How to speed up first unique character lookup












0















I'm solving 387. First Unique Character in a String LeetCode problem defined as:




Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1.



Examples:



s = "leetcode"
return 0.

s = "loveleetcode",
return 2.


Note: You may assume the string contain only lowercase letters.




Taking advantage of the input being fully lowercase ASCII I created two bit vectors to track when we encounter a character for the first and second time.



Can below code be improved further? LeetCode says that below code is better than 94.33% solutions. What else could have been done by the last 5.67% solutions that they were better?



class Solution {

public int firstUniqChar(String s) {
int firstSpot = 0;
int secondSpot = 0;
char chars = s.toCharArray();

for (char c : chars) {
int mask = 1 << c - 'a';
if ((firstSpot & mask) == 0) {
firstSpot |= mask;
} else if ((secondSpot & mask) == 0) {
secondSpot |= mask;
}
}

int i = 0;
for (char c : chars) {
int mask = 1 << c - 'a';
if ((secondSpot & mask) == 0) {
return i;
}
i++;
}
return -1;
}

}


LeetCode score



Are there tricks that can be done to improve the LeetCode score? It seems that enhanced for-each loop performs better than standard for loop but I can't prove it, It's an observation based on few of my previous submissions.










share|improve this question


















  • 3





    This question is better asked on codereview.stackexchange.com

    – Andreas
    Nov 19 '18 at 23:03








  • 1





    Your post could get a better reception at Code Review site.

    – PM 77-1
    Nov 19 '18 at 23:04






  • 1





    Re: "LeetCode says that below code is better than 94.33% solutions": My impression is that LeetCode is doing an unscientific microbenchmark that might depend on random factors (such as the other load on their servers at submission-time). So I'm not sure this figure is meant to be taken very seriously.

    – ruakh
    Nov 20 '18 at 0:25
















0















I'm solving 387. First Unique Character in a String LeetCode problem defined as:




Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1.



Examples:



s = "leetcode"
return 0.

s = "loveleetcode",
return 2.


Note: You may assume the string contain only lowercase letters.




Taking advantage of the input being fully lowercase ASCII I created two bit vectors to track when we encounter a character for the first and second time.



Can below code be improved further? LeetCode says that below code is better than 94.33% solutions. What else could have been done by the last 5.67% solutions that they were better?



class Solution {

public int firstUniqChar(String s) {
int firstSpot = 0;
int secondSpot = 0;
char chars = s.toCharArray();

for (char c : chars) {
int mask = 1 << c - 'a';
if ((firstSpot & mask) == 0) {
firstSpot |= mask;
} else if ((secondSpot & mask) == 0) {
secondSpot |= mask;
}
}

int i = 0;
for (char c : chars) {
int mask = 1 << c - 'a';
if ((secondSpot & mask) == 0) {
return i;
}
i++;
}
return -1;
}

}


LeetCode score



Are there tricks that can be done to improve the LeetCode score? It seems that enhanced for-each loop performs better than standard for loop but I can't prove it, It's an observation based on few of my previous submissions.










share|improve this question


















  • 3





    This question is better asked on codereview.stackexchange.com

    – Andreas
    Nov 19 '18 at 23:03








  • 1





    Your post could get a better reception at Code Review site.

    – PM 77-1
    Nov 19 '18 at 23:04






  • 1





    Re: "LeetCode says that below code is better than 94.33% solutions": My impression is that LeetCode is doing an unscientific microbenchmark that might depend on random factors (such as the other load on their servers at submission-time). So I'm not sure this figure is meant to be taken very seriously.

    – ruakh
    Nov 20 '18 at 0:25














0












0








0


0






I'm solving 387. First Unique Character in a String LeetCode problem defined as:




Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1.



Examples:



s = "leetcode"
return 0.

s = "loveleetcode",
return 2.


Note: You may assume the string contain only lowercase letters.




Taking advantage of the input being fully lowercase ASCII I created two bit vectors to track when we encounter a character for the first and second time.



Can below code be improved further? LeetCode says that below code is better than 94.33% solutions. What else could have been done by the last 5.67% solutions that they were better?



class Solution {

public int firstUniqChar(String s) {
int firstSpot = 0;
int secondSpot = 0;
char chars = s.toCharArray();

for (char c : chars) {
int mask = 1 << c - 'a';
if ((firstSpot & mask) == 0) {
firstSpot |= mask;
} else if ((secondSpot & mask) == 0) {
secondSpot |= mask;
}
}

int i = 0;
for (char c : chars) {
int mask = 1 << c - 'a';
if ((secondSpot & mask) == 0) {
return i;
}
i++;
}
return -1;
}

}


LeetCode score



Are there tricks that can be done to improve the LeetCode score? It seems that enhanced for-each loop performs better than standard for loop but I can't prove it, It's an observation based on few of my previous submissions.










share|improve this question














I'm solving 387. First Unique Character in a String LeetCode problem defined as:




Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1.



