Counting number of pairs in an integer array












0












$begingroup$


So, I wanted to have the count for number of pairs of same integer there are in an array for which I used a HashMap



Input contains two lines:






  1. Integer n:



    Signifies the size of input



  2. Input numbers separated by white space character:

    E.g 4 6 6 7 7 8 7 6 4





For the same input,




Expected Output:



3




import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class SockMerchant {

public int returnsockArr() {
Scanner sc = new Scanner(System.in);
int arraySize = sc.nextInt();
int sockArr = new int[arraySize];
for(int i = 0;i <= arraySize-1; i++) {
sockArr[i] = sc.nextInt();
}
sc.close();
return sockArr;
}

public int returnCount(Map<String, Object> mapCount) {
int count = 0;
for(String keyName : mapCount.keySet()) {
int value = (int)mapCount.get(keyName);
count = count+(value/2);
}
return count;
}

public int setMapAndReturnCount(int sockArr) {
Map<String, Object> mapCount = new HashMap<String, Object>();
for(int j = 0;j<= sockArr.length-1;j++) {
if(mapCount.containsKey(Integer.toString(sockArr[j]))) {
mapCount.put(Integer.toString(sockArr[j]), (int)mapCount.get(Integer.toString(sockArr[j]))+1);
}
else {
mapCount.put(Integer.toString(sockArr[j]), 1);
}
}
return returnCount(mapCount);
}

public static void main(String args) {
SockMerchant sm = new SockMerchant();
int sockArr = sm.returnsockArr();
int finalCount = sm.setMapAndReturnCount(sockArr);
System.out.println(finalCount);
}
}


I think that I have complicated the solution a bit too much



Is there a better approach to do the same thing?










share|improve this question









$endgroup$

















    0












    $begingroup$


    So, I wanted to have the count for number of pairs of same integer there are in an array for which I used a HashMap



    Input contains two lines:






    1. Integer n:



      Signifies the size of input



    2. Input numbers separated by white space character:

      E.g 4 6 6 7 7 8 7 6 4





    For the same input,




    Expected Output:



    3




    import java.util.HashMap;
    import java.util.Map;
    import java.util.Scanner;

    public class SockMerchant {

    public int returnsockArr() {
    Scanner sc = new Scanner(System.in);
    int arraySize = sc.nextInt();
    int sockArr = new int[arraySize];
    for(int i = 0;i <= arraySize-1; i++) {
    sockArr[i] = sc.nextInt();
    }
    sc.close();
    return sockArr;
    }

    public int returnCount(Map<String, Object> mapCount) {
    int count = 0;
    for(String keyName : mapCount.keySet()) {
    int value = (int)mapCount.get(keyName);
    count = count+(value/2);
    }
    return count;
    }

    public int setMapAndReturnCount(int sockArr) {
    Map<String, Object> mapCount = new HashMap<String, Object>();
    for(int j = 0;j<= sockArr.length-1;j++) {
    if(mapCount.containsKey(Integer.toString(sockArr[j]))) {
    mapCount.put(Integer.toString(sockArr[j]), (int)mapCount.get(Integer.toString(sockArr[j]))+1);
    }
    else {
    mapCount.put(Integer.toString(sockArr[j]), 1);
    }
    }
    return returnCount(mapCount);
    }

    public static void main(String args) {
    SockMerchant sm = new SockMerchant();
    int sockArr = sm.returnsockArr();
    int finalCount = sm.setMapAndReturnCount(sockArr);
    System.out.println(finalCount);
    }
    }


    I think that I have complicated the solution a bit too much



    Is there a better approach to do the same thing?










    share|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      So, I wanted to have the count for number of pairs of same integer there are in an array for which I used a HashMap



      Input contains two lines:






      1. Integer n:



        Signifies the size of input



      2. Input numbers separated by white space character:

        E.g 4 6 6 7 7 8 7 6 4





      For the same input,




      Expected Output:



      3




      import java.util.HashMap;
      import java.util.Map;
      import java.util.Scanner;

      public class SockMerchant {

      public int returnsockArr() {
      Scanner sc = new Scanner(System.in);
      int arraySize = sc.nextInt();
      int sockArr = new int[arraySize];
      for(int i = 0;i <= arraySize-1; i++) {
      sockArr[i] = sc.nextInt();
      }
      sc.close();
      return sockArr;
      }

      public int returnCount(Map<String, Object> mapCount) {
      int count = 0;
      for(String keyName : mapCount.keySet()) {
      int value = (int)mapCount.get(keyName);
      count = count+(value/2);
      }
      return count;
      }

      public int setMapAndReturnCount(int sockArr) {
      Map<String, Object> mapCount = new HashMap<String, Object>();
      for(int j = 0;j<= sockArr.length-1;j++) {
      if(mapCount.containsKey(Integer.toString(sockArr[j]))) {
      mapCount.put(Integer.toString(sockArr[j]), (int)mapCount.get(Integer.toString(sockArr[j]))+1);
      }
      else {
      mapCount.put(Integer.toString(sockArr[j]), 1);
      }
      }
      return returnCount(mapCount);
      }

      public static void main(String args) {
      SockMerchant sm = new SockMerchant();
      int sockArr = sm.returnsockArr();
      int finalCount = sm.setMapAndReturnCount(sockArr);
      System.out.println(finalCount);
      }
      }


      I think that I have complicated the solution a bit too much



      Is there a better approach to do the same thing?










      share|improve this question









      $endgroup$




      So, I wanted to have the count for number of pairs of same integer there are in an array for which I used a HashMap



      Input contains two lines:






      1. Integer n:



        Signifies the size of input



      2. Input numbers separated by white space character:

        E.g 4 6 6 7 7 8 7 6 4





      For the same input,




      Expected Output:



      3




      import java.util.HashMap;
      import java.util.Map;
      import java.util.Scanner;

      public class SockMerchant {

      public int returnsockArr() {
      Scanner sc = new Scanner(System.in);
      int arraySize = sc.nextInt();
      int sockArr = new int[arraySize];
      for(int i = 0;i <= arraySize-1; i++) {
      sockArr[i] = sc.nextInt();
      }
      sc.close();
      return sockArr;
      }

      public int returnCount(Map<String, Object> mapCount) {
      int count = 0;
      for(String keyName : mapCount.keySet()) {
      int value = (int)mapCount.get(keyName);
      count = count+(value/2);
      }
      return count;
      }

      public int setMapAndReturnCount(int sockArr) {
      Map<String, Object> mapCount = new HashMap<String, Object>();
      for(int j = 0;j<= sockArr.length-1;j++) {
      if(mapCount.containsKey(Integer.toString(sockArr[j]))) {
      mapCount.put(Integer.toString(sockArr[j]), (int)mapCount.get(Integer.toString(sockArr[j]))+1);
      }
      else {
      mapCount.put(Integer.toString(sockArr[j]), 1);
      }
      }
      return returnCount(mapCount);
      }

      public static void main(String args) {
      SockMerchant sm = new SockMerchant();
      int sockArr = sm.returnsockArr();
      int finalCount = sm.setMapAndReturnCount(sockArr);
      System.out.println(finalCount);
      }
      }


      I think that I have complicated the solution a bit too much



      Is there a better approach to do the same thing?







      java beginner array






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 16 hours ago









      Akash SinghAkash Singh

      132




      132






















          1 Answer
          1






          active

          oldest

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          0












          $begingroup$

          There are couple of improvement possible in your code.



          In this method setMapAndReturnCount you should change the declaration of map from,



          Map<String, Object> mapCount = new HashMap<String, Object>();


          to



          Map<Integer, Integer> mapCount = new HashMap<Integer, Integer>();


          As you know, you want to store the input array numbers as key in Map and value is the count of occurence of a particular number in the array.



          By using Object class as value in Map, you are defeating one of the main purpose of having generics implementation in Java. If you use Object then you need to uselessly cast, and otherwise don't have to.



