How to implement duplicates in QuickSelect









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2
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I have made the quick select algorithm, which is to find the kth smallest number in an array. My problem is, it only works with an array without duplicates.
If I have an array




arr = 1,2,2,3,5,5,8,2,4,8,8




It will say that the third smallest number is 2, but it is actually 3.



I am stuck on what to do, here are my two methods quickSelect and Partition:



private int quickselect(int array, int leftIndex, int rightIndex, int kthSmallest) 

if(kthSmallest > array.length - 1)
System.out.print("Number does not exist. Please enter a number less than: ");
return array.length - 1;


if (leftIndex == rightIndex)
return array[leftIndex];


int indexOfPivot = generatePivot(leftIndex, rightIndex);

indexOfPivot = quickSelectPartition(array, leftIndex, rightIndex, indexOfPivot);

if (kthSmallest == indexOfPivot)

return array[kthSmallest];

else if (kthSmallest < indexOfPivot)

return quickselect(array, leftIndex, indexOfPivot - 1, kthSmallest);

else

return quickselect(array, indexOfPivot + 1, rightIndex, kthSmallest);




private int quickSelectPartition(int array, int left, int right, int pivotIndex)

int pivotValue = array[pivotIndex];

swapIndexes(array, pivotIndex, right);

int firstPointer = left;

for(int secondPointer = left; secondPointer < right; secondPointer++)

if(array[secondPointer] < pivotValue)

swapIndexes(array, firstPointer, secondPointer);

firstPointer++;



swapIndexes(array, right, firstPointer);

return firstPointer;










share|improve this question





















  • I think that is impossible to do in O(n) best case any more. You can use hash tables, but that gives you only expected time.
    – Yola
    Nov 11 at 15:58














up vote
2
down vote

favorite












I have made the quick select algorithm, which is to find the kth smallest number in an array. My problem is, it only works with an array without duplicates.
If I have an array




arr = 1,2,2,3,5,5,8,2,4,8,8




It will say that the third smallest number is 2, but it is actually 3.



I am stuck on what to do, here are my two methods quickSelect and Partition:



private int quickselect(int array, int leftIndex, int rightIndex, int kthSmallest) 

if(kthSmallest > array.length - 1)
System.out.print("Number does not exist. Please enter a number less than: ");
return array.length - 1;


if (leftIndex == rightIndex)
return array[leftIndex];


int indexOfPivot = generatePivot(leftIndex, rightIndex);

indexOfPivot = quickSelectPartition(array, leftIndex, rightIndex, indexOfPivot);

if (kthSmallest == indexOfPivot)

return array[kthSmallest];

else if (kthSmallest < indexOfPivot)

return quickselect(array, leftIndex, indexOfPivot - 1, kthSmallest);

else

return quickselect(array, indexOfPivot + 1, rightIndex, kthSmallest);




private int quickSelectPartition(int array, int left, int right, int pivotIndex)

int pivotValue = array[pivotIndex];

swapIndexes(array, pivotIndex, right);

int firstPointer = left;

for(int secondPointer = left; secondPointer < right; secondPointer++)

if(array[secondPointer] < pivotValue)

swapIndexes(array, firstPointer, secondPointer);

firstPointer++;



swapIndexes(array, right, firstPointer);

return firstPointer;










share|improve this question





















  • I think that is impossible to do in O(n) best case any more. You can use hash tables, but that gives you only expected time.
    – Yola
    Nov 11 at 15:58












up vote
2
down vote

favorite









up vote
2
down vote

favorite











I have made the quick select algorithm, which is to find the kth smallest number in an array. My problem is, it only works with an array without duplicates.
If I have an array




arr = 1,2,2,3,5,5,8,2,4,8,8




It will say that the third smallest number is 2, but it is actually 3.



