Checking whether given array is sorted by divide-and-conquer
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3
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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
add a comment |
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)
java algorithm array recursion divide-and-conquer
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 on3, 4, 1, 2
.
– vnp
Nov 25 at 6:43
@vnpisSorted(list, 0, 3)
will callisSorted(arr, 1, 3)
which will callisSorted(arr, 1, 2)
which will returnfalse
. That test case is fine.
– AJNeufeld
Nov 25 at 18:02
add a comment |
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)
java algorithm array recursion divide-and-conquer
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 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 on3, 4, 1, 2
.
– vnp
Nov 25 at 6:43
@vnpisSorted(list, 0, 3)
will callisSorted(arr, 1, 3)
which will callisSorted(arr, 1, 2)
which will returnfalse
. That test case is fine.
– AJNeufeld
Nov 25 at 18:02
add a comment |
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 on3, 4, 1, 2
.
– vnp
Nov 25 at 6:43
@vnpisSorted(list, 0, 3)
will callisSorted(arr, 1, 3)
which will callisSorted(arr, 1, 2)
which will returnfalse
. 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
add a comment |
1 Answer
1
active
oldest
votes
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]
.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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]
.
add a comment |
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]
.
add a comment |
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]
.
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 at 20:09
AJNeufeld
3,869317
3,869317
add a comment |
add a comment |
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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 callisSorted(arr, 1, 3)
which will callisSorted(arr, 1, 2)
which will returnfalse
. That test case is fine.– AJNeufeld
Nov 25 at 18:02