Checking whether given array is sorted by divide-and-conquer











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I've written a code that I try to use divide and conquer approach to determine if the given array is sorted. I wonder whether I apply the approach accurately.



public static boolean isSorted(List<Integer> arr, int start, int end) {
if (end - start == 1) // base case to compare two elements
return arr.get(end) > arr.get(start);

int middle = (end + start) >>> 1; // division by two
boolean leftPart = isSorted(arr, start, middle);
boolean rightPart = isSorted(arr, middle, end);
return leftPart && rightPart;
}


To use call,



isSorted(list, 0, list.size() - 1)









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  • 3




    Please do not update the code in your question to incorporate feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers.
    – Martin R
    Nov 24 at 22:24










  • Test your code on 3, 4, 1, 2.
    – vnp
    Nov 25 at 6:43










  • @vnp isSorted(list, 0, 3) will call isSorted(arr, 1, 3) which will call isSorted(arr, 1, 2) which will return false. That test case is fine.
    – AJNeufeld
    Nov 25 at 18:02















up vote
3
down vote

favorite












I've written a code that I try to use divide and conquer approach to determine if the given array is sorted. I wonder whether I apply the approach accurately.



public static boolean isSorted(List<Integer> arr, int start, int end) {
if (end - start == 1) // base case to compare two elements
return arr.get(end) > arr.get(start);

int middle = (end + start) >>> 1; // division by two
boolean leftPart = isSorted(arr, start, middle);
boolean rightPart = isSorted(arr, middle, end);
return leftPart && rightPart;
}


To use call,



isSorted(list, 0, list.size() - 1)









share|improve this question




















  • 3




    Please do not update the code in your question to incorporate feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers.
    – Martin R
    Nov 24 at 22:24










  • Test your code on 3, 4, 1, 2.
    – vnp
    Nov 25 at 6:43










  • @vnp isSorted(list, 0, 3) will call isSorted(arr, 1, 3) which will call isSorted(arr, 1, 2) which will return false. That test case is fine.
    – AJNeufeld
    Nov 25 at 18:02













up vote
3
down vote

favorite









up vote
3
down vote

favorite











I've written a code that I try to use divide and conquer approach to determine if the given array is sorted. I wonder whether I apply the approach accurately.



public static boolean isSorted(List<Integer> arr, int start, int end) {
if (end - start == 1) // base case to compare two elements
return arr.get(end) > arr.get(start);

int middle = (end + start) >>> 1; // division by two
boolean leftPart = isSorted(arr, start, middle);
boolean rightPart = isSorted(arr, middle, end);
return leftPart && rightPart;
}


To use call,



isSorted(list, 0, list.size() - 1)









share|improve this question















I've written a code that I try to use divide and conquer approach to determine if the given array is sorted. I wonder whether I apply the approach accurately.



public static boolean isSorted(List<Integer> arr, int start, int end) {
if (end - start == 1) // base case to compare two elements
return arr.get(end) > arr.get(start);

int middle = (end + start) >>> 1; // division by two
boolean leftPart = isSorted(arr, start, middle);
boolean rightPart = isSorted(arr, middle, end);
return leftPart && rightPart;
}


To use call,



isSorted(list, 0, list.size() - 1)






java algorithm array recursion divide-and-conquer






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edited Nov 24 at 22:24









Martin R

15.4k12264




15.4k12264










asked Nov 24 at 19:24









itsnotmyrealname

1525




1525








  • 3




    Please do not update the code in your question to incorporate feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers.
    – Martin R
    Nov 24 at 22:24










  • Test your code on 3, 4, 1, 2.
    – vnp
    Nov 25 at 6:43










  • @vnp isSorted(list, 0, 3) will call isSorted(arr, 1, 3) which will call isSorted(arr, 1, 2) which will return false. That test case is fine.
    – AJNeufeld
    Nov 25 at 18:02














  • 3




    Please do not update the code in your question to incorporate feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers.
    – Martin R
    Nov 24 at 22:24










