is there a reason why my merge sort does not work for array length 10How does the Java 'for each' loop work?Why does Java have transient fields?Java & Merge SortWhy is it faster to process a sorted array than an unsorted array?Why does this code using random strings print “hello world”?merge sort not working when copying the result of merge operation directly to auxiliary arrayMerge Sort : need of copying elements from temporary arrayhow to make bottom up merge sort workImplementing Merge SortMergesort algorithm implementation

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is there a reason why my merge sort does not work for array length 10


How does the Java 'for each' loop work?Why does Java have transