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Median Algorithm for 4 sorted arrays


Find median in four (individually) sorted arrays with O(1) spaceSort a Map<Key, Value> by valuesCreate ArrayList from arrayHow do I check if an array includes an object in JavaScript?What is the best algorithm for an overridden System.Object.GetHashCode?How do I create a URL shortener?Fastest sort of fixed length 6 int arrayEasy interview question got harder: given numbers 1..100, find the missing number(s)Image Processing: Algorithm Improvement for 'Coca-Cola Can' RecognitionWhy is it faster to process a sorted array than an unsorted array?What is the optimal algorithm for the game 2048?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;








-2















I need to write an Algorithm for my course, to find the middle value of 4 sorted arrays different sizes in O(n), and i'm not allowed to create an array to store the data.
how should I approach the problem? I thought about running on a loop with 4 indexes as if i'm sorting the arrays into a big array but instead just run without storing the data. the loop will stop at n/2 and it should provide me the middle value.
writing it seems complex and very messy (i need to check for 4 of the arrays if i'm out of bound), is there a better way to approach this?










share|improve this question

















  • 3





    Possible duplicate of Find median in four (individually) sorted arrays with O(1) space

    – Matt Timmermans
    Mar 21 at 23:38











  • The question linked dropped the explicit mention of different sizes. It has no useful answer as of this comment.

    – greybeard
    Mar 22 at 6:52

















-2















I need to write an Algorithm for my course, to find the middle value of 4 sorted arrays different sizes in O(n), and i'm not allowed to create an array to store the data.
how should I approach the problem? I thought about running on a loop with 4 indexes as if i'm sorting the arrays into a big array but instead just run without storing the data. the loop will stop at n/2 and it should provide me the middle value.
writing it seems complex and very messy (i need to check for 4 of the arrays if i'm out of bound), is there a better way to approach this?










share|improve this question

















  • 3





    Possible duplicate of Find median in four (individually) sorted arrays with O(1) space

    – Matt Timmermans
    Mar 21 at 23:38











  • The question linked dropped the explicit mention of different sizes. It has no useful answer as of this comment.

    – greybeard
    Mar 22 at 6:52













-2












-2








-2








I need to write an Algorithm for my course, to find the middle value of 4 sorted arrays different sizes in O(n), and i'm not allowed to create an array to store the data.
how should I approach the problem? I thought about running on a loop with 4 indexes as if i'm sorting the arrays into a big array but instead just run without storing the data. the loop will stop at n/2 and it should provide me the middle value.
writing it seems complex and very messy (i need to check for 4 of the arrays if i'm out of bound), is there a better way to approach this?










share|improve this question














I need to write an Algorithm for my course, to find the middle value of 4 sorted arrays different sizes in O(n), and i'm not allowed to create an array to store the data.
how should I approach the problem? I thought about running on a loop with 4 indexes as if i'm sorting the arrays into a big array but instead just run without storing the data. the loop will stop at n/2 and it should provide me the middle value.
writing it seems complex and very messy (i need to check for 4 of the arrays if i'm out of bound), is there a better way to approach this?







java algorithm median






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Mar 21 at 22:58









Arik ShapiroArik Shapiro

1




1







  • 3





    Possible duplicate of Find median in four (individually) sorted arrays with O(1) space

    – Matt Timmermans
    Mar 21 at 23:38











  • The question linked dropped the explicit mention of different sizes. It has no useful answer as of this comment.

    – greybeard
    Mar 22 at 6:52












  • 3





    Possible duplicate of Find median in four (individually) sorted arrays with O(1) space

    – Matt Timmermans
    Mar 21 at 23:38











  • The question linked dropped the explicit mention of different sizes. It has no useful answer as of this comment.

    – greybeard
    Mar 22 at 6:52







3




3





Possible duplicate of Find median in four (individually) sorted arrays with O(1) space

– Matt Timmermans
Mar 21 at 23:38





Possible duplicate of Find median in four (individually) sorted arrays with O(1) space

– Matt Timmermans
Mar 21 at 23:38













The question linked dropped the explicit mention of different sizes. It has no useful answer as of this comment.

– greybeard
Mar 22 at 6:52





The question linked dropped the explicit mention of different sizes. It has no useful answer as of this comment.

– greybeard
Mar 22 at 6:52












2 Answers
2






active

oldest

votes


















0














I think you're on to the key idea: the median value of all the values in four arrays is just the median of all the values, so if we get half way through all the values then whatever is next is the median. I would suggest structuring as follows:



int firstIndex = 0;
int secondIndex = 0;
int thirdIndex = 0;
int fourthIndex = 0;
double current;

for (int i = 0; i < n/2; i++)
// 1.) Find the value out of the four at firstIndex, secondIndex, ...
// which is smallest, and assign it to current
// 2.) Increment whichever of the four indices belongs to that element

// whatever is in current at the end of the loop is the middle element


You probably want a function findMin(int index1, int index2, int index3, int index4). This method could also be responsible for the out-of-bounds checks, so the main loop could just rely on it to be pointed in the right direction, and not care if it's run out of elements in any given array.



Does this make sense? I've tried to leave enough ambiguity to let you handle most of the real implementation work :)






share|improve this answer























  • I already had this much figured out.. the problem is findMin cannot be responsible for out-of-bounds checks because you'd need to be giving it the arrays too or atleast at array's length. the second problem is if you make a findMin function once you find the min value and send it back how will you know which index gave you that number. you'd need a global index but that's prohibited.

    – Arik Shapiro
    Mar 22 at 10:26











  • “You’d need to be giving it the arrays too” yes, or make the arrays part of state which findMin() can access (i.e. global variables). “How will you know which index gave you that number” since you’re keeping track of firstIndex, secondIndex, etc, why should findMin() return the min value at all?

    – MyStackRunnethOver
    Mar 23 at 3:34


















0














Think of a single unsorted array



Consider thinking about the 4 arrays as a single unsorted array, broken into 4 parts. If you can modify the arrays, you can sort all 4 arrays as if 1 by swapping values between them (some optimizations can be made since you know the 4 arrays are sorted). Once you've sorted the arrays up to n/2 (where n is the aggregate length of the 4 arrays), just return the middle value of all 4.



Some Code



The implementation below begins to make multiple arrays function like a single one. I've implemented get, set, and length methods, the basis for any array. All that needs to happen now is for the class' data to be sorted (possibly up to n/2) using get(int), set(int,int), and length(), and a method which returns the median value median().



There is also room for further optimization by sorting only up to n/2 within the median method, also when caching (i,j) pairs for each element when doing so.



int median( int[] a1, int[] a2, int[] a3, int[] a4 ) 
MultiIntArray array = new MultiIntArray( a1, a2, a3, a4 );
array.sort();
return array.get( array.length() / 2 );



public class MultiIntArray 

private int[][] data;

public MultiIntArray( int[]... data )
this.data = data;


public void sort()
// FOR YOU TO IMPLEMENT


public int length()
int length = 0;
for ( int[] array : data )
length += array.length;

return length;


public int get( int index )
int i = 0;
while ( index >= data[i].length )
index -= data[i].length;
i += 1;

return data[i][index];


public void set( int index, int value )
int i = 0;
while ( index >= data[i].length )
index -= data[i].length;
i += 1;

data[i][index] = value;








share|improve this answer

























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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    I think you're on to the key idea: the median value of all the values in four arrays is just the median of all the values, so if we get half way through all the values then whatever is next is the median. I would suggest structuring as follows:



    int firstIndex = 0;
    int secondIndex = 0;
    int thirdIndex = 0;
    int fourthIndex = 0;
    double current;

    for (int i = 0; i < n/2; i++)
    // 1.) Find the value out of the four at firstIndex, secondIndex, ...
    // which is smallest, and assign it to current
    // 2.) Increment whichever of the four indices belongs to that element

    // whatever is in current at the end of the loop is the middle element


    You probably want a function findMin(int index1, int index2, int index3, int index4). This method could also be responsible for the out-of-bounds checks, so the main loop could just rely on it to be pointed in the right direction, and not care if it's run out of elements in any given array.



    Does this make sense? I've tried to leave enough ambiguity to let you handle most of the real implementation work :)






    share|improve this answer























    • I already had this much figured out.. the problem is findMin cannot be responsible for out-of-bounds checks because you'd need to be giving it the arrays too or atleast at array's length. the second problem is if you make a findMin function once you find the min value and send it back how will you know which index gave you that number. you'd need a global index but that's prohibited.

