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Home/ Questions/Q 612133
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T17:51:36+00:00 2026-05-13T17:51:36+00:00

I came across a interview question that asked to remove the repeated char from

  • 0

I came across a interview question that asked to remove the repeated char from a given string, in-place.
So if the input was “hi there” the output expected was “hi ter”. It was also told to consider only alphabetic repititions and all the
alphabets were lower case. I came up with the following program. I have comments to make my logic clear. But the program does not work as expectd for some inputs. If the input is “hii” it works, but if its “hi there” it fails. Please help.

#include <stdio.h>
int main() 
{
    char str[] = "programming is really cool"; // original string.
    char hash[26] = {0}; // hash table.
    int i,j; // loop counter.
// iterate through the input string char by char.
for(i=0,j=0;str[i];)
{
    // if the char is not hashed.
    if(!hash[str[i] - 'a'])
    {
        // hash it.
        hash[str[i] - 'a'] = 1;
        // copy the char at index i to index j.
        str[j++] = str[i++];
    }
    else
    {
        // move to next char of the original string.
        // do not increment j, so that later we can over-write the repeated char.
        i++;
    }
}

// add a null char.
str[j] = 0;

// print it.
printf("%s\n",str); // "progamin s ely c" expected.

return 0;

}

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  1. Editorial Team
    Editorial Team
    2026-05-13T17:51:37+00:00Added an answer on May 13, 2026 at 5:51 pm

    when str[i] is a non-alphabet, say a space and when you do:

    hash[str[i] - 'a']
    

    your program can blow.

    ASCII value of space is 32 and that of a is 97 so you are effectively accessing array hash with a negative index.

    To solve this you can ignore non-alphabets by doing :

    if(! isalpha(str[i]) {
        str[j++] = str[i++]; // copy the char.
        continue;  // ignore rest of the loop.
    }
    
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