I've nearly finished my anagram solver program where I input two strings and get the result of whether they are anagrams of each other. For this example i'm using 'Payment received' and 'Every cent paid me'.
The problem i'm getting is when I output the letterCount arrays, letterCount1 is incorrect (it doesn't think there is a character 'd' but there is.) but letterCount2 is correct.
Can anyone see a problem with this because i'm completely baffled?
#include <stdio.h>
#include <string.h>
int checkAnagram(char string1[], char string2[])
{
int i;
int count = 0, count2 = 0;
int letterCount1[26] = {0};
int letterCount2[26] = {0};
for(i = 0; i < strlen(string1); i++)
{
if(!isspace(string1[i]))
{
string1[i] = tolower(string1[i]);
count++;
}
}
for(i = 0; i < strlen(string2); i++)
{
if(!isspace(string2[i]))
{
string2[i] = tolower(string2[i]);
count2++;
}
}
if(count == count2)
{
for(i = 0; i < count; i++)
{
if(string1[i] >='a' && string1[i] <= 'z')
{
letterCount1[string1[i] - 'a'] ++;
}
if(string2[i] >='a' && string2[i] <= 'z')
{
letterCount2[string2[i] - 'a'] ++;
}
}
printf("%s\n", string1);
for(i = 0; i < 26; i++)
{
printf("%d ", letterCount1[i]);
printf("%d ", letterCount2[i]);
}
}
}
main()
{
char string1[100];
char string2[100];
gets(string1);
gets(string2);
if(checkAnagram(string1, string2) == 1)
{
printf("%s", "Yes");
} else
{
printf("%s", "No");
}
}
That's because your count holds the count of non-space characters, but you keep the strings with the spaces.
For example, the string "hello world" has 11 characters, but if you run it through the loops your count will be 10 (you don't count the space). However, when you later go over the strings and count the appearance of each letter, you will go over the first 10 characters, therefore completely ignoring the last character - a 'd'.
To fix it, you need to go over all characters of the string, and only count the alphanumeric ones.
I fixed it for you:
#include <stdio.h>
#include <string.h>
int checkAnagram(char string1[], char string2[])
{
int i;
int count = 0, count2 = 0;
int letterCount1[26] = {0};
int letterCount2[26] = {0};
int len1 = strlen(string1);
int len2 = strlen(string2);
for(i = 0; i < len1; i++)
{
if(!isspace(string1[i]))
{
string1[i] = tolower(string1[i]);
count++;
}
}
for(i = 0; i < len2; i++)
{
if(!isspace(string2[i]))
{
string2[i] = tolower(string2[i]);
count2++;
}
}
if(count == count2)
{
for (i=0; i<len1; i++)
if (!isspace(string1[i]))
letterCount1[string1[i]-'a']++;
for (i=0; i<len2; i++)
if (!isspace(string2[i]))
letterCount2[string2[i]-'a']++;
int flag = 1;
for(i = 0; flag && i < 26; i++)
if (letterCount1[i] != letterCount2[i])
flag = 0;
return flag;
}
return 0;
}
main()
{
char string1[100];
char string2[100];
gets(string1);
gets(string2);
if(checkAnagram(string1, string2) == 1)
{
printf("%s", "Yes");
} else
{
printf("%s", "No");
}
}
First, don't calculate an string's length inside a loop. I extracted them into len1 and len2 variables.
Second, your loop was wrong! You shouldn't go up to count, you should go up to that string's length.
Third, you didn't return anything from checkAnagram function.
Related
I am trying to get the most frequent letter in a string with only uppercased letters and no spaces between words. to do that I used a function maxArray(array, sizeof array) that gives the biggest number in the array in order to count how many times the letter is repeated and store the info in another array in the same position of each letter in the string. But in the two algorithms I have came up with it doesn't work.
PS: I am just a beginner.
Here is the code:
int maxArray(int *tab, int n) {
int i, tmp;
tmp = tab[0];
for (i = 1; i < n; i++) {
if (tmp < tab[i])
tmp = tab[i];
}
return tmp;
}
//first algo(didn't finish it)
char occurencedelettre(char *string) {
int *array;
int i, j, compt, max;
for (i = 0; string[i] !='\0'; i++) {
compt = 0;
for (j = 0; string[j] !='\0'; j++) {
if (string[i] == string[j])
compt++;
}
array[i] = compt;
}
return array;
}
//second one
char occurencedelettre(char *string) {
int count[25] = { 0 };
int x = 0;
char result;
for (int i = 0; string[i] != '\0'; i++) {
count[string[i]]++;
if (x < count[string[i]]) {
x = count[string[i]];
result = string[i];
}
}
return result;
}
The second approach is almost correct except for these problems:
the array should have a length of 26
you must subtract 'A' from the value of the letter to get an index between 0 and 25, assuming the word only contains uppercase letters in ASCII.
you must intialize result to return 0 (or any other specific value) for an empty word.
Here is a modified version:
char occurencedelettre(const char *string) {
size_t count[26] = { 0 };
size_t x = 0;
char result = '\0';
for (size_t i = 0; string[i] != '\0'; i++) {
count[string[i] - 'A']++;
if (x < count[string[i]]) {
x = count[string[i]];
result = string[i];
}
}
return result;
}
The first approach is more cumbersome and slower, but also more generic as it may work for any word contents. Here is a modified version:
char occurencedelettre(const char *string) {
char result = 0;
size_t max_count = 0;
for (size_t i = 0; string[i] != '\0'; i++) {
size_t count = 1;
for (size_t j = i + 1; string[j] != '\0'; j++) {
if (string[i] == string[j])
count++;
}
if (max_count < count) {
max_count = count;
result = string[i];
}
}
return result;
}
Note that you can adapt the first approach for any word contents, assuming 8-bit bytes:
char occurencedelettre(const char *string) {
size_t count[256] = { 0 };
size_t x = 0;
char result = '\0';
for (size_t i = 0; string[i] != '\0'; i++) {
count[(unsigned char)string[i]]++;
if (x < count[(unsigned char)string[i]]) {
x = count[(unsigned char)string[i]];
result = string[i];
}
}
return result;
}
I want to insert string to the array until I type "ok". Why I am getting just "ok" and original array at the output?
int main(void)
{
char b[20];
char* str[10] = { "1","2" };
int i = 2;
while (1) {
gets(b);
if (strcmp(b, "ok") == 0) break;
str[i] = b;
i++;
}
for (int j = 0; j < i; j++)
printf("%s ", str[j]);
return 0;
}
You need to allocate a string on each iteration:
int main(void)
{
char* b;
char* str[10] = { "1","2" };
int i = 2;
while (1) {
b = malloc(20);
gets(b);
if (strcmp(b, "ok") == 0) break;
str[i] = b;
i++;
}
for (int j = 0; j < i; j++)
printf("%s ", str[j]);
// free allocated strings
while (i > 2)
free(str[--i]);
return 0;
}
They all point to b, which gets overwritten in each iteration.