          Another point in same method, you can write your for loop in a better way like this, instead of your current code which does double work of first checking through containsKey and then uses mapCount.get to again pick the value of key.



          public int setMapAndReturnCount(int sockArr) {
          Map<Integer, Integer> mapCount = new HashMap<Integer, Integer>();
          for (int j = 0; j < sockArr.length; j++) {
          Integer count = mapCount.get(sockArr[j]);
          if (count == null) {
          mapCount.put(sockArr[j], 1);
          } else {
          mapCount.put(sockArr[j], count + 1);
          }
          }
          return returnCount(mapCount);
          }


          Similarly, you can change returnCount method to this,



          public int returnCount(Map<Integer, Integer> mapCount) {
          int count = 0;
          for (Integer value : mapCount.values()) {
          count += (value / 2);
          }
          return count;
          }


          As you don't need to first iterate on keySet and then retrieve the values with get which will be slower, and instead just iterate it on values as that is what you need for calculating number of pairs, which will be relatively faster.



          Also, if you are using Java-8 and above,



          You can change your returnCount to one liner like this,



          public int returnCount(Map<Integer, Integer> mapCount) {
          return mapCount.values().stream().mapToInt(x -> x / 2).sum();
          }





          share|improve this answer











          $endgroup$













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            1 Answer
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            1 Answer
            1






            active

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            active

            oldest

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            0












            $begingroup$

            There are couple of improvement possible in your code.



            In this method setMapAndReturnCount you should change the declaration of map from,



            Map<String, Object> mapCount = new HashMap<String, Object>();


            to



            Map<Integer, Integer> mapCount = new HashMap<Integer, Integer>();


            As you know, you want to store the input array numbers as key in Map and value is the count of occurence of a particular number in the array.



            By using Object class as value in Map, you are defeating one of the main purpose of having generics implementation in Java. If you use Object then you need to uselessly cast, and otherwise don't have to.



            Another point in same method, you can write your for loop in a better way like this, instead of your current code which does double work of first checking through containsKey and then uses mapCount.get to again pick the value of key.



            public int setMapAndReturnCount(int sockArr) {
            Map<Integer, Integer> mapCount = new HashMap<Integer, Integer>();
            for (int j = 0; j < sockArr.length; j++) {
            Integer count = mapCount.get(sockArr[j]);
            if (count == null) {
            mapCount.put(sockArr[j], 1);
            } else {
            mapCount.put(sockArr[j], count + 1);
            }
            }
            return returnCount(mapCount);
            }


            Similarly, you can change returnCount method to this,



            public int returnCount(Map<Integer, Integer> mapCount) {
            int count = 0;
            for (Integer value : mapCount.values()) {
            count += (value / 2);
            }
            return count;
            }


            As you don't need to first iterate on keySet and then retrieve the values with get which will be slower, and instead just iterate it on values as that is what you need for calculating number of pairs, which will be relatively faster.



            Also, if you are using Java-8 and above,



            You can change your returnCount to one liner like this,



            public int returnCount(Map<Integer, Integer> mapCount) {
            return mapCount.values().stream().mapToInt(x -> x / 2).sum();
            }





            share|improve this answer











            $endgroup$


















              0












              $begingroup$

              There are couple of improvement possible in your code.



              In this method setMapAndReturnCount you should change the declaration of map from,



              Map<String, Object> mapCount = new HashMap<String, Object>();


              to



              Map<Integer, Integer> mapCount = new HashMap<Integer, Integer>();


              As you know, you want to store the input array numbers as key in Map and value is the count of occurence of a particular number in the array.



              By using Object class as value in Map, you are defeating one of the main purpose of having generics implementation in Java. If you use Object then you need to uselessly cast, and otherwise don't have to.



              Another point in same method, you can write your for loop in a better way like this, instead of your current code which does double work of first checking through containsKey and then uses mapCount.get to again pick the value of key.



              public int setMapAndReturnCount(int sockArr) {
              Map<Integer, Integer> mapCount = new HashMap<Integer, Integer>();
              for (int j = 0; j < sockArr.length; j++) {
              Integer count = mapCount.get(sockArr[j]);
              if (count == null) {
              mapCount.put(sockArr[j], 1);
              } else {
              mapCount.put(sockArr[j], count + 1);
              }
              }
              return returnCount(mapCount);
              }


              Similarly, you can change returnCount method to this,



              public int returnCount(Map<Integer, Integer> mapCount) {
              int count = 0;
              for (Integer value : mapCount.values()) {
              count += (value / 2);
              }
              return count;
              }


              As you don't need to first iterate on keySet and then retrieve the values with get which will be slower, and instead just iterate it on values as that is what you need for calculating number of pairs, which will be relatively faster.



              Also, if you are using Java-8 and above,



              You can change your returnCount to one liner like this,



              public int returnCount(Map<Integer, Integer> mapCount) {
              return mapCount.values().stream().mapToInt(x -> x / 2).sum();
              }





              share|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                There are couple of improvement possible in your code.



                In this method setMapAndReturnCount you should change the declaration of map from,



                Map<String, Object> mapCount = new HashMap<String, Object>();


                to



                Map<Integer, Integer> mapCount = new HashMap<Integer, Integer>();


                As you know, you want to store the input array numbers as key in Map and value is the count of occurence of a particular number in the array.



                By using Object class as value in Map, you are defeating one of the main purpose of having generics implementation in Java. If you use Object then you need to uselessly cast, and otherwise don't have to.



                Another point in same method, you can write your for loop in a better way like this, instead of your current code which does double work of first checking through containsKey and then uses mapCount.get to again pick the value of key.



                public int setMapAndReturnCount(int sockArr) {
                Map<Integer, Integer> mapCount = new HashMap<Integer, Integer>();
                for (int j = 0; j < sockArr.length; j++) {
                Integer count = mapCount.get(sockArr[j]);
                if (count == null) {
                mapCount.put(sockArr[j], 1);
                } else {
                mapCount.put(sockArr[j], count + 1);
                }
                }
                return returnCount(mapCount);
                }


                Similarly, you can change returnCount method to this,



                public int returnCount(Map<Integer, Integer> mapCount) {
                int count = 0;
                for (Integer value : mapCount.values()) {
                count += (value / 2);
                }
                return count;
                }


                As you don't need to first iterate on keySet and then retrieve the values with get which will be slower, and instead just iterate it on values as that is what you need for calculating number of pairs, which will be relatively faster.



                Also, if you are using Java-8 and above,



                You can change your returnCount to one liner like this,



                public int returnCount(Map<Integer, Integer> mapCount) {
                return mapCount.values().stream().mapToInt(x -> x / 2).sum();
                }





                share|improve this answer











                $endgroup$



                There are couple of improvement possible in your code.



                In this method setMapAndReturnCount you should change the declaration of map from,



                Map<String, Object> mapCount = new HashMap<String, Object>();


                to



                Map<Integer, Integer> mapCount = new HashMap<Integer, Integer>();


                As you know, you want to store the input array numbers as key in Map and value is the count of occurence of a particular number in the array.



                By using Object class as value in Map, you are defeating one of the main purpose of having generics implementation in Java. If you use Object then you need to uselessly cast, and otherwise don't have to.



                Another point in same method, you can write your for loop in a better way like this, instead of your current code which does double work of first checking through containsKey and then uses mapCount.get to again pick the value of key.



                public int setMapAndReturnCount(int sockArr) {
                Map<Integer, Integer> mapCount = new HashMap<Integer, Integer>();
                for (int j = 0; j < sockArr.length; j++) {
                Integer count = mapCount.get(sockArr[j]);
                if (count == null) {
                mapCount.put(sockArr[j], 1);
                } else {
                mapCount.put(sockArr[j], count + 1);
                }
                }
                return returnCount(mapCount);
                }


                Similarly, you can change returnCount method to this,



                public int returnCount(Map<Integer, Integer> mapCount) {
                int count = 0;
                for (Integer value : mapCount.values()) {
                count += (value / 2);
                }
                return count;
                }


                As you don't need to first iterate on keySet and then retrieve the values with get which will be slower, and instead just iterate it on values as that is what you need for calculating number of pairs, which will be relatively faster.



                Also, if you are using Java-8 and above,



                You can change your returnCount to one liner like this,



                public int returnCount(Map<Integer, Integer> mapCount) {
                return mapCount.values().stream().mapToInt(x -> x / 2).sum();
                }






                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 15 hours ago

























                answered 15 hours ago









                Pushpesh Kumar RajwanshiPushpesh Kumar Rajwanshi

                1413




                1413






























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