I am stuck on what to do, here are my two methods quickSelect and Partition:



private int quickselect(int array, int leftIndex, int rightIndex, int kthSmallest) 

if(kthSmallest > array.length - 1)
System.out.print("Number does not exist. Please enter a number less than: ");
return array.length - 1;


if (leftIndex == rightIndex)
return array[leftIndex];


int indexOfPivot = generatePivot(leftIndex, rightIndex);

indexOfPivot = quickSelectPartition(array, leftIndex, rightIndex, indexOfPivot);

if (kthSmallest == indexOfPivot)

return array[kthSmallest];

else if (kthSmallest < indexOfPivot)

return quickselect(array, leftIndex, indexOfPivot - 1, kthSmallest);

else

return quickselect(array, indexOfPivot + 1, rightIndex, kthSmallest);




private int quickSelectPartition(int array, int left, int right, int pivotIndex)

int pivotValue = array[pivotIndex];

swapIndexes(array, pivotIndex, right);

int firstPointer = left;

for(int secondPointer = left; secondPointer < right; secondPointer++)

if(array[secondPointer] < pivotValue)

swapIndexes(array, firstPointer, secondPointer);

firstPointer++;



swapIndexes(array, right, firstPointer);

return firstPointer;










share|improve this question













I have made the quick select algorithm, which is to find the kth smallest number in an array. My problem is, it only works with an array without duplicates.
If I have an array




arr = 1,2,2,3,5,5,8,2,4,8,8




It will say that the third smallest number is 2, but it is actually 3.



I am stuck on what to do, here are my two methods quickSelect and Partition:



private int quickselect(int array, int leftIndex, int rightIndex, int kthSmallest) 

if(kthSmallest > array.length - 1)
System.out.print("Number does not exist. Please enter a number less than: ");
return array.length - 1;


if (leftIndex == rightIndex)
return array[leftIndex];


int indexOfPivot = generatePivot(leftIndex, rightIndex);

indexOfPivot = quickSelectPartition(array, leftIndex, rightIndex, indexOfPivot);

if (kthSmallest == indexOfPivot)

return array[kthSmallest];

else if (kthSmallest < indexOfPivot)

return quickselect(array, leftIndex, indexOfPivot - 1, kthSmallest);

else

return quickselect(array, indexOfPivot + 1, rightIndex, kthSmallest);




private int quickSelectPartition(int array, int left, int right, int pivotIndex)

int pivotValue = array[pivotIndex];

swapIndexes(array, pivotIndex, right);

int firstPointer = left;

for(int secondPointer = left; secondPointer < right; secondPointer++)

if(array[secondPointer] < pivotValue)

swapIndexes(array, firstPointer, secondPointer);

firstPointer++;



swapIndexes(array, right, firstPointer);

return firstPointer;







java algorithm big-o quickselect






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asked Nov 11 at 10:39







user7722505


















  • I think that is impossible to do in O(n) best case any more. You can use hash tables, but that gives you only expected time.
    – Yola
    Nov 11 at 15:58
















  • I think that is impossible to do in O(n) best case any more. You can use hash tables, but that gives you only expected time.
    – Yola
    Nov 11 at 15:58















I think that is impossible to do in O(n) best case any more. You can use hash tables, but that gives you only expected time.
– Yola
Nov 11 at 15:58




I think that is impossible to do in O(n) best case any more. You can use hash tables, but that gives you only expected time.
– Yola
Nov 11 at 15:58












1 Answer
1






active

oldest

votes

















up vote
0
down vote



accepted










If adding 2xN to the running time is acceptable, you could start by copying distinct elements into a new array:



private int getDistinct(int array) 
Set<Integer> distinct = new HashSet<>();
int endIdx = 0;

for (int element : array)

if (distinct.add(element))
array[endIdx++] = element;



return Arrays.copyOfRange(array, 0, endIdx);



and then do quickselect on that:



int arr = new int 1, 2, 2, 3, 5, 5, 8, 2, 4, 8, 8;
int kthSmallest = 6;

int distinctArray = getDistinct(arr);
int kthSmallestElement = quickselect(distinctArray, 0, distinctArray.length - 1, kthSmallest - 1);

System.out.println(kthSmallestElement);


Output:



8





share|improve this answer






















  • The running time of QuickSelect is best case and average case O(n), so adding 2N would still be O(n) I suppose?
    – user7722505
    Nov 11 at 16:12