  • Test your code on 3, 4, 1, 2.
    – vnp
    Nov 25 at 6:43










  • @vnp isSorted(list, 0, 3) will call isSorted(arr, 1, 3) which will call isSorted(arr, 1, 2) which will return false. That test case is fine.
    – AJNeufeld
    Nov 25 at 18:02








3




3




Please do not update the code in your question to incorporate feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers.
– Martin R
Nov 24 at 22:24




Please do not update the code in your question to incorporate feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers.
– Martin R
Nov 24 at 22:24












Test your code on 3, 4, 1, 2.
– vnp
Nov 25 at 6:43




Test your code on 3, 4, 1, 2.
– vnp
Nov 25 at 6:43












@vnp isSorted(list, 0, 3) will call isSorted(arr, 1, 3) which will call isSorted(arr, 1, 2) which will return false. That test case is fine.
– AJNeufeld
Nov 25 at 18:02




@vnp isSorted(list, 0, 3) will call isSorted(arr, 1, 3) which will call isSorted(arr, 1, 2) which will return false. That test case is fine.
– AJNeufeld
Nov 25 at 18:02










1 Answer
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up vote
2
down vote













It looks like your are doing it mostly right. You have problems with length zero and length 1 arrays, but you should be able to fix those pretty quick.



You may be doing more work than necessary. If an array is not sorted, you might find leftPart is false, but you unconditionally go on to determine the value of rightPart anyway, despite it not mattering. The simplest way to avoid that is to combine that recursive calls and the && operation. Ie:



return isSorted(arr, start, middle) && isSorted(arr, middle, end);


Lastly, if the array contains duplicates, can it still be considered sorted? You return false for [1, 2, 2, 3].






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    1 Answer
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    1 Answer
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    up vote
    2
    down vote













    It looks like your are doing it mostly right. You have problems with length zero and length 1 arrays, but you should be able to fix those pretty quick.



    You may be doing more work than necessary. If an array is not sorted, you might find leftPart is false, but you unconditionally go on to determine the value of rightPart anyway, despite it not mattering. The simplest way to avoid that is to combine that recursive calls and the && operation. Ie:



    return isSorted(arr, start, middle) && isSorted(arr, middle, end);


    Lastly, if the array contains duplicates, can it still be considered sorted? You return false for [1, 2, 2, 3].






    share|improve this answer

























      up vote
      2
      down vote













      It looks like your are doing it mostly right. You have problems with length zero and length 1 arrays, but you should be able to fix those pretty quick.



      You may be doing more work than necessary. If an array is not sorted, you might find leftPart is false, but you unconditionally go on to determine the value of rightPart anyway, despite it not mattering. The simplest way to avoid that is to combine that recursive calls and the && operation. Ie:



      return isSorted(arr, start, middle) && isSorted(arr, middle, end);


      Lastly, if the array contains duplicates, can it still be considered sorted? You return false for [1, 2, 2, 3].






      share|improve this answer























        up vote
        2
        down vote










        up vote
        2
        down vote









        It looks like your are doing it mostly right. You have problems with length zero and length 1 arrays, but you should be able to fix those pretty quick.



        You may be doing more work than necessary. If an array is not sorted, you might find leftPart is false, but you unconditionally go on to determine the value of rightPart anyway, despite it not mattering. The simplest way to avoid that is to combine that recursive calls and the && operation. Ie:



        return isSorted(arr, start, middle) && isSorted(arr, middle, end);


        Lastly, if the array contains duplicates, can it still be considered sorted? You return false for [1, 2, 2, 3].






        share|improve this answer












        It looks like your are doing it mostly right. You have problems with length zero and length 1 arrays, but you should be able to fix those pretty quick.



        You may be doing more work than necessary. If an array is not sorted, you might find leftPart is false, but you unconditionally go on to determine the value of rightPart anyway, despite it not mattering. The simplest way to avoid that is to combine that recursive calls and the && operation. Ie:



        return isSorted(arr, start, middle) && isSorted(arr, middle, end);


        Lastly, if the array contains duplicates, can it still be considered sorted? You return false for [1, 2, 2, 3].







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 24 at 20:09









        AJNeufeld

        3,784317




        3,784317






























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