You need to create a copy of the string when you assign it:
str[i] = strdup(b);
You also may consider using fgets instead of gets; however, you will need to remove the newline:
size_t size;
fgets(str, 20, stdin);
size = strlen(str);
if(str[size-1] == '\n')
str[size-1] = '\0';
Also, print a newline at the end of your program, so it won't interfere with the shell:
putchar('\n');
Full code:
int main(void)
{
char b[20];
char* str[10] = { "1","2" };
int i = 2;
while (1) {
size_t size;
fgets(str, 20, stdin);
size = strlen(str);
if(str[size-1] == '\n')
str[size-1] = '\0';
if (strcmp(b, "ok") == 0)
break;
str[i] = strdup(b);
i++;
}
for (int j = 0; j < i; j++)
printf("%s ", str[j]);
putchar('\n');
return 0;
}
You need to make a copy of the input string, then save a pointer to the copy of the input string in your array. Something like:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
char b[20];
char *str[10] = { "1","2" };
int i = 2;
char *p;
size_t lenb;
for(i = 2 ; i < 10 ; ++i)
{
fgets(b, sizeof(b), stdin);
lenb = strlen(b);
if(lenb > 0 && *(b+lenb-1) == '\n')
{
*(b+lenb-1) = '\0'; /* overwrite the trailing \n */
lenb = strlen(b);
}
if (strcmp(b, "ok") == 0)
break;
p = malloc(lenb+1);
strcpy(p, b);
str[i] = p;
}
for (int j = 0; j < i; j++)
printf("%s\n", str[j]);
return 0;
}
hii guys i need a serious help
i m trying to write a code for finding anagrams in input sentence
but when the if function is getting strcmp it stops and its not accepting the condition. any body know why is that happening
Basically my code supposed to do two things one is taking a sentence from the user and making the words appear in the Backwoods order two Its need to take the whole sentence and look for anagrams ( anagram means that there is the same letters but in a different order for example this and shit are anagrams) thank you very much for your help :)
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
void main()
{
int index_for_word_start, words_num = 1,amount_of_letters;
int i, j, k;
char inpot_Sentence[1001], temp_letters;
char **words,**sorting_words;
int counter = 0,counter_max_for_anegram=0;
printf_s("Please enter the sentence, and then press Enter:\n");
gets(inpot_Sentence);
/////////////////////////////makeing the sentence backwards///////////////////////
for (i = 0; inpot_Sentence[i] != '\0'; i++) //loop for counting how many words(it will be use to know how many pointer we need)
{
if (inpot_Sentence[i] == ' ')
{
words_num++;
}
}
words = (char **)malloc(sizeof(char *)*words_num); //malloc for pointers that point on the pointer of the word
index_for_word_start = 0;
for (j = 0; j<words_num; j++)
{
for (i = index_for_word_start; inpot_Sentence[i] != ' '; i++)
{
if (!inpot_Sentence[i]) //if the user didnt put any word(break)
{
break;
}
}
words[j] = (char*)malloc(sizeof(char)*(i - index_for_word_start + 1)); //malloc of pointers that point on each word
strncpy_s(words[j], i - index_for_word_start+1, &inpot_Sentence[index_for_word_start], i - index_for_word_start); //copy the words from inpot sentence to array
words[j][i - index_for_word_start] = 0; //puts '\0' after the word copy ends
index_for_word_start = i + 1;
}
printf_s("\nThe reverse sentence is:\n");
for (i = words_num - 1; i >= 0; i--) //print the words in backwards Sequence
{
printf("%s ", words[i]);
}
putchar('\n');
i = 0;
/////////////////////anegrams check///////////////////////
for (j = 0; j < words_num; j++) //loops that Arrange the array by haski value
{
amount_of_letters = strlen(words[j]);
for ( i = 0; i < amount_of_letters; i++)
{
for (k = 0; k < amount_of_letters; k++)
{
if (words[j][i]<words[j][k])
{
temp_letters = words[j][i];
words[j][i] = words[j][k];
words[j][k] = temp_letters;
}
}
}
printf_s("this is words %s\n", words[j]);
}i = 0;
for ( j = 0; j < words_num-1; j++)
{
for ( i = 0; i < words_num-1; i++)
{
if (!strcmp(words[j],words[i]) && (i!=j) && (strcmp(words[j],"\0")))
{
counter++;
words[i] = 0;
}
else
{
break;
}
}
if (counter>counter_max_for_anegram)
{
counter_max_for_anegram = counter;
}
counter = 0;
}
printf_s("%d\n", counter_max_for_anegram);
for ( j = 0; j < words_num; j++)
{
free(words[j]);
}
free(words);
}
#include <stdio.h>
#include <string.h>
int check_anagram(char[],char[]);
int main()
{
char a[100],b[100];
int flag;
puts("Enter the first string");
fgets(a,100,stdin);
a[strcspn(a, "\r\n")] = '\0';
puts("Enter the second string");
fgets(b,100,stdin);
b[strcspn(b, "\r\n")] = '\0';
flag=check_anagram(a,b);
if(flag)
printf("%s and %s are anagrams",a,b);
else
printf("%s and %s are not anagrams",a,b);
}
int check_anagram(char a[], char b[])
{
int first[26]={0},second[26]={0},c=0;
while(a[c]!='\0')
{
first[a[c]-'a']++;
c++;
}
c=0;
while(b[c]!='\0')
{
second[b[c]-'a']++;
c++;
}
for(c=0;c<26;c++)
{
if(first[c]!=second[c])
return 0;
}
return 1;
}
So I have an assignment where I should delete a character if it has duplicates in a string. Right now it does that but also prints out trash values at the end. Im not sure why it does that, so any help would be nice.