  • Yes, that is correct.
    – runcoderun
    Nov 11 at 16:16










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










If adding 2xN to the running time is acceptable, you could start by copying distinct elements into a new array:



private int getDistinct(int array) 
Set<Integer> distinct = new HashSet<>();
int endIdx = 0;

for (int element : array)

if (distinct.add(element))
array[endIdx++] = element;



return Arrays.copyOfRange(array, 0, endIdx);



and then do quickselect on that:



int arr = new int 1, 2, 2, 3, 5, 5, 8, 2, 4, 8, 8;
int kthSmallest = 6;

int distinctArray = getDistinct(arr);
int kthSmallestElement = quickselect(distinctArray, 0, distinctArray.length - 1, kthSmallest - 1);

System.out.println(kthSmallestElement);


Output:



8





share|improve this answer






















  • The running time of QuickSelect is best case and average case O(n), so adding 2N would still be O(n) I suppose?
    – user7722505
    Nov 11 at 16:12










  • Yes, that is correct.
    – runcoderun
    Nov 11 at 16:16














up vote
0
down vote



accepted










If adding 2xN to the running time is acceptable, you could start by copying distinct elements into a new array:



private int getDistinct(int array) 
Set<Integer> distinct = new HashSet<>();
int endIdx = 0;

for (int element : array)

if (distinct.add(element))
array[endIdx++] = element;



return Arrays.copyOfRange(array, 0, endIdx);



and then do quickselect on that:



int arr = new int 1, 2, 2, 3, 5, 5, 8, 2, 4, 8, 8;
int kthSmallest = 6;

int distinctArray = getDistinct(arr);
int kthSmallestElement = quickselect(distinctArray, 0, distinctArray.length - 1, kthSmallest - 1);

System.out.println(kthSmallestElement);


Output:



8





share|improve this answer






















  • The running time of QuickSelect is best case and average case O(n), so adding 2N would still be O(n) I suppose?
    – user7722505
    Nov 11 at 16:12










  • Yes, that is correct.
    – runcoderun
    Nov 11 at 16:16












up vote
0
down vote



accepted







up vote
0
down vote



accepted






If adding 2xN to the running time is acceptable, you could start by copying distinct elements into a new array:



private int getDistinct(int array) 
Set<Integer> distinct = new HashSet<>();
int endIdx = 0;

for (int element : array)

if (distinct.add(element))
array[endIdx++] = element;



return Arrays.copyOfRange(array, 0, endIdx);



and then do quickselect on that:



int arr = new int 1, 2, 2, 3, 5, 5, 8, 2, 4, 8, 8;
int kthSmallest = 6;

int distinctArray = getDistinct(arr);
int kthSmallestElement = quickselect(distinctArray, 0, distinctArray.length - 1, kthSmallest - 1);

System.out.println(kthSmallestElement);


Output:



8





share|improve this answer














If adding 2xN to the running time is acceptable, you could start by copying distinct elements into a new array:



private int getDistinct(int array) 
Set<Integer> distinct = new HashSet<>();
int endIdx = 0;

for (int element : array)

if (distinct.add(element))
array[endIdx++] = element;



return Arrays.copyOfRange(array, 0, endIdx);



and then do quickselect on that:



int arr = new int 1, 2, 2, 3, 5, 5, 8, 2, 4, 8, 8;
int kthSmallest = 6;

int distinctArray = getDistinct(arr);
int kthSmallestElement = quickselect(distinctArray, 0, distinctArray.length - 1, kthSmallest - 1);

System.out.println(kthSmallestElement);


Output:



8






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 11 at 16:14

























answered Nov 11 at 15:56









runcoderun

36417




36417











  • The running time of QuickSelect is best case and average case O(n), so adding 2N would still be O(n) I suppose?
    – user7722505
    Nov 11 at 16:12










  • Yes, that is correct.
    – runcoderun
    Nov 11 at 16:16
















  • The running time of QuickSelect is best case and average case O(n), so adding 2N would still be O(n) I suppose?
    – user7722505
    Nov 11 at 16:12










  • Yes, that is correct.
    – runcoderun
    Nov 11 at 16:16















The running time of QuickSelect is best case and average case O(n), so adding 2N would still be O(n) I suppose?
– user7722505
Nov 11 at 16:12




The running time of QuickSelect is best case and average case O(n), so adding 2N would still be O(n) I suppose?
– user7722505
Nov 11 at 16:12












Yes, that is correct.
– runcoderun
Nov 11 at 16:16




Yes, that is correct.
– runcoderun
Nov 11 at 16:16

















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