strcat(result, output[l]);
It needs memory:
char *result = malloc(501);//or more space if needed.
if(result)
{
//now use strcat or strcpy to add content
I want to replace all occurrences in an array (string) with another array.
I have a code that:
stores the string in an array in which the replacing is to take place output[],
another array that stores the string to be searched for as replace[] and a third array called toBeReplacedBy and the replacing of the first occurrence works just fine but it skips the other occurrences in the output
for example:
replace[]:
abc
toBeReplacedBy[]:
xyz
output[]:
abcdefabc
becomes
xyzdefabc
but it should become:
xyzdefxyz
I suspect the problem lies with the replacer code :
//the replacer
for (i = 0; i<80; i++) {
if (output[i] == replace[i])
output[i] = toBeReplacedBy[i];
}
//debug purpose
puts("output[]:\n");
puts(output);
return 0;
}
What have I done wrong here and how could I get it to replace all occurrences in the array.
please be aware that I only wish to use stdio.h to do this
thabks in advance
Never iterate further than the array length. This leads to undefined and possibly dangerous behaviour. If you only expect strings, use something like:
int i = 0;
while(output[i] != '\0')
{
// your logic here
i++;
}
Additionally you want to check for concurrent appearances of the same characters. But in your code you only check the first three characters. Everything after that is undefinded behaviour, because you cannot know what replace[3] returns.
Something similar to this could work:
int i = 0;
int j = 0;
int k;
while(output[i] != '\0')
{
if (output[i] == replace[j])
j++;
else
j = 0;
// replace 3 with the array length of the replace[] array
if (j == 3)
{
for(k = i; j >= 0; k-- )
{
output[k] = toBeReplacedBy[j]
j--
}
j = 0;
}
i++;
}
But please check the array boundaries.
edit: Additionally as Nellie states using a debugger would help you to understand what went wrong. Go through your program step by step and look how and when values change.
First advice is to try to debug your program if it does not work.
for (i = 0; i<80; i++) {
if (output[i] == replace[i])
output[i] = toBeReplacedBy[i];
}
There are two problems in this loop.
The first is that are iterating until i is 80. Let's look what happens when i becomes 3. output[3] in case of abcdefabc is d, but what is replace[3]? Your replacement array had only 3 letters, so you have to go back in the replacement array once you finish with one occurrence of it in the original string.
The second is that you check letter by letter.
Say you original array, which you named output somehow was abkdefabc, first three letters do not match your replacement string, but you will check the first two letters they will match with the replacement's first two letters and you will incorrectly change them.
So you need to first check that the whole replacement string is there and only then replace.
You should use strlen() to know length of your array or iterate until you reach the end of a your array ('\0').
'\0' and strlen are only available for array of char.
Your loop should looks like this :
int i = 0;
int len = strlen(my_string);
while (i < len)
{
//logic here
i = i + 1;
}
OR
int i = 0;
while (my_string[i] != '\0')
{
// logic here
i = i + 1;
}
Here is the question:
Write a solution that only iterates over the string once and uses O(1) additional memory, since this is what you would be asked to do during a real interview.
Given a string s, find and return the first instance of a non-repeating character in it. If there is no such character, return '_'.
And here is my code:
char firstNotRepeatingCharacter(char * s) {
int count;
for (int i=0;i<strlen(s);i++){
count=0;
char temp=s[i];
s[i]="_";
char *find= strchr(s,temp);
s[i]=temp;
if (find!=NULL) count++;
else return s[i];
}
if (count!=0) return '_';
}
I dont know what's wrong but when given an input:
s: "abcdefghijklmnopqrstuvwxyziflskecznslkjfabe"
the output is for my code is "g" instead of "d".
I thought the code should have escaped the loop and return "d" soon as "d" was found.
Thx in advance!!!
In your program, problem is in this statement-
s[i]="_";
You are assigning a string to a character type variable s[i]. Change it to -
s[i]='_';
At the bottom of your firstNotRepeatingCharacter() function, the return statement is under the if condition and compiler must be giving a warning for this as the function is supposed to return a char. Moreover, count variable is not needed. You could do something like:
char firstNotRepeatingCharacter(char * s) {
for (int i=0;i<strlen(s);i++){
char temp=s[i];
s[i]='_';
char *find= strchr(s,temp);
s[i]=temp;
if (find==NULL)
return s[i];
}
return '_';
}
But this code is using strchr inside the loop which iterates over the string so, this is not the exact solution of your problem as you have a condition that - the program should iterates over the string once only. You need to reconsider the solution for the problem.
May you use recursion to achieve your goal, something like - iterate the string using recursion and, somehow, identify the repetitive characters and while the stack winding up identify the first instance of a non-repeating character in the string. It's implementation -
#include <stdio.h>
int ascii_arr[256] = {0};
char firstNotRepeatingCharacter(char * s) {
char result = '-';
if (*s == '\0')
return result;
ascii_arr[*s] += 1;
result = firstNotRepeatingCharacter(s+1);
if (ascii_arr[*s] == 1)
result = *s;
return result;
}
int main()
{
char a[] = "abcdefghijklmnopqrstuvwxyziflskecznslkjfabe";
printf ("First non repeating character: %c\n", firstNotRepeatingCharacter(a));
return 0;
}
In the above code, firstNotRepeatingCharacter() function iterates over the string only once using recursion and during winding up of the stack it identifies the first non-repetitive character. I am using a global int array ascii_arr of length 256 to keep the track of non-repetitive character.
Java Solution:
Time Complexity: O(n)
Space Complexity: with constant space as it will only use more 26 elements array to maintain count of chars in the input
Using Java inbuilt utilities : but for inbuilt utilities time complexity is more than O(n)
char solution(String s) {
char[] c = s.toCharArray();
for (int i = 0; i < s.length(); i++) {
if (s.indexOf(c[i]) == s.lastIndexOf(c[i]))
return c[i];
}
return '_';
}
Using simple arrays. O(n)
char solution(String s) {
// maintain count of the chars in a constant space
int[] base = new int[26];
// convert string to char array
char[] input = s.toCharArray();
// linear loop to get count of all
for(int i=0; i< input.length; i++){
int index = input[i] - 'a';
base[index]++;
}
// just find first element in the input that is not repeated.
for(int j=0; j<input.length; j++){
int inputIndex = input[j]-'a';
if(base[inputIndex]==1){
System.out.println(j);
return input[j];
}
}
return '_';
}
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.