Examples:



s = "leetcode"
return 0.

s = "loveleetcode",
return 2.


Note: You may assume the string contain only lowercase letters.




Taking advantage of the input being fully lowercase ASCII I created two bit vectors to track when we encounter a character for the first and second time.



Can below code be improved further? LeetCode says that below code is better than 94.33% solutions. What else could have been done by the last 5.67% solutions that they were better?



class Solution {

public int firstUniqChar(String s) {
int firstSpot = 0;
int secondSpot = 0;
char chars = s.toCharArray();

for (char c : chars) {
int mask = 1 << c - 'a';
if ((firstSpot & mask) == 0) {
firstSpot |= mask;
} else if ((secondSpot & mask) == 0) {
secondSpot |= mask;
}
}

int i = 0;
for (char c : chars) {
int mask = 1 << c - 'a';
if ((secondSpot & mask) == 0) {
return i;
}
i++;
}
return -1;
}

}


LeetCode score



Are there tricks that can be done to improve the LeetCode score? It seems that enhanced for-each loop performs better than standard for loop but I can't prove it, It's an observation based on few of my previous submissions.







java algorithm






share|improve this question













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asked Nov 19 '18 at 23:00









Karol DowbeckiKarol Dowbecki

21.5k93154




21.5k93154








  • 3





    This question is better asked on codereview.stackexchange.com

    – Andreas
    Nov 19 '18 at 23:03








  • 1





    Your post could get a better reception at Code Review site.

    – PM 77-1
    Nov 19 '18 at 23:04






  • 1





    Re: "LeetCode says that below code is better than 94.33% solutions": My impression is that LeetCode is doing an unscientific microbenchmark that might depend on random factors (such as the other load on their servers at submission-time). So I'm not sure this figure is meant to be taken very seriously.

    – ruakh
    Nov 20 '18 at 0:25














  • 3





    This question is better asked on codereview.stackexchange.com

    – Andreas
    Nov 19 '18 at 23:03








  • 1





    Your post could get a better reception at Code Review site.

    – PM 77-1
    Nov 19 '18 at 23:04






  • 1





    Re: "LeetCode says that below code is better than 94.33% solutions": My impression is that LeetCode is doing an unscientific microbenchmark that might depend on random factors (such as the other load on their servers at submission-time). So I'm not sure this figure is meant to be taken very seriously.

    – ruakh
    Nov 20 '18 at 0:25








3




3





This question is better asked on codereview.stackexchange.com

– Andreas
Nov 19 '18 at 23:03







This question is better asked on codereview.stackexchange.com

– Andreas
Nov 19 '18 at 23:03






1




1





Your post could get a better reception at Code Review site.

– PM 77-1
Nov 19 '18 at 23:04





Your post could get a better reception at Code Review site.

– PM 77-1
Nov 19 '18 at 23:04




1




1





Re: "LeetCode says that below code is better than 94.33% solutions": My impression is that LeetCode is doing an unscientific microbenchmark that might depend on random factors (such as the other load on their servers at submission-time). So I'm not sure this figure is meant to be taken very seriously.

– ruakh
Nov 20 '18 at 0:25





Re: "LeetCode says that below code is better than 94.33% solutions": My impression is that LeetCode is doing an unscientific microbenchmark that might depend on random factors (such as the other load on their servers at submission-time). So I'm not sure this figure is meant to be taken very seriously.

– ruakh
Nov 20 '18 at 0:25












3 Answers
3






active

oldest

votes


















2














I got 98.58% with this:-



public int firstUniqChar(String s) {
int count = new int[122 - 96];
final char chars = s.toCharArray();
for (int i = 0; i < chars.length; i++) {
count[chars[i] - 97]++;
}
for (int i = 0; i < chars.length; i++) {
if (count[chars[i] - 97] == 1)
return i;
}
return -1;
}


enter image description here






share|improve this answer
























  • You can get it to 100% by adding int setup=( {cin.tie(0);ios_base::sync_with_stdio(false);return 0;})(); to the beginning of the code.

