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    <p>You can do this in one pass with something like:</p> <pre><code>#include &lt;stdio.h&gt; #include &lt;string.h&gt; int main (void) { char str[] = "p123h12p97h62p32h"; // Input string. int found[256]; // Assumes 8-bit char. char *src, *dst; // Source and destination pointers. // Output initial value and set found flags to false. printf ("Before: [%s]\n", str); memset (found, 0, sizeof(found)); // One loop, processing each source character in string. for (src = dst = str; *src != '\0'; src++) { // If not yet found, flag it found and transfer it, else do nothing. if (!found[(unsigned int)(*src)]) { found[(unsigned int)(*src)] = 1; *dst++ = *src; } } // Insert new end of string, then output it. *dst = '\0'; printf (" After: [%s]\n", str); return 0; } </code></pre> <p>This removes duplicates in one single pass, using two pointers that advance independently through the string:</p> <pre><code>src | v p123h12p97h62p32h ^ | dst </code></pre> <p>The <code>src</code> pointer advances by one each iteration of the loop. A character is copied from <code>src</code> to <code>dst</code> and the <code>dst</code> pointer is advanced only if the character has not been seen before (using the <code>found</code> array). The output is:</p> <pre><code>Before: [p123h12p97h62p32h] After: [p123h976] </code></pre> <p>Note the <code>"assumes 8-bit char"</code> comment - this is fine where you know the character set is ASCII (or any other 8-bit character set) but it may not be portable to more exotic implementations. You just have to ensure that there are enough elements in the <code>found</code> array for all of your characters. For example, a 10-bit <code>char</code> type would need 1024 elements (<code>2<sup>10</sup> = 1024</code>).</p>
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    1. POHow to optimize this piece of code In c
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    1. COStill two for loops are there!!
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      1. USAmit Singh Tomar
    2. COThe first is just zeroing his flag array. Replace with memset if you prefer.
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      1. USGazTheDestroyer
    3. COor just use initialize way of C.... Like I just edited. Beside, this is two loops one after each other, still O(n) while the first way was loop in loop which is O(n^2)
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      1. USRoee Gavirel
 

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