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;
}
Related
this code firstly takes two words for user. Then, it check those length with a function. Then the code send the words and their length to another function for it to find commen letters in these words. The same function also print that how many commen letters they have.(The user enter just lower letter or just capital letters)
for example:
input1: emirhan
input2: celek
output: The words that you entered have 2 common letters
output shouldn't be 3 because input2 (celek) has 2 'e' and the code count just one of them.
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define MAX 40
int length(char word[30]) {
int uzun;
uzun = strlen(word) - 1;
return uzun;
}
int equation(char x[MAX], char y[MAX], int length1, int length2) {
int i, j, k, l;
int sayac = 0; // sayac means counter
if (length1 >= length2) {
for (i = 0; i < length1; i++) {
for (j = 0; j < length2; i++) {
if (x[i] == y[j]) { // if the code find a common letter,it will enter if
sayac++;
for (k = i + 1; k < length1; k++) { //This 'for' check that does first word same letter or not?
if (x[k] == x[i]) {
sayac--;
}
}
for (l = j + 1; l < length2; l++) { //This 'for' check that does second word same letter or not?
if (y[l] == y[j]) {
sayac--;
}
}
}
}
}
}
if (length1 < length2) {
for (i = 0; i < length2; i++) {
for (j = 0; j < length1; i++) {
if (x[i] == y[j]) {
sayac++;
for (k = i + 1; k < length2; k++) {
if (x[k] == x[i]) {
sayac--;
}
}
for (l = j + 1; l < length1; l++) {
if (y[l] == y[j]) {
sayac--;
}
}
}
}
}
}
printf("The words that you entered have %d common letters", sayac);
}
int main() {
char x[MAX]; //input1
char y[MAX]; //input2
int length1; // length of first word
int length2; // length of second word
printf("PLS, enter the first word:");
fgets(x, sizeof(x), stdin);
printf("PLS, enter the second word:");
fgets(y, sizeof(y), stdin);
length1 = length(x);
length2 = length(y);
equation(x, y, length1, length2);
return 0;
}
I edited the my code again and I solved the problem. I had a mistake in loops. I coded loops again.
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define MAX 40
int length(char word[30]) {
int uzun;
uzun = strlen(word) - 1;
return uzun;
}
void equation(char x[MAX], char y[MAX], int length1, int length2) {
int i, j, k, l;
int sayac = 0; // sayac means counter
for (i = 0; i < length1; i++) {
for (j = 0; j < length2; j++) {
if (x[i] == y[j]) { // if the code find a common letter,it will enter if
sayac++;
for (k = i + 1; k < length1; k++) { //This 'for' check that does first word same letter or not?
if (x[k] == x[i]) {
sayac--;
}
}
for (l = j + 1; l < length2; l++) { //This 'for' check that does second word same letter or not?
if (y[l] == y[j]) {
sayac--;
}
}
}
}
}
printf("The words that you entered have %d common letters", sayac);
}
int main() {
char x[MAX]; //input1
char y[MAX]; //input2
int length1; // length of first word
int length2; // length of second word
printf("PLS, enter the first word:");
fgets(x, sizeof(x), stdin);
printf("PLS, enter the second word:");
fgets(y, sizeof(y), stdin);
length1 = length(x);
length2 = length(y);
equation(x, y, length1, length2);
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;
}
how to code a user defined function that searches and replaces a character occurrences of any of the character contained in another string with a character string.
Cannot used any string variable in the code, has to be a user defined function.
Thanks
This is what i have tried so far
#define _CRT_SECURE_NO_WARNINGS
#include
#include
void s1();
void s2();
int main(void)
{
int i=0;
s1();
s2();
printf("c = {'$'} ");
}//main
void s1(){
int i = 0;
while (i <= 40){
printf("%c", (rand() % 25) + 'A');
i++;
}
}
void s2(){
char s2[20];
printf("\nEnter a string of minimum 2 and maximum 20 characters= ");
gets(s2);
puts(s2);
}
/*
I just need to make another function that searches s1 and replaces any occurrence of any of the character contained is s2 with a character that can be anything(e.g. '$')
*/
//If I have understood your question then this should be answer
char *replace(char [] a, char b[], int lower, int upper){
char c[100];
int j = 0;
for(int i = 0; i < lower; i++){
c[j] = a[i];
j++;
}
for(int i = 0; i < strlen(b); i++){
c[j] = b[i];
j++;
}
for(int i = upper; i < strlen(a); i++){
c[j] = a[i];
j++;
}
c[j] = '\0'
for(int i = 0; i < strlen(c); i++){
a[i]= c[i];
}
a[i] = '\0';
return a;
}
Can someone advise why the loop in the main dies after the fifth iteration never completing
it's intended goal of reducing the character array down to 1 final element? I've gotten it this
far and am completely consumed as their should be 11 iterations as returned by the call
size_t strlen( char const *str )
{
int length = 0;
while (*str++ !='\0')
{
length += 1;
}
return length;
}
void abracadabra( char *word )
{
int i, c;
int len = strlen(word)-1;
for (i = 0; i <= len; i++)
{
putchar(*word);
putchar(' ');
*(word++);
}
}
int main()
{
char word[250];
int i, j;
printf ("enter your word:\n");
scanf ("%[^\n]s", &word);
for (i = 0; i <= strlen(word)-1; i++)
{
abracadabra(word);
putchar('\0');
printf("\n");
for (j = 0; j <= i; j++)
{
putchar('\0');
}
word[strlen(word) - 1] = '\0';
}
word[strlen(word)-1] = '\0';
printf("\n");
system("pause");
return 0;
}
Each time you execute the outer for loop in main the size of the string decreases by 1. The counter i is also increasing by 1 each time. This causes you to run the loop half of the times that you intend to.
int size = strlen(word);
for (i = 0; i < size; i++) {
\\same inner code
}
Subbing the above code for the outer for loop in main resolves the issue.
The is mistake in using the variable i and strlen in for loop. i is keep increasing and word length is decreasing. So in the mid, loop is terminated due to i>strlen (word)
int main()
{
char word[250];
int i, j;
printf ("enter your word:\n");
scanf ("%[^\n]s", &word);
// for (i = 0; i <= strlen(word)-1; i++)
while ( strlen(word) )
{
abracadabra(word);
putchar('\0');
printf("\n");
word[strlen(word) - 1] = '\0';
}
word[strlen(word)-1] = '\0';
printf("\n");
system("pause");
return 0;
}
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.