Also im not sure how I should print out the length of the new string.
This is my main.c file:
#include <stdio.h>
#include <string.h>
#include "functions.h"
int main() {
char string[256];
int length;
printf("Enter char array size of string(counting with backslash 0): \n");
/*
Example: The word aabc will get a size of 5.
a = 0
a = 1
b = 2
c = 3
/0 = 4
Total 5 slots to allocate */
scanf("%d", &length);
printf("Enter string you wish to remove duplicates from: \n");
for (int i = 0; i < length; i++)
{
scanf("%c", &string[i]);
}
deleteDuplicates(string, length);
//String output after removing duplicates. Prints out trash values!
for (int i = 0; i < length; i++) {
printf("%c", string[i]);
}
//Length of new string. The length is also wrong!
printf("\tLength: %d\n", length);
printf("\n\n");
getchar();
return 0;
}
The output from the printf("%c", string[i]); prints out trash values at the end of the string which is not correct.
The deleteDuplicates function looks like this in the functions.c file:
void deleteDuplicates(char string[], int length)
{
for (int i = 0; i < length; i++)
{
for (int j = i + 1; j < length;)
{
if (string[j] == string[i])
{
for (int k = j; k < length; k++)
{
string[k] = string[k + 1];
}
length--;
}
else
{
j++;
}
}
}
}
There is a more efficent and secure way to do the exercise:
#include <stdio.h>
#include <string.h>
void deleteDuplicates(char string[], int *length)
{
int p = 1; //current
int f = 0; //flag found
for (int i = 1; i < *length; i++)
{
f = 0;
for (int j = 0; j < i; j++)
{
if (string[j] == string[i])
{
f = 1;
break;
}
}
if (!f)
string[p++] = string[i];
}
string[p] = '\0';
*length = p;
}
int main() {
char aux[100] = "asdñkzzcvjhasdkljjh";
int l = strlen(aux);
deleteDuplicates(aux, &l);
printf("result: %s -> %d", aux, l);
}
You can see the results here:
http://codepad.org/wECjIonL
Or even a more refined way can be found here:
http://codepad.org/BXksElIG
Functions in C are pass by value by default, not pass by reference. So your deleteDuplicates function is not modifying the length in your main function. If you modify your function to pass by reference, your length will be modified.
Here's an example using your code.
The function call would be:
deleteDuplicates(string, &length);
The function would be:
void deleteDuplicates(char string[], int *length)
{
for (int i = 0; i < *length; i++)
{
for (int j = i + 1; j < *length;)
{
if (string[j] == string[i])
{
for (int k = j; k < *length; k++)
{
string[k] = string[k + 1];
}
*length--;
}
else
{
j++;
}
}
}
}
You can achieve an O(n) solution by hashing the characters in an array.
However, the other answers posted will help you solve your current problem in your code. I decided to show you a more efficient way to do this.
You can create a hash array like this:
int hashing[256] = {0};
Which sets all the values to be 0 in the array. Then you can check if the slot has a 0, which means that the character has not been visited. Everytime 0 is found, add the character to the string, and mark that slot as 1. This guarantees that no duplicate characters can be added, as they are only added if a 0 is found.
This is a common algorithm that is used everywhere, and it will help make your code more efficient.
Also it is better to use fgets for reading input from user, instead of scanf().
