C - Find most frequent element in char array - c

i'm developing a little function to display the most frequent character in a (char) array.
This is what I've accomplished so far, but I think i'm on the wrong way.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char test[10] = "ciaociaoci";
max_caratt(test, 10);
}
int max_caratt(char input[], int size)
{
int i;
char max[300];
max[0] = input[0];
for (i=0; i<size; i++)
{
if(strncmp(input,input[i],1) == 1)
{
printf("occourrence found");
max[i] = input[i];
}
}
}
Any help?

Actually, the correct code is this.
It's just a corrected version of IntermediateHacker's below snippet.
void main()
{
int array[255] = {0}; // initialize all elements to 0
char str[] = "thequickbrownfoxjumpedoverthelazydog";
int i, max, index;
for(i = 0; str[i] != 0; i++)
{
++array[str[i]];
}
// Find the letter that was used the most
max = array[0];
index = 0;
for(i = 0; str[i] != 0; i++)
{
if( array[str[i]] > max)
{
max = array[str[i]];
index = i;
}
}
printf("The max character is: %c \n", str[index]);
}

The easiest way to find the most common character is to create an int array of 255 and just increment the arraly element that corresponds to the character. For example: if the charcter is 'A', then increment the 'A'th element (if you look at any ascii table you will see that the letter 'A' has a decimal value of 65)
int array[255] = {0}; // initialize all elements to 0
char str[] = "The quick brown fox jumped over the lazy dog.";
int i, max, index;
// Now count all the letters in the sentence
for(i = 0; str[i] != 0; i++)
{
++array[str[i]];
}
// Find the letter that was used the most
max = array[0];
index = 0;
for(i = 0; str[i] != 0; i++)
{
if( array[i] > max)
{
max = array[i];
index = i;
}
}
printf("The max character is: %c \n", (char)index);

You're passing a (almost) string and a char to strncmp(). strncmp() takes two strings (and an integer). Your program shouldn't even compile!
Suggestion: increase the warning level of your compiler and mind the warnings.
You may want to look at strchr() ...

Assuming an input array of 0-127, the following should get you the most common character in a single pass through the string. Note, if you want to worry about negative numbers, shift everything up by +127 as needed...
char mostCommonChar(char *str) {
/* we are making the assumption that the string passed in has values
* between 0 and 127.
*/
int cnt[128], max = 0;
char *idx = str;
/* clear counts */
memset((void *)cnt, 0, sizeof(int) * 128);
/* collect info */
while(*idx) {
cnt[*idx]++;
if(cnt[*idx] > cnt[max]) {
max = *idx;
}
idx++;
}
/* we know the max */
return max;
}

If you don't need to preserve the input array, you could sort the input array first, then find the longest contiguous run of a single character. This approach is slower, but uses less space.

I made a working version using structs. It works fine, I guess, but I think there's a MUCH better way to write this algorithm.
#include <stdio.h>
#include <stdlib.h>
struct alphabet {
char letter;
int times;
};
typedef struct alphabet Alphabet;
void main() {
char string[300];
gets(string);
Alphabet Alph[300];
int i=0, j=0;
while (i<=strlen(string)) {
while(j<=300) {
if(string[i] != Alph[j].letter) {
Alph[i].letter = string[i];
Alph[i].times = 1;
}
else {
Alph[j].times++;
}
j++;
}
j=0;
i++;
}
int y,max=0;
char letter_max[0];
for (y=0; y<strlen(string); y++) {
printf("Letter: %c, Times: %d \n", Alph[y].letter, Alph[y].times);
if(Alph[y].times>max) {
max=Alph[y].times;
letter_max[0]=Alph[y].letter;
}
}
printf("\n\n\t\tMost frequent letter: %c - %d times \n\n", letter_max[0], max);
}

I saw you all creating big arrays and "complex" stuff so here I have easy and simple code xD
char most_used_char (char s[]) {
int i; //array's index
int v; //auxiliary index for counting characters
char c_aux; //auxiliary character
int sum = 0; //auxiliary character's occurrence
char c_max; //most used character
int max = 0; //most used character's occurrence
for (i = 0; s[i]; i++) {
c_aux = s[i];
for (v = 0; s[v]; v++)
if (c_aux == s[v]) sum++; /* responsible cycle for counting
character occurrence */
if (sum > max) { //checks if new character is the most used
max = sum;
c_max = c_aux;
}
sum = 0; /* reset counting variable so it can counts new
characters occurrence */
}
return c_max; //this is the most used character!
}

