how to find duplicate string in an array of strings

Question

I have an array of string from which i have to find duplicate string and then remove that duplicate string like i have string

 char aa[50]="Amit Hanish Mahesh Amit"

Now Amit is duplicate and have to remove it from string .

#include "string.h"
main()
{
  char x[100] = "Amit Hanish Mahesh Amit";
  char y[3][100];
  int i = 0, k = 0, j = 0, c = 0, end, t;
  int current = 1;
  while (x[i] != '\0') {
    if (x[i] != ' ') {
      y[k][j] = x[i];
      j++;
      i++;
    } else {
      // c = c + 1;
      i++;
      k++;
      j = 0;
    }
    y[k][j] = '\0';
  }

  for (end = 1; end <= 3; end++) {
    for (t = 0; t < end; t++) {
      if (strcmp(y[end], y[t]) == 0) break;
    }
    if (end == t) {
      strcpy(y[current],y[t]);
       current++;
    }
  }
  y[current] = 0;
  printf("%s",y);
}

I have written a smalll routine for it .Does not seems to be worked .Any one have any suggestion where i am going wrong?

you seem like you didnt use strstr function that makes your task little more efficient try to write a function using strstr thats an advice, I got some work to do when im out of it i will try to create a small function for you Thanks — niko
– niko, Commented Oct 14, 2011 at 7:25
Your approach is flawed because you don't know in advance the number of word occurence in the initial string. So it is not possible to have y as a fixed array. You take the risk to have a memory fault (which in your case might even be true, as you will have 4 elements instead of 3). — Huygens
– Huygens, Commented Oct 14, 2011 at 8:29
@Huygens do you have any other idea to accomplish this task? — Amit Singh Tomar
– Amit Singh Tomar, Commented Oct 14, 2011 at 8:55

uscjeremy · Accepted Answer · 2011-10-15 01:16:30Z

4

The other answers you got work fine for a small number strings (your example code only has 4). But, if you're comparing a large number this will be quite slow since you're doing n^2 comparisons. I'd suggest first splitting the string into an array of strings, then sorting the array using qsort(). In a sorted array all duplicates are guaranteed to be adjacent. This reduces the time from n^2 to n log n -- the time required to sort.

answered Oct 15, 2011 at 1:16

uscjeremy

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1 Comment

jfs Over a year ago

I've posted the answer that uses that approach.

Huygens · Accepted Answer · 2012-04-19 12:27:19Z

2

I would split the string array using strtok (see the man page).

So I would have something like this

char x[100]="Amit Hanish Mahesh Amit";

/* Preparing the result string */
size_t sz_result = sizeof(char) * (strlen(x) + 1);
char* result = (char*) malloc( sz_result );
result[0] = '\0';

/* Parsing the string from one element to the other */
char* elm = strtok(x, " ");
while( (elm = strtok(NULL, " ")) != NULL )
{
  ...

You will have each element of the string to verify if they are unique.

Then I would use something like a hashmap (you can use the one from the glib) or I would put the read string element in a new string only if it is not already in.

Here is an example for the second solution:

  ...
  /* Is the element already in the result string? */
  if ( strstr(result, elm) == NULL )
  {
    /* Then we should add it */
    snprintf( result, sz_result - 1, "%s %s", result, elm );
  }
}

In the end if you want x to be modified, you simply copy result in x:

strncpy( x, result, 99 );

Here is a sample code (not optimised, not using the strn* primitives, etc.)

#include <unistd.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>


int main()
{
  char x[100]="Amit Hanish Mahesh Amit";

  /* Preparing the result string */
  size_t sz_result = sizeof(char) * (strlen(x) + 1);
  char* result = (char*) malloc( sz_result );
  result[0] = '\0';

  /* Parsing the string from one element to the other */
  char* elm = strtok(x, " ");
  if (elm != NULL) strcpy(result, elm);
  while( (elm = strtok(NULL, " ")) != NULL )
  {
    /* Is the element already in the result string? */
    if ( strstr(result, elm) == NULL )
    {
      /* Then we should add it */
      strcat( result, " " );
      strcat( result, elm );
    }
  }

  strcpy( x, result );

  fprintf( stdout, "Result: %s\n", x );
}

edited Apr 19, 2012 at 12:27

answered Oct 14, 2011 at 7:42

Huygens

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4 Comments

Huygens Over a year ago

It's way too complicated for doing what you want :-) and I don't have enough time, my apologies.

Amit Singh Tomar Over a year ago

With your solution output comes out to be Hanish Mahesh Amit but it should be Amit Hanish Mahesh .Any way @Huygens thanks for your time!!

Huygens Over a year ago

This should not be the case. I've made a small test program (quick n'dirty), replaced the snprintf by strcat. And it outputs the result in the order you would like. I'm going to edit my answer to add this sample complete code.