Here is some modified code I wrote a while ago which shows this idea of hashing:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#define NUMCHAR 256
char *remove_dups(char *string);
int main(void) {
char string[NUMCHAR], temp;
char *result;
size_t len, i;
int ch;
printf("Enter char array size of string(counting with backslash 0): \n");
if (scanf("%zu", &len) != 1) {
printf("invalid length entered\n");
exit(EXIT_FAILURE);
}
ch = getchar();
while (ch != '\n' && ch != EOF);
if (len >= NUMCHAR) {
printf("Length specified is longer than buffer size of %d\n", NUMCHAR);
exit(EXIT_FAILURE);
}
printf("Enter string you wish to remove duplicates from: \n");
for (i = 0; i < len; i++) {
if (scanf("%c", &temp) != 1) {
printf("invalid character entered\n");
exit(EXIT_FAILURE);
}
if (isspace(temp)) {
break;
}
string[i] = temp;
}
string[i] = '\0';
printf("Original string: %s Length: %zu\n", string, strlen(string));
result = remove_dups(string);
printf("Duplicates removed: %s Length: %zu\n", result, strlen(result));
return 0;
}
char *remove_dups(char *str) {
int hash[NUMCHAR] = {0};
size_t count = 0, i;
char temp;
for (i = 0; str[i]; i++) {
temp = str[i];
if (hash[(unsigned char)temp] == 0) {
hash[(unsigned char)temp] = 1;
str[count++] = str[i];
}
}
str[count] = '\0';
return str;
}
Example input:
Enter char array size of string(counting with backslash 0):
20
Enter string you wish to remove duplicates from:
hellotherefriend
Output:
Original string: hellotherefriend Length: 16
Duplicates removed: helotrfind Length: 10
C program for removal of duplicate characters from the given string. It uses the O(n2) can we do it in O(n) order. Please comment on this program.
int main()
{
char a[100],b[100],temp='\0';
int i,n,j,count=0,p=0,k=0;
printf("ENTRE THE STRING \n");
scanf("%s",a);
n = strlen(a);
i=0;
while(i < n)
{
count=0;
temp = a[i];
for(j = i ; j < n ; j++ )
{
if(temp==a[j])
{
count++;
}
}
if(count<2)
{
b[k] = temp;
k++;
}
i++;
}
b[k]='\0';
printf("THE RESULTED STRING IS \n");
for(p = 0 ; p < k ; p++)
printf("%c ",b[p]);
printf("\n");
return 0;
}
You can create a O(n) algorithm for this.
Steps:
Create another array bucket[] with size 255. (Should adjust all the characters)
Initialise every element in bucket[] to 0.
Run a loop and increment the bucket[] at the index a[i].
Now, run another loop through the bucket[], if bucket[i] > 0, append the (char) i to the b[] array.
Code:
#include <stdio.h>
#include <string.h>
int main()
{
char a[100], b[100];
int bucket[256] = {0};
int i;
printf("Enter the string:");
scanf("%s",a);
int n = strlen(a);
for(i = 0; i < n; ++i)
{
//Incrementing the character count of each character.
bucket[a[i]]++;
}
//Keep track of the index where the next character is to be appended.
int b_pos = 0;
for (i = 0; i < 256; ++i)
{
//Character occurs in a[], we don't care if it occurs once
//or twice, we just need one instance of it.
if (bucket[i] > 0)
{
b[b_pos] = (char) i;
b_pos++;
}
}
b[b_pos] = '\0';
printf("Modified string : %s",b);
}
Take a look at this:
int main()
{
char a[100],b[100];
int i,n,j,count=0,k=0;
printf("ENTRE THE STRING \n");
scanf("%s",a);
n = strlen(a);
b[0] = a[0];
k = 1;
for(i=1;i<n;i++)
{
for(j=0;j<i;j++)
{
if(a[i] == b[j])
{
count = 1;
break;
}
}
if(count == 0)
{
b[k] = a[i];
k++;
}
else
{
count = 0;
}
}
b[k] = 0;
printf("RESULT %s",b);
return 0;
}