Write an efficient algorithm to find the first non-repeated character in a string defined

Replies

• 

  Ankita Katdare

  A simple algorithm should be:
  
  Step 1. Scanning the string from left to right and then storing the values in a count array.
  Step 2. Scanning the string from left to right again and checking for count of each character, if an element who's count is 1 is found, return it.
• 

  PraveenKumar Purushothaman

  AbraKaDabra

  A simple algorithm should be:
  
  Step 1. Scanning the string from left to right and then storing the values in a count array.
  Step 2. Scanning the string from left to right again and checking for count of each character, if an element who's count is 1 is found, return it.

  AKD, the algorithm of yours seems to be quadratic time complexity!
• 

  Ankita Katdare

  Praveen-Kumar

  AKD, the algorithm of yours seems to be quadratic time complexity!

  Hmm, yes. O(n2) #-Link-Snipped-# write one algo here for us all.
• 

  PraveenKumar Purushothaman

  AbraKaDabra

  Hmm, yes. O(n2) #-Link-Snipped-# write one algo here for us all.

  Er... Will try... If PHP I guess I can easily say... There is a concept of associative arrays, where you can do this way: count["char"] = n and so on.
  
  For the algorithm,
  1. Define an empty associative array and initialize to 0. def count[] = 0;
  2. For each character, check the occurrence and increment the particular array node. count["s"]++;
  3. Loop through the array and get the first value, which is 1!
  

  foreach (val as count)
    if val is 1
      print the index

  The complexity is O(n).
• 

  silverscorpion

  A simple algorithm to do it in only one pass of the string is to maintain an array of count of the letters, and then scanning through the count array instead of the string itself again.
  
  The algorithm is like:
  
  

  initialize count[256][2] to all zeros
  read S
  for i in range 0 to length(S):
      if count[S[i]][0] == 0:
          count[S[i]][1] = i
      count[S[i]][0]++
  Index = length(S)
  for i in range 1 to 255:
      if count[i][0] == 1:
          if count[i][1] < Index:
              Index = count[i][1]
  return S(Index)

  I hope the pseudo code is clear..
  
  The running time is O(n+c) where c is 256 here. So, essentially it is O(n)
• 

  silverscorpion

  Praveen-Kumar

  Er... Will try... If PHP I guess I can easily say... There is a concept of associative arrays, where you can do this way: count["char"] = n and so on.
  
  For the algorithm,
  1. Define an empty associative array and initialize to 0. def count[] = 0;
  2. For each character, check the occurrence and increment the particular array node. count["s"]++;
  3. Loop through the array and get the first value, which is 1!
  

  foreach (val as count)
    if val is 1
      print the index

  The complexity is O(n).

  I think that won't work.. If you are using associative arrays, you need to specify the count of the character as well as the index when it was first encountered.
  
  For example, take the string "bbzaccddbcd"
  If you fill an associative array with this string, with the key as the characters, you would get
  
  array['b']=3, array['c']=3, array['a']=1, etc.. and array['z']=1
  
  Now you will scan through the array and the first element that has the value 1 is 'a'. So, you will return a, while the answer is z.
  
  That's why, you also need to store the index where each character is first encountered. Then compare that value also.. Just checking for the count being 1 is not enough.
• 

  PraveenKumar Purushothaman

  silverscorpion

  I think that won't work.. If you are using associative arrays, you need to specify the count of the character as well as the index when it was first encountered.
  
  For example, take the string "bbzaccddbcd"
  If you fill an associative array with this string, with the key as the characters, you would get
  
  array['b']=3, array['c']=3, array['a']=1, etc.. and array['z']=1
  
  Now you will scan through the array and the first element that has the value 1 is 'a'. So, you will return a, while the answer is z.
  
  That's why, you also need to store the index where each character is first encountered. Then compare that value also.. Just checking for the count being 1 is not enough.

  Hey wait... Coming to what you said, consider the string "bbzaccddbcd":
  

  count["b"]++;
  count = {
      "b" = 1
  }
  count["b"]++;
  count = {
      "b" = 2
  count["z"]++;
  }
  count = {
      "b" = 2,
      "z" = 1
  }
  ...
  ...
  ...
  ...
  count["d"]++;
  count = {
      "b" = 2,
      "z" = 1,
      "a" = 1,
      "c" = 3,
      "d" = 3
  }

  When the loop encounters the first value of 1, it returns "z" and not "a". Hope you are clear.