Related

Every k-th digit cyclic problem using strings in C

Given some number in a form of string, I want to extract every k-th number from it. Then I go through the remaining string and extract every k-th number again. The thing I get as a result should be the number formed by these extracted ones(in a proper order). Example: 123456789, k = 3 --> 369485271
My algorithm is as follows: While the lenght of the string allows extracting every k-th number, I go through the string and store every k-th element in another string. Then I delete the extracted elements from the original string by tracking the proper index of an element and proceed forvard while the lenght of my str is sufficient.
I can't figure out what's the problem with my code. And maybe my approach isn't that good and there are some better/simpler ways of diong this?
#include <stdio.h>
#include <string.h>
void remove(char *str, unsigned int index) {
char *src;
for (src = str+index; *src != '\0'; *src = *(src+1),++src) ;
*src = '\0';
}
int main() {
char number[100];
char result[100];
int k;
printf("Enter a string: ");
scanf("%s",number);
printf("Enter a key: ");
scanf("%d",&k);
while (strlen(number)>k-1) {
for (int i = 0, p = 0; number[i] != '\0'; i++) {
if (i % k == (k-1)) {
result[p] = number[i];
p++;
}
}
for (int j = 0; number[j] != '\0'; j++){
if (j % k == (k-1)) {
remove(number, j);
j+=1; /*since the index was shifted due to removing an element*/
}
}
}
puts(result);
return 0;
}
You some issues:
You start writing your output from scratch again in each iteration of your while loop.
You do not handle the last digits
You do not treat the input as a cyclic input.
You do not terminate your output string.
remove is already a name of standard library function.
A shorter version could be this (untested):
#include <stdio.h>
#include <string.h>
void remove_digit(char *str, unsigned int index) {
char *src;
for (src = str+index; *src != '\0'; *src = *(src+1),++src)
;
}
int main() {
char number[100];
char result[100];
int k;
printf("Enter a string: ");
scanf("%s",number);
printf("Enter a key: ");
scanf("%d",&k);
int p = 0;
int i = 0;
int skip = k-1; // We remove 1 digit and skip k-1 digits
while (number[0] != 0) {
i = (i + skip) % strlen(number);
result[p] = number[i];
p++;
remove_digit(number, i);
}
number[p] = 0;
puts(result);
return 0;
}
The following code seems to be what you want:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void remove_(char *str, unsigned int index) {
char *src;
for (src = str+index; *src != '\0'; *src = *(src+1),++src) ;
*src = '\0';
}
int main(int argc, const char * argv[]) {
char number[100];
char result[100];
int tmp[100];
int k;
printf("Enter a string: ");
scanf("%s",number);
printf("Enter a key: ");
scanf("%d",&k);
int p = 0;
for (int tp = 0; strlen(number) > k-1; tp = 0) {
for (int i = 0; number[i] != '\0'; i++)
if (i % k == (k-1))result[p++] = number[i];
for (int j = 0; number[j] != '\0'; j++)
if (j % k == (k-1)) tmp[tp++] = j;
for (; tp; --tp) remove_(number, tmp[tp-1]);
}
// The newly added code
for (int index; strlen(number); ) {
index = (k-1) % strlen(number);
result[p++] = number[index];
remove_(number, index);
}
puts(result);
return 0;
}
The most important thing is that every while loop, you need to remove the elements in number at once. While ensuring the integrity of your original code, I made some changes. Unfortunately, the main idea of ​​the original code is wrong.
It should circulate from the tail (including the rest) to the head after one round. But I found that the function of the code you provided is that after each round, the next round starts from the 0th element of the head.
By the way, your algorithm is similar to the Josephus problem

How do I compare an integer with a character in a string in C?

I want to compare the integers in a string with integers (0-9) and I wrote this -
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
char num[100];
int count = 0;
scanf("%s", num);
int len = strlen(num);
for (int i = 0; i <= 9; i++)
{
for (int j = 0; j <= len; j++)
{
if (i == (num[j] - '0'))
{
count++;
}
}
printf("%d ", count);
count = 0;
}
return 0;
}
No problems with this (works in most cases but it is failing in few cases). So can you please give me alternate and best idea to do this?
Thanks in advance
Complete pic -
The root cause is not in char comparison, but in the under-allocated buffer:
char num[100];
The assignment constraint is:
1 <= len(num) <= 1000
After increasing the buffer size, all the tests pass.
Besides a too small input buffer (i.e. 100 instead of 1001), I think your approach is too complex.
Instead of a nested loop, I'll suggest an array to count the frequency, i.e. an array with 10 elements so that you have a counter for each digit.
int main() {
char num[1001]; // 1000 chars + 1 zero termination
int count[10] = {0}; // Array of 10 zero initialized counters, one for each digit
scanf("%1000s", num); // At max accept 1000 chars input
char* p = num;
while (*p)
{
if (isdigit(*p) ++count[*p - '0'];
++p;
}
for (int i = 0; i < 10; ++i) printf("%d ", count[i]);
puts("");
return 0;
}
If you don't want to use isdigit you can instead do:
if (*p >= '0' && *p <= '9') ++count[*p - '0'];