      – Arik Shapiro
      Mar 22 at 10:26











    • “You’d need to be giving it the arrays too” yes, or make the arrays part of state which findMin() can access (i.e. global variables). “How will you know which index gave you that number” since you’re keeping track of firstIndex, secondIndex, etc, why should findMin() return the min value at all?

      – MyStackRunnethOver
      Mar 23 at 3:34















    0














    I think you're on to the key idea: the median value of all the values in four arrays is just the median of all the values, so if we get half way through all the values then whatever is next is the median. I would suggest structuring as follows:



    int firstIndex = 0;
    int secondIndex = 0;
    int thirdIndex = 0;
    int fourthIndex = 0;
    double current;

    for (int i = 0; i < n/2; i++)
    // 1.) Find the value out of the four at firstIndex, secondIndex, ...
    // which is smallest, and assign it to current
    // 2.) Increment whichever of the four indices belongs to that element

    // whatever is in current at the end of the loop is the middle element


    You probably want a function findMin(int index1, int index2, int index3, int index4). This method could also be responsible for the out-of-bounds checks, so the main loop could just rely on it to be pointed in the right direction, and not care if it's run out of elements in any given array.



    Does this make sense? I've tried to leave enough ambiguity to let you handle most of the real implementation work :)






    share|improve this answer























    • I already had this much figured out.. the problem is findMin cannot be responsible for out-of-bounds checks because you'd need to be giving it the arrays too or atleast at array's length. the second problem is if you make a findMin function once you find the min value and send it back how will you know which index gave you that number. you'd need a global index but that's prohibited.

      – Arik Shapiro
      Mar 22 at 10:26











    • “You’d need to be giving it the arrays too” yes, or make the arrays part of state which findMin() can access (i.e. global variables). “How will you know which index gave you that number” since you’re keeping track of firstIndex, secondIndex, etc, why should findMin() return the min value at all?

      – MyStackRunnethOver
      Mar 23 at 3:34













    0












    0








    0







    I think you're on to the key idea: the median value of all the values in four arrays is just the median of all the values, so if we get half way through all the values then whatever is next is the median. I would suggest structuring as follows:



    int firstIndex = 0;
    int secondIndex = 0;
    int thirdIndex = 0;
    int fourthIndex = 0;
    double current;

    for (int i = 0; i < n/2; i++)
    // 1.) Find the value out of the four at firstIndex, secondIndex, ...
    // which is smallest, and assign it to current
    // 2.) Increment whichever of the four indices belongs to that element

    // whatever is in current at the end of the loop is the middle element


    You probably want a function findMin(int index1, int index2, int index3, int index4). This method could also be responsible for the out-of-bounds checks, so the main loop could just rely on it to be pointed in the right direction, and not care if it's run out of elements in any given array.



    Does this make sense? I've tried to leave enough ambiguity to let you handle most of the real implementation work :)






    share|improve this answer













    I think you're on to the key idea: the median value of all the values in four arrays is just the median of all the values, so if we get half way through all the values then whatever is next is the median. I would suggest structuring as follows:



    int firstIndex = 0;
    int secondIndex = 0;
    int thirdIndex = 0;
    int fourthIndex = 0;
    double current;

    for (int i = 0; i < n/2; i++)
    // 1.) Find the value out of the four at firstIndex, secondIndex, ...
    // which is smallest, and assign it to current
    // 2.) Increment whichever of the four indices belongs to that element

    // whatever is in current at the end of the loop is the middle element


    You probably want a function findMin(int index1, int index2, int index3, int index4). This method could also be responsible for the out-of-bounds checks, so the main loop could just rely on it to be pointed in the right direction, and not care if it's run out of elements in any given array.



    Does this make sense? I've tried to leave enough ambiguity to let you handle most of the real implementation work :)







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Mar 21 at 23:40









    MyStackRunnethOverMyStackRunnethOver

    1,011819




    1,011819












    • I already had this much figured out.. the problem is findMin cannot be responsible for out-of-bounds checks because you'd need to be giving it the arrays too or atleast at array's length. the second problem is if you make a findMin function once you find the min value and send it back how will you know which index gave you that number. you'd need a global index but that's prohibited.

      – Arik Shapiro
      Mar 22 at 10:26











    • “You’d need to be giving it the arrays too” yes, or make the arrays part of state which findMin() can access (i.e. global variables). “How will you know which index gave you that number” since you’re keeping track of firstIndex, secondIndex, etc, why should findMin() return the min value at all?

