Given a string, find its first non-repeating character | GeeksforGeeks

when it comes to finding the first non-repeater, we just have to scan the count array, instead of the string

struct countIndex {

   int count;

   int index;

};

/* Returns an array of above structure type. The size of

   array is NO_OF_CHARS */

struct countIndex *getCharCountArray(char *str)

{

   struct countIndex *count =

        (struct countIndex *)calloc(sizeof(countIndex), NO_OF_CHARS);

   int i;

   for (i = 0; *(str+i);  i++)

   {

      (count[*(str+i)].count)++;

      // If it's first occurrence, then store the index

      if (count[*(str+i)].count == 1)

         count[*(str+i)].index = i;

   }

   return count;

}

/* The function returns index of the first non-repeating

    character in a string. If all characters are repeating

    then reurns INT_MAX */

int firstNonRepeating(char *str)

{

  struct countIndex *count = getCharCountArray(str);

  int result = INT_MAX, i;

  for (i = 0; i < NO_OF_CHARS;  i++)

  {

    // If this character occurs only once and appears

    // before the current result, then update the result

    if (count[i].count == 1 && result > count[i].index)

       result = count[i].index;

  }

  free(count); // To avoid memory leak

  return result;

}

http://hehejun.blogspot.com/2015/01/algorithmfirst-non-repeating-character.html

private class Counter {

private int count;

private int index;

public Counter(int index) {

count = 1;

this.index = index;

}

}

public char find(String s) {

if (s == null)

return ' ';

int len = s.length();

HashMap<Character, Counter> map = new HashMap<Character, Counter>();

for (int i = 0; i < len; i++) {

if (!map.containsKey(s.charAt(i)))

map.put(s.charAt(i), new Counter(i));

else 

map.get(s.charAt(i)).count++;

}

Iterator<Entry<Character, Counter>> iter = map.entrySet().iterator();

int index = Integer.MAX_VALUE;

char res = ' ';

while (iter.hasNext()) {

Entry<Character, Counter> entry = iter.next();

if (entry.getValue().count == 1 && entry.getValue().index < index) {

index = entry.getValue().index;

res = entry.getKey();

}

}

return res;

}

We can use string characters as index and build a count array. Following is the algorithm.

1) Scan the string from left to right and construct the count array.
2) Again, scan the string from left to right and check for count of each
 character, if you find an element who's count is 1, return it.

http://yuanhsh.iteye.com/blog/2185878

1. int *getCharCountArray(char *str) {  
2.    int *count = (int *)calloc(sizeof(int), NO_OF_CHARS);  
3.    int i;  
4.    for (i = 0; *(str+i);  i++)  
5.       count[*(str+i)]++;  
6.    return count;  
7. }  
8.    
9. /* The function returns index of first non-repeating 
10.    character in a string. If all characters are repeating  
11.    then returns -1 */  
12. int firstNonRepeating(char *str) {  
13.   int *count = getCharCountArray(str);  
14.   int index = -1, i;  
15.    
16.   for (i = 0; *(str+i);  i++) {  
17.     if (count[*(str+i)] == 1) {  
18.       index = i;  
19.       break;  
20.     }     
21.   }    
22.      
23.   free(count); // To avoid memory leak  
24.   return index;  
25. }

Related: http://massivealgorithms.blogspot.com/2015/07/finding-first-non-repeating-character.html
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