Comparing elements between 2 strings

Hello, let's say I got 2 strings, "Today is a nice day" and "ao". I want to delete the chars of the 2nd string that appear in the 1st one.
This is my issue:
char c[20];
char p[10];
int i,j;
int l1,l2;
printf("Enter a string \n");
scanf("%s",cd);
printf("Enter another string \n");
scanf("%s",car);
len1 = strlen(cd);
len2 = strlen(car);
for (i=0;i<len1;i++){
for (j=0;j<len2;j++){
if (cd[i]==car[j]){
cd[i]="";
}
}
}
What I want is the 1st string to be like "Tdy is nice dy". So I empty the positions where the elements are the same to reposition it later.
Apparently "cd[i]==car[j]" can't be done on C, I got "Invalid conversion from 'const char*' to 'char'.
So i'm pretty much stuck. I'll thank any help.
1) This is a solution matching your algorithm as close as possible.
All what you need is an extra loop and to replace cd[i]=""; which cannot be compiled with cd[i]=0;. The error given by the compiler relates to expression cd[i]=""; cd[i] is a character type and you cannot assign string "" which has a type const char * to char variable. cd[i] is a character "" is a pointer.
The operation cd[i]=0; gives you want you wanted: I empty the positions where the elements are the same to reposition it later. It replaces the unwanted characters with 0.
#include <stdio.h>
#include <string.h>
int main()
{
char cd[] = "Today is a nice day";
char tmp[] = "Today is a nice day";
char car[] = "ao";
int i;
int j;
int k;
int len1 = strlen(cd);
int len2 = strlen(car);
for (i=0;i<len1;i++){
for (j=0;j<len2;j++){
if (cd[i] == car[j]){
cd[i]=0;
}
}
}
k = 0;
for (i=0; i<len1; i++)
{
if(cd[i] == 0)
{
}
else
{
tmp[k] = cd[i];
k++;
}
}
tmp[k] = 0; /* remember to terminate the tmp */
printf("%s\n", tmp);
strcpy(cd,tmp);
printf("%s\n", cd);
return 0;
}
OUTPUT:
Tdy is nice dy
Tdy is nice dy
Alternatively, instead of clearing unwanted character with 0 you could just skip it. This solution is given below:
#include <stdio.h>
#include <string.h>
int main()
{
char cd[] = "Today is a nice day";
char car[] = "ao";
int i;
int j;
int k = 0;
int skip = 0;
int len1 = strlen(cd);
int len2 = strlen(car);
for (i=0; i<len1; i++)
{
for (j=0; j<len2; j++)
{
if (cd[i] == car[j])
{
skip++; // make note that this character is not needed
}
}
if(skip == 0)
{
cd[k] = cd[i]; // copy the character
k++; // increase the position index
}
else
{
// skip the copy of charcter; clear the skip marker
skip = 0;
}
}
cd[k] = 0; // remember to terminate the new ck string!
printf("%s\n", cd);
return 0;
}
OUTPUT:
Tdy is nice dy