0xff0000 Over a year ago

This answer is not exactly correct. strstr is not sensitive to tokens(spaces), and for an input of Amit Manish Mahesh Amit Mah , it strips out the last Mah , even though Mah is not the same as Mahesh.

jfs · Accepted Answer · 2011-10-15 02:33:31Z

To remove duplicates from an array without preserving the order of elements:

sort the array
copy unique elements to the beginning of the array
remove the tail with duplicate elements

int remove_duplicates(StringArray array) {
  if (! (array and array->items)) return 0; // empty array or NULL

  StringArray_sort(array); // sort

  // unique_copy()
  String result = array->items, last = array->items + array->size;
  for (String first = array->items; first != last; ++result) {
    String_copy(result, first); // copy first to result
    for (String prev = first; ++first != last and String_cmp(prev, first) == 0;)
      { /* skip adjacent equal items */ }
  }
  // shrink
  return StringArray_remove(array, result, last);
}

Example

int main() {
  char text[] = "Mahesh Amit  Hanish Amit";
  StringArray array = split(text, sizeof(text));
  StringArray_dump(array, "<"); // print array before removing duplicates
  if (remove_duplicates(array) < 0)
    perror("error remove_duplicates(), OS error if any");
  StringArray_dump(array, ">"); // print it after
  StringArray_destroy(array);
  return 0;
}

Where split() is:

StringArray split(const char* text, size_t size) {
  if (! (text and text[size-1] == '\0')) return NULL;

  StringArray array = StringArray_create();
  if (! array) return NULL;

  size_t n = -1;
  for (const char* p = text; p != text+size; p += n+1) {
    n = strcspn(p, " \t\n"); // find index of the next whitespace
    if (n == 0) continue; // skip consecutive whitespace

    // append characters in range [p, p+n)
    // as a string to the array
    const String string = String_create(p, n);
    if (StringArray_append(array, string) < 0) {
      String_destroy(string);
      StringArray_destroy(array);
      return NULL;
    }
    String_destroy(string);
   }  
  return array;
}

Output

Mahesh<Amit<Hanish<Amit<
Amit>Hanish>Mahesh>

Full source code

ktf · Accepted Answer · 2011-10-14 07:09:50Z

0

I'm pretty sure, that the following line is not intended (assignment, not comparison)

 if (end = t) {

See what happens, if you code a == and come back, if you still have problems.

Hint: Always code blanks around operators, so expressions are easier to read.

answered Oct 14, 2011 at 7:09

ktf

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6 Comments

Amit Singh Tomar Over a year ago

oh its my mistake but still i am not achieve what am looking for

ktf Over a year ago

OK - looked at your code at more detail - 2nd hint: You are using an array of strings y[3][100] to store the found words. This array has room for 3 words - but the input line contains 4 words...

ktf Over a year ago

3rd hint: you're assuming that each word is delimited by one blank - but between the 1st and the 2nd word are 2 blanks - this eats up one entry in array y.

Amit Singh Tomar Over a year ago

Yes its delimited by one space only thats not my concern

ktf Over a year ago

I'd remove, rethink and then rewrite the whole 2nd part where you're searching the array y for duplicates. The loops make no sense this way - and the strcpy(y[current],y[i]); goes way behind your array of strings y (remember, that you used i as index for the string x. y[current] = 0; makes also no sense (y[current] is a string) and the last printf lacks a subscript on array y.

|

Engin · Accepted Answer · 2011-10-14 09:28:45Z

It's always fun to try to solve this kind of simple problems in C as exercise. Here's my take.

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

char* strstrn(const char *haystack, const char *needle, size_t needle_len)
{
    while((haystack = strchr(haystack, *needle)))
    {
        if (strncmp(haystack, needle, needle_len) == 0)
            return (char *) haystack;
        haystack++;
    }
    return NULL;
 }

char* find_duplicate(const char* str, size_t len, size_t dup_len)
{
    for(size_t i = 0; i < (len - dup_len); i++)
    {
        char* r = strstrn(str + i + 1, str + i, dup_len);
        if(r) return r;
    }
    return NULL;
}

int main(int argc, char** argv)
{
    if(argc < 3)
    {
        fprintf(stderr, "Usage: %s haystack dup_size\n", argv[0]);
        return 1;
    }
    char* haystack = argv[1];
    size_t len = atoi(argv[2]);
    char* r;
    while((r = find_duplicate(haystack, strlen(haystack), len)))
    {
        strcpy(r, r + len);
    }
    puts(haystack);
    return 0;
}

Rahul Sahu · Accepted Answer · 2013-12-27 13:50:43Z

/* 
 * C Program to Find the Frequency of  Every Word in a 
 * given String
 */
#include <stdio.h>
#include <string.h>

void main()
{
    int count = 0, c = 0, i, j = 0, k, space = 0;
    char str[100], p[50][100], str1[20], ptr1[50][100];
    printf("Enter the string\n");
    scanf(" %[^\n]s", str);
    printf("string length is %d\n", strlen(str));
    for (i = 0;i<strlen(str);i++)
    {
        if ((str[i] == ' ')||(str[i] == ', ')||(str[i] == '.'))
        {
            space++;
        }
    }
    for (i = 0, j = 0, k = 0;j < strlen(str);j++)
    {
        if ((str[j] == ' ')||(str[j] == 44)||(str[j] == 46))  
        {    
            p[i][k] = '\0';
            i++;
            k = 0;
        }        
        else
             p[i][k++] = str[j];
    }
    k = 0;
    for (i = 0;i <= space;i++)
    {
        for (j = 0;j <= space;j++)
        {
            if (i == j)
            {
                strcpy(ptr1[k], p[i]);
                k++;
                count++;
                break;
            }
            else
            {
                if (strcmp(ptr1[j], p[i]) != 0)
                    continue;
                else
                    break;
            }
        }
    }
    for (i = 0;i < count;i++) 
    {
        for (j = 0;j <= space;j++)
        {
            if (strcmp(ptr1[i], p[j]) == 0)
                c++;
        }
        printf("%s -> %d times\n", ptr1[i], c);
        c = 0;
    }
}

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