    – 0x499602D2
    Nov 20 '18 at 2:44











  • @0x499602D2 java equivalent please

    – Kartik
    Nov 20 '18 at 2:45











  • Right this is Java not C++. Sorry.

    – 0x499602D2
    Nov 20 '18 at 2:45



















1














As @Ruakh says in a comment, the precise timings produced by leetcode are subject to a certain amount of randomness, so they should be taken with a grain of salt:




My impression is that LeetCode is doing an unscientific microbenchmark that might depend on random factors.




Still, it is possible to speed your first loop up quite a bit by getting rid of the tests. The following loop is functionally equivalent; although it sets the variables more often, changing the value of a local integer variable costs less than testing whether it is necessary to change:



for (char c : chars) {
int mask = 1 << c - 'a';
secondSpot |= mask & firstSpot;
firstSpot |= mask;
}


(It's important that the assignment be in that order, so that the first one does nothing if the character hasn't yet been seen.)






share|improve this answer































    1














    Use this:



    import java.util.Arrays;

    class Solution {
    // 0x60 < 'a' < 'z' < 0x80
    private final byte bCounter = new byte[0x80];
    private final int iCounter = new int[0x80];

    public int firstUniqChar(String s) {
    int len = s.length();
    if ((len & 0xff) == len) {
    Arrays.fill(bCounter, 0x60, 0x80, (byte)-1);
    for (int i = 0; i < len; i++)
    bCounter[s.charAt(i)]++;
    for (int i = 0; i < len; i++)
    if (bCounter[s.charAt(i)] == 0)
    return i;
    } else {
    Arrays.fill(iCounter, 0x60, 0x80, -1);
    for (int i = 0; i < len; i++)
    iCounter[s.charAt(i)]++;
    for (int i = 0; i < len; i++)
    if (iCounter[s.charAt(i)] == 0)
    return i;
    }
    return -1;
    }
    }





    share|improve this answer

























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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2














      I got 98.58% with this:-



      public int firstUniqChar(String s) {
      int count = new int[122 - 96];
      final char chars = s.toCharArray();
      for (int i = 0; i < chars.length; i++) {
      count[chars[i] - 97]++;
      }
      for (int i = 0; i < chars.length; i++) {
      if (count[chars[i] - 97] == 1)
      return i;
      }
      return -1;
      }


      enter image description here






      share|improve this answer
























      • You can get it to 100% by adding int setup=( {cin.tie(0);ios_base::sync_with_stdio(false);return 0;})(); to the beginning of the code.

        – 0x499602D2
        Nov 20 '18 at 2:44











      • @0x499602D2 java equivalent please

        – Kartik
        Nov 20 '18 at 2:45











      • Right this is Java not C++. Sorry.

        – 0x499602D2
        Nov 20 '18 at 2:45
















      2














      I got 98.58% with this:-



      public int firstUniqChar(String s) {
      int count = new int[122 - 96];
      final char chars = s.toCharArray();
      for (int i = 0; i < chars.length; i++) {
      count[chars[i] - 97]++;
      }
      for (int i = 0; i < chars.length; i++) {
      if (count[chars[i] - 97] == 1)
      return i;
      }
      return -1;
      }


      enter image description here






      share|improve this answer
























      • You can get it to 100% by adding int setup=( {cin.tie(0);ios_base::sync_with_stdio(false);return 0;})(); to the beginning of the code.

        – 0x499602D2
        Nov 20 '18 at 2:44











      • @0x499602D2 java equivalent please

        – Kartik
        Nov 20 '18 at 2:45











      • Right this is Java not C++. Sorry.

        – 0x499602D2
        Nov 20 '18 at 2:45














      2












      2








      2







      I got 98.58% with this:-



      public int firstUniqChar(String s) {
      int count = new int[122 - 96];
      final char chars = s.toCharArray();
      for (int i = 0; i < chars.length; i++) {
      count[chars[i] - 97]++;
      }
      for (int i = 0; i < chars.length; i++) {
      if (count[chars[i] - 97] == 1)
      return i;
      }
      return -1;
      }


      enter image description here






      share|improve this answer













      I got 98.58% with this:-



      public int firstUniqChar(String s) {
      int count = new int[122 - 96];
      final char chars = s.toCharArray();
      for (int i = 0; i < chars.length; i++) {
      count[chars[i] - 97]++;
      }
      for (int i = 0; i < chars.length; i++) {
      if (count[chars[i] - 97] == 1)
      return i;
      }
      return -1;
      }


      enter image description here







      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered Nov 20 '18 at 0:19









      KartikKartik

      3,74231435




      3,74231435













      • You can get it to 100% by adding int setup=( {cin.tie(0);ios_base::sync_with_stdio(false);return 0;})(); to the beginning of the code.

        – 0x499602D2
        Nov 20 '18 at 2:44











      • @0x499602D2 java equivalent please

        – Kartik
        Nov 20 '18 at 2:45











      • Right this is Java not C++. Sorry.

        – 0x499602D2
        Nov 20 '18 at 2:45



















      • You can get it to 100% by adding int setup=( {cin.tie(0);ios_base::sync_with_stdio(false);return 0;})(); to the beginning of the code.

        – 0x499602D2
        Nov 20 '18 at 2:44











      • @0x499602D2 java equivalent please

        – Kartik
        Nov 20 '18 at 2:45











      • Right this is Java not C++. Sorry.

        – 0x499602D2
        Nov 20 '18 at 2:45

















      You can get it to 100% by adding int setup=( {cin.tie(0);ios_base::sync_with_stdio(false);return 0;})(); to the beginning of the code.

      – 0x499602D2
      Nov 20 '18 at 2:44





      You can get it to 100% by adding int setup=( {cin.tie(0);ios_base::sync_with_stdio(false);return 0;})(); to the beginning of the code.

      – 0x499602D2
      Nov 20 '18 at 2:44













      @0x499602D2 java equivalent please

      – Kartik
      Nov 20 '18 at 2:45





      @0x499602D2 java equivalent please

      – Kartik
      Nov 20 '18 at 2:45













      Right this is Java not C++. Sorry.

      – 0x499602D2
      Nov 20 '18 at 2:45





      Right this is Java not C++. Sorry.

      – 0x499602D2
      Nov 20 '18 at 2:45













      1














      As @Ruakh says in a comment, the precise timings produced by leetcode are subject to a certain amount of randomness, so they should be taken with a grain of salt:




      My impression is that LeetCode is doing an unscientific microbenchmark that might depend on random factors.




      Still, it is possible to speed your first loop up quite a bit by getting rid of the tests. The following loop is functionally equivalent; although it sets the variables more often, changing the value of a local integer variable costs less than testing whether it is necessary to change:



      for (char c : chars) {
      int mask = 1 << c - 'a';
      secondSpot |= mask & firstSpot;
      firstSpot |= mask;
      }


      (It's important that the assignment be in that order, so that the first one does nothing if the character hasn't yet been seen.)






      share|improve this answer




























        1














        As @Ruakh says in a comment, the precise timings produced by leetcode are subject to a certain amount of randomness, so they should be taken with a grain of salt:




        My impression is that LeetCode is doing an unscientific microbenchmark that might depend on random factors.




        Still, it is possible to speed your first loop up quite a bit by getting rid of the tests. The following loop is functionally equivalent; although it sets the variables more often, changing the value of a local integer variable costs less than testing whether it is necessary to change:



        for (char c : chars) {
        int mask = 1 << c - 'a';
        secondSpot |= mask & firstSpot;
        firstSpot |= mask;
        }


        (It's important that the assignment be in that order, so that the first one does nothing if the character hasn't yet been seen.)






        share|improve this answer


























          1












          1








          1







          As @Ruakh says in a comment, the precise timings produced by leetcode are subject to a certain amount of randomness, so they should be taken with a grain of salt:




          My impression is that LeetCode is doing an unscientific microbenchmark that might depend on random factors.




          Still, it is possible to speed your first loop up quite a bit by getting rid of the tests. The following loop is functionally equivalent; although it sets the variables more often, changing the value of a local integer variable costs less than testing whether it is necessary to change:



          for (char c : chars) {
          int mask = 1 << c - 'a';
          secondSpot |= mask & firstSpot;
          firstSpot |= mask;
          }


          (It's important that the assignment be in that order, so that the first one does nothing if the character hasn't yet been seen.)






          share|improve this answer













          As @Ruakh says in a comment, the precise timings produced by leetcode are subject to a certain amount of randomness, so they should be taken with a grain of salt:




          My impression is that LeetCode is doing an unscientific microbenchmark that might depend on random factors.