Count and get integers from a string using C

I am self teaching C programming.
I am trying to count number of int present in given string which are separated by space.
exp:
input str = "1 2 11 84384 0 212"
output should be: 1, 2, 11, 84384, 0, 212
total int = 6
When I try. It gives me all the digits as output which make sense since I am not using a right approach here.
I know in python I can use str.split (" ") function which can do my job very quickly.
But I want to try something similar in C. Trying to create my own split method.
#include <stdio.h>
#include <string.h>
void count_get_ints(const char *data) {
int buf[10000];
int cnt = 0, j=0;
for (int i=0; i<strlen(data); i++) {
if (isspace(data[i] == false)
buf[j] = data[i]-'0';
j++;
}
printf("%d", j);
}
// when I check the buffer it includes all the digits of the numbers.
// i.e for my example.
// buf = {1,2,1,1,8,4,3,8,4,0,2,1,2}
// I want buf to be following
// buf = {1,2,11,84384,0,212}
I know this is not a right approach to solve this problem. One way to keep track of prev and dynamically create a memory using number of non space digits encountered.
But I am not sure if that approach helps.
You want to build your number incrementally until you hit a space, then put that into the array. You can do this by multiplying by 10 then adding the next digit each time.
void count_get_ints(const char *data) {
int buf[10000];
int j = 0;
int current_number = 0;
// Move this outside the loop to eliminate recalculating the length each time
int total_length = strlen(data);
for (int i=0; i <= total_length; i++) {
// Go up to 1 character past the length so you
// capture the last number as well
if (i == total_length || isspace(data[i])) {
// Save the number, and reset it
buf[j++] = current_number;
current_number = 0;
}
else {
current_number *= 10;
current_number += data[i] - '0';
}
}
}
I think strtok will provide a cleaner solution, unless you really want to iterate over every char in the string. It has been a while since I did C, so please excuse any errors in the code below, hopefully it will give you the right idea.
#include <stdio.h>
#include <stdlib.h>
int main() {
char str[19] = "1 2 11 84384 0 212";
const char s[2] = " ";
char *token;
int total;
total = 0;
token = strtok(str, s);
while (token != NULL) {
printf("%s\n", token);
total += atoi(token);
token = strtok(NULL, s);
}
printf("%d\n", total);
return 0;
}
You can check the ascii value of each character by doing c-'0'. If it's between [0,9], then it's an integer. By having a state variable, when you're inside an integer by checking if a given character is a number of space, you can keep track of the count by ignoring white space. Plus you don't need a buffer, what happens if data is larger than 10,000, and you write pass the end of the buffer?, undefined behavior will happen. This solution doesn't require a buffer.
Edit, the solution now prints the integers that are in the string
void count_get_ints(const char *data) {
int count = 0;
int state = 0;
int start = 0;
int end = 0;
for(int i = 0; i<strlen(data); i++){
int ascii = data[i]-'0';
if(ascii >= 0 && ascii <= 9){
if(state == 0){
start = i;
}
state = 1;
}else{
//Detected a whitespace
if(state == 1){
count++;
state = 0;
end = i;
//Print the integer from the start to end spot in data
for(int j = start; j<end; j++){
printf("%c",data[j]);
}
printf(" ");
}
}
}
//Check end
if(state == 1){
count++;
for(int j = start; j<strlen(data); j++){
printf("%c",data[j]);
}
printf(" ");
}
printf("Number of integers %d\n",count);
}
I believe the standard way of doing this would be using sscanf using the %n format specifier to keep track of how much of the string is read.
You can start with a large array to read into -
int array[100];
Then you can keep reading integers from the string till you can't read anymore or you are done reading 100.
int total = 0;
int cont = 0;
int ret = 1;
while(ret == 1 && total < 100) {
ret = sscanf(input, "%d%n", &array[total++], &cont);
input += cont;
}
total--;
printf("Total read = %d\n", total);
and array contains all the numbers read.
Here is the DEMO
Example using strtol
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <errno.h>
#include <ctype.h>
int count_get_ints(int output[], int output_size, const char *input) {
const char *p = input;
int cnt;
for(cnt = 0; cnt < output_size && *p; ++cnt){
char *endp;
long n;
errno = 0;
n = strtol(p, &endp, 10);
if(errno == 0 && (isspace((unsigned char)*endp) || !*endp) && INT_MIN <= n && n <= INT_MAX){
output[cnt] = n;
while(isspace((unsigned char)*endp))
++endp;//skip spaces
p = endp;//next parse point
} else {
fprintf(stderr, "invalid input '%s' in %s\n", p, __func__);
break;
}
}
return cnt;
}
int main(void) {
const char *input = "1 2 11 84384 0 212";
int data[10000];
int n = sizeof(data)/sizeof(*data);//number of elements of data
n = count_get_ints(data, n, input);
for(int i = 0; i < n; ++i){
if(i)
printf(", ");
printf("%d", data[i]);
}
puts("");
}
Assuming you don't have any non-numbers in your string, you can just count the number of spaces + 1 to find the number of integers in the string like so in this pseudo code:
for(i = 0; i < length of string; i++) {
if (string x[i] == " ") {
Add y to the list of strings
string y = "";
counter++;
}
string y += string x[i]
}
numberOfIntegers = counter + 1;
Also, this reads the data between the white spaces. Keep in mind this is pseudo code, so the syntax is different.

