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)
java algorithm array recursion divide-and-conquer
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add a comment |
$begingroup$
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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3
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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.
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– Martin R
Nov 24 '18 at 22:24
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Test your code on3, 4, 1, 2
.
$endgroup$
– vnp
Nov 25 '18 at 6:43
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@vnpisSorted(list, 0, 3)
will callisSorted(arr, 1, 3)
which will callisSorted(arr, 1, 2)
which will returnfalse
. That test case is fine.
$endgroup$
– AJNeufeld
Nov 25 '18 at 18:02
add a comment |
$begingroup$
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
$endgroup$
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
java algorithm array recursion divide-and-conquer
edited Nov 24 '18 at 22:24
Martin R
16.1k12364
16.1k12364
asked Nov 24 '18 at 19:24
itsnotmyrealnameitsnotmyrealname
1625
1625
3
$begingroup$
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.
$endgroup$
– Martin R
Nov 24 '18 at 22:24
$begingroup$
Test your code on3, 4, 1, 2
.
$endgroup$
– vnp
Nov 25 '18 at 6:43
$begingroup$
@vnpisSorted(list, 0, 3)
will callisSorted(arr, 1, 3)
which will callisSorted(arr, 1, 2)
which will returnfalse
. That test case is fine.
$endgroup$
– AJNeufeld
Nov 25 '18 at 18:02
add a comment |
3
$begingroup$
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.
$endgroup$
– Martin R
Nov 24 '18 at 22:24
$begingroup$
Test your code on3, 4, 1, 2
.
$endgroup$
– vnp
Nov 25 '18 at 6:43
$begingroup$
@vnpisSorted(list, 0, 3)
will callisSorted(arr, 1, 3)
which will callisSorted(arr, 1, 2)
which will returnfalse
. That test case is fine.
$endgroup$
– AJNeufeld
Nov 25 '18 at 18:02
3
3
$begingroup$
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.
$endgroup$
– Martin R
Nov 24 '18 at 22:24
$begingroup$
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.
$endgroup$
– Martin R
Nov 24 '18 at 22:24
$begingroup$
Test your code on
3, 4, 1, 2
.$endgroup$
– vnp
Nov 25 '18 at 6:43
$begingroup$
Test your code on
3, 4, 1, 2
.$endgroup$
– vnp
Nov 25 '18 at 6:43
$begingroup$
@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.$endgroup$
– AJNeufeld
Nov 25 '18 at 18:02
$begingroup$
@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.$endgroup$
– AJNeufeld
Nov 25 '18 at 18:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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]
.
$endgroup$
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
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votes
$begingroup$
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]
.
$endgroup$
add a comment |
$begingroup$
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]
.
$endgroup$
add a comment |
$begingroup$
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]
.
$endgroup$
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]
.
answered Nov 24 '18 at 20:09
AJNeufeldAJNeufeld
5,9751520
5,9751520
add a comment |
add a comment |
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3
$begingroup$
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.
$endgroup$
– Martin R
Nov 24 '18 at 22:24
$begingroup$
Test your code on
3, 4, 1, 2
.$endgroup$
– vnp
Nov 25 '18 at 6:43
$begingroup$
@vnp
isSorted(list, 0, 3)
will callisSorted(arr, 1, 3)
which will callisSorted(arr, 1, 2)
which will returnfalse
. That test case is fine.$endgroup$
– AJNeufeld
Nov 25 '18 at 18:02