          Still, it is possible to speed your first loop up quite a bit by getting rid of the tests. The following loop is functionally equivalent; although it sets the variables more often, changing the value of a local integer variable costs less than testing whether it is necessary to change:



          for (char c : chars) {
          int mask = 1 << c - 'a';
          secondSpot |= mask & firstSpot;
          firstSpot |= mask;
          }


          (It's important that the assignment be in that order, so that the first one does nothing if the character hasn't yet been seen.)







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 20 '18 at 14:34









          ricirici

          155k19135202




          155k19135202























              1














              Use this:



              import java.util.Arrays;

              class Solution {
              // 0x60 < 'a' < 'z' < 0x80
              private final byte bCounter = new byte[0x80];
              private final int iCounter = new int[0x80];

              public int firstUniqChar(String s) {
              int len = s.length();
              if ((len & 0xff) == len) {
              Arrays.fill(bCounter, 0x60, 0x80, (byte)-1);
              for (int i = 0; i < len; i++)
              bCounter[s.charAt(i)]++;
              for (int i = 0; i < len; i++)
              if (bCounter[s.charAt(i)] == 0)
              return i;
              } else {
              Arrays.fill(iCounter, 0x60, 0x80, -1);
              for (int i = 0; i < len; i++)
              iCounter[s.charAt(i)]++;
              for (int i = 0; i < len; i++)
              if (iCounter[s.charAt(i)] == 0)
              return i;
              }
              return -1;
              }
              }





              share|improve this answer






























                1














                Use this:



                import java.util.Arrays;

                class Solution {
                // 0x60 < 'a' < 'z' < 0x80
                private final byte bCounter = new byte[0x80];
                private final int iCounter = new int[0x80];

                public int firstUniqChar(String s) {
                int len = s.length();
                if ((len & 0xff) == len) {
                Arrays.fill(bCounter, 0x60, 0x80, (byte)-1);
                for (int i = 0; i < len; i++)
                bCounter[s.charAt(i)]++;
                for (int i = 0; i < len; i++)
                if (bCounter[s.charAt(i)] == 0)
                return i;
                } else {
                Arrays.fill(iCounter, 0x60, 0x80, -1);
                for (int i = 0; i < len; i++)
                iCounter[s.charAt(i)]++;
                for (int i = 0; i < len; i++)
                if (iCounter[s.charAt(i)] == 0)
                return i;
                }
                return -1;
                }
                }





                share|improve this answer




























                  1












                  1








                  1







                  Use this:



                  import java.util.Arrays;

                  class Solution {
                  // 0x60 < 'a' < 'z' < 0x80
                  private final byte bCounter = new byte[0x80];
                  private final int iCounter = new int[0x80];

                  public int firstUniqChar(String s) {
                  int len = s.length();
                  if ((len & 0xff) == len) {
                  Arrays.fill(bCounter, 0x60, 0x80, (byte)-1);
                  for (int i = 0; i < len; i++)
                  bCounter[s.charAt(i)]++;
                  for (int i = 0; i < len; i++)
                  if (bCounter[s.charAt(i)] == 0)
                  return i;
                  } else {
                  Arrays.fill(iCounter, 0x60, 0x80, -1);
                  for (int i = 0; i < len; i++)
                  iCounter[s.charAt(i)]++;
                  for (int i = 0; i < len; i++)
                  if (iCounter[s.charAt(i)] == 0)
                  return i;
                  }
                  return -1;
                  }
                  }





                  share|improve this answer















                  Use this:



                  import java.util.Arrays;

                  class Solution {
                  // 0x60 < 'a' < 'z' < 0x80
                  private final byte bCounter = new byte[0x80];
                  private final int iCounter = new int[0x80];

                  public int firstUniqChar(String s) {
                  int len = s.length();
                  if ((len & 0xff) == len) {
                  Arrays.fill(bCounter, 0x60, 0x80, (byte)-1);
                  for (int i = 0; i < len; i++)
                  bCounter[s.charAt(i)]++;
                  for (int i = 0; i < len; i++)
                  if (bCounter[s.charAt(i)] == 0)
                  return i;
                  } else {
                  Arrays.fill(iCounter, 0x60, 0x80, -1);
                  for (int i = 0; i < len; i++)
                  iCounter[s.charAt(i)]++;
                  for (int i = 0; i < len; i++)
                  if (iCounter[s.charAt(i)] == 0)
                  return i;
                  }
                  return -1;
                  }
                  }






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                  edited Nov 23 '18 at 7:25

























                  answered Nov 23 '18 at 6:23









                  John McClaneJohn McClane

                  1,5612419




                  1,5612419






























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