I am trying to compare string literals and I want to remove repeated literals I want to do it without using POINTERS.,

I am trying to compare string literals and I want to remove repeated literals I want to do it without using POINTERS.
This is my code:
char str[30];
printf("Enter strings : ");
fgets(str,29,stdin);
char tem[30];
int count , county;
for(count = 0 ; count < strlen(str)-1 ; count++) {
for(county = 1 ; county < strlen(str) ; county++) {
if(str[count] != str[county]) {
tem[count] = str[count];
}
}
}
//PRINT
for(count = 0 ;count < strlen(str) -1 ; count++) {
printf("%c",tem[count]);
}
Input: happen
Expected output: hapen
Correcting your code:
#include <stdio.h>
#include <string.h>
int main()
{
char str[30];
printf("Enter strings : ");
fgets(str,30,stdin);
char tem[30];
size_t count;
size_t county=0;;
for(count = 0 ; count < strlen(str)-1 ; count++) {
if(str[count] != str[count+1]) {
tem[county++] = str[count];
}
}
tem[county] = '\0';
printf("%s\n", tem);
return 0;
}
Take note that this code remove double chars if this char have +1 displacement in the sting.
EDIT
To have mspi as output of entered string mississippi
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
int main()
{
char str[30];
printf("Enter strings : ");
fgets(str,30,stdin);
char tem[30];
size_t count;
size_t county;
size_t tem_index = 0;
size_t size_of_string = strlen(str);
bool found;
for(count = 0 ; count < size_of_string-1; count++)
{
found = false;
county = count+1;
while ((found == false) && (county<size_of_string))
{
if(str[count] == str[county])
{
found = true;
}
county++;
}
if (found == false)
{
tem[tem_index++] = str[count];
}
}
tem[tem_index] = '\0';
printf("%s\n", tem);
return 0;
}
EDIT 2
To have misp as output of entered string mississippi
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
int main()
{
char str[30];
printf("Enter strings : ");
fgets(str,30,stdin);
char tem[30];
size_t count;
size_t county;
size_t tem_index = 0;
size_t size_of_string = strlen(str);
bool found;
for(count = 0 ; count < size_of_string-1; count++)
{
found = false;
county = 0;
while ((found == false) && (county<tem_index))
{
if(str[count] == tem[county])
{
found = true;
}
county++;
}
if (found == false)
{
tem[tem_index++] = str[count];
}
}
tem[tem_index] = '\0';
printf("%s\n", tem);
return 0;
}
Take note that all these solutions are case sensitive.
Your code to remove allduplicates isn't quite there yet:
You use the same indices for the new and the old string, but you need two different indices, because the new string is as long or shorter than the old string. If the old string is "aaab", your index for the old string is 3 when you see the "b", but the index for the new string is only 1. (By skipping indices you leave uninitialised gaps in your string.)
You look forward to find other occurrences of the same letter, but you append to the new string for every letter that doesn't atch. You must look at all folllowing letters, but you must append to the new string only once. That is, you must make your decision whether the letter is duplicate or not after the loop, based on the information that you've found in the loop.
When you look forward, you shouldn't start a 1, but at the letter after he current letter. If you start at one, you will find duplicates for every letter after the first, because you check each letter with itself.
This is not an arror, but it's not a good idea to call strlen repeatedly in a loop. The length of the input string doesn't change, so you can determine the string length beforehand. If you just want to use it as your termination condition, you can test whethet the current letter is the null terminator.
Below is a solution that uses your logic, albeit by looking backwards, not forward. (If you look forward, you will copy the last occurence of a letter, if you look back, you'll copy the first occurrence. It may make a difference in the order of the letters. For Mississippi, you'll get "Mspi" or "Misp" depending on which strategy you use.)
The program overwrites the same string. This is possible, because you are filtering out letters and the new index is equal to the old index or smaller:
#include <stdlib.h>
#include <stdio.h>
void remdup(char *str)
{
int i = 0; // index into old string
int j = 0; // index into new string
for (i = 0; str[i]; i++) {
int k = 0;
int dup = 0;
for (k = 0; k < i; k++) {
if (str[i] == str[k]) {
dup = 1;
break;
}
}
if (dup == 0) str[j++] = str[i];
}
str[j] = '\0';
}
int main()
{
char str[] = "Mississippi";
puts(str);
remdup(str);
puts(str);
return 0;
}
This solution doesn't scale for large strings. A more effective method would be to keep a table of which of the 256 possible characters have already been used.

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