      – MyStackRunnethOver
      Mar 23 at 3:34

















    • I already had this much figured out.. the problem is findMin cannot be responsible for out-of-bounds checks because you'd need to be giving it the arrays too or atleast at array's length. the second problem is if you make a findMin function once you find the min value and send it back how will you know which index gave you that number. you'd need a global index but that's prohibited.

      – Arik Shapiro
      Mar 22 at 10:26











    • “You’d need to be giving it the arrays too” yes, or make the arrays part of state which findMin() can access (i.e. global variables). “How will you know which index gave you that number” since you’re keeping track of firstIndex, secondIndex, etc, why should findMin() return the min value at all?

      – MyStackRunnethOver
      Mar 23 at 3:34
















    I already had this much figured out.. the problem is findMin cannot be responsible for out-of-bounds checks because you'd need to be giving it the arrays too or atleast at array's length. the second problem is if you make a findMin function once you find the min value and send it back how will you know which index gave you that number. you'd need a global index but that's prohibited.

    – Arik Shapiro
    Mar 22 at 10:26





    I already had this much figured out.. the problem is findMin cannot be responsible for out-of-bounds checks because you'd need to be giving it the arrays too or atleast at array's length. the second problem is if you make a findMin function once you find the min value and send it back how will you know which index gave you that number. you'd need a global index but that's prohibited.

    – Arik Shapiro
    Mar 22 at 10:26













    “You’d need to be giving it the arrays too” yes, or make the arrays part of state which findMin() can access (i.e. global variables). “How will you know which index gave you that number” since you’re keeping track of firstIndex, secondIndex, etc, why should findMin() return the min value at all?

    – MyStackRunnethOver
    Mar 23 at 3:34





    “You’d need to be giving it the arrays too” yes, or make the arrays part of state which findMin() can access (i.e. global variables). “How will you know which index gave you that number” since you’re keeping track of firstIndex, secondIndex, etc, why should findMin() return the min value at all?

    – MyStackRunnethOver
    Mar 23 at 3:34













    0














    Think of a single unsorted array



    Consider thinking about the 4 arrays as a single unsorted array, broken into 4 parts. If you can modify the arrays, you can sort all 4 arrays as if 1 by swapping values between them (some optimizations can be made since you know the 4 arrays are sorted). Once you've sorted the arrays up to n/2 (where n is the aggregate length of the 4 arrays), just return the middle value of all 4.



    Some Code



    The implementation below begins to make multiple arrays function like a single one. I've implemented get, set, and length methods, the basis for any array. All that needs to happen now is for the class' data to be sorted (possibly up to n/2) using get(int), set(int,int), and length(), and a method which returns the median value median().



    There is also room for further optimization by sorting only up to n/2 within the median method, also when caching (i,j) pairs for each element when doing so.



    int median( int[] a1, int[] a2, int[] a3, int[] a4 ) 
    MultiIntArray array = new MultiIntArray( a1, a2, a3, a4 );
    array.sort();
    return array.get( array.length() / 2 );



    public class MultiIntArray 

    private int[][] data;

    public MultiIntArray( int[]... data )
    this.data = data;


    public void sort()
    // FOR YOU TO IMPLEMENT


    public int length()
    int length = 0;
    for ( int[] array : data )
    length += array.length;

    return length;


    public int get( int index )
    int i = 0;
    while ( index >= data[i].length )
    index -= data[i].length;
    i += 1;

    return data[i][index];


    public void set( int index, int value )
    int i = 0;
    while ( index >= data[i].length )
    index -= data[i].length;
    i += 1;

    data[i][index] = value;








    share|improve this answer





























      0














      Think of a single unsorted array



      Consider thinking about the 4 arrays as a single unsorted array, broken into 4 parts. If you can modify the arrays, you can sort all 4 arrays as if 1 by swapping values between them (some optimizations can be made since you know the 4 arrays are sorted). Once you've sorted the arrays up to n/2 (where n is the aggregate length of the 4 arrays), just return the middle value of all 4.



      Some Code



      The implementation below begins to make multiple arrays function like a single one. I've implemented get, set, and length methods, the basis for any array. All that needs to happen now is for the class' data to be sorted (possibly up to n/2) using get(int), set(int,int), and length(), and a method which returns the median value median().



      There is also room for further optimization by sorting only up to n/2 within the median method, also when caching (i,j) pairs for each element when doing so.



      int median( int[] a1, int[] a2, int[] a3, int[] a4 ) 
      MultiIntArray array = new MultiIntArray( a1, a2, a3, a4 );
      array.sort();
      return array.get( array.length() / 2 );



      public class MultiIntArray 

      private int[][] data;

      public MultiIntArray( int[]... data )
      this.data = data;


      public void sort()
      // FOR YOU TO IMPLEMENT


      public int length()
      int length = 0;
      for ( int[] array : data )
      length += array.length;

      return length;


      public int get( int index )
      int i = 0;
      while ( index >= data[i].length )
      index -= data[i].length;
      i += 1;

      return data[i][index];


      public void set( int index, int value )
      int i = 0;
      while ( index >= data[i].length )
      index -= data[i].length;
      i += 1;

      data[i][index] = value;








      share|improve this answer



























        0












        0








        0







        Think of a single unsorted array



        Consider thinking about the 4 arrays as a single unsorted array, broken into 4 parts. If you can modify the arrays, you can sort all 4 arrays as if 1 by swapping values between them (some optimizations can be made since you know the 4 arrays are sorted). Once you've sorted the arrays up to n/2 (where n is the aggregate length of the 4 arrays), just return the middle value of all 4.



        Some Code



        The implementation below begins to make multiple arrays function like a single one. I've implemented get, set, and length methods, the basis for any array. All that needs to happen now is for the class' data to be sorted (possibly up to n/2) using get(int), set(int,int), and length(), and a method which returns the median value median().



        There is also room for further optimization by sorting only up to n/2 within the median method, also when caching (i,j) pairs for each element when doing so.



        int median( int[] a1, int[] a2, int[] a3, int[] a4 ) 
        MultiIntArray array = new MultiIntArray( a1, a2, a3, a4 );
        array.sort();
        return array.get( array.length() / 2 );



        public class MultiIntArray 

        private int[][] data;

        public MultiIntArray( int[]... data )
        this.data = data;


        public void sort()
        // FOR YOU TO IMPLEMENT


        public int length()
        int length = 0;
        for ( int[] array : data )
        length += array.length;

        return length;


        public int get( int index )
        int i = 0;
        while ( index >= data[i].length )
        index -= data[i].length;
        i += 1;

        return data[i][index];


        public void set( int index, int value )
        int i = 0;
        while ( index >= data[i].length )
        index -= data[i].length;
        i += 1;

        data[i][index] = value;








        share|improve this answer















        Think of a single unsorted array



        Consider thinking about the 4 arrays as a single unsorted array, broken into 4 parts. If you can modify the arrays, you can sort all 4 arrays as if 1 by swapping values between them (some optimizations can be made since you know the 4 arrays are sorted). Once you've sorted the arrays up to n/2 (where n is the aggregate length of the 4 arrays), just return the middle value of all 4.



        Some Code



        The implementation below begins to make multiple arrays function like a single one. I've implemented get, set, and length methods, the basis for any array. All that needs to happen now is for the class' data to be sorted (possibly up to n/2) using get(int), set(int,int), and length(), and a method which returns the median value median().



        There is also room for further optimization by sorting only up to n/2 within the median method, also when caching (i,j) pairs for each element when doing so.



        int median( int[] a1, int[] a2, int[] a3, int[] a4 ) 
        MultiIntArray array = new MultiIntArray( a1, a2, a3, a4 );
        array.sort();
        return array.get( array.length() / 2 );



        public class MultiIntArray 

        private int[][] data;

        public MultiIntArray( int[]... data )
        this.data = data;


        public void sort()
        // FOR YOU TO IMPLEMENT


        public int length()
        int length = 0;
        for ( int[] array : data )
        length += array.length;

        return length;


        public int get( int index )
        int i = 0;
        while ( index >= data[i].length )
        index -= data[i].length;
        i += 1;

        return data[i][index];


        public void set( int index, int value )
        int i = 0;
        while ( index >= data[i].length )
        index -= data[i].length;
        i += 1;

        data[i][index] = value;









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        edited Mar 22 at 1:22

























        answered Mar 22 at 0:53









        Robert E FryRobert E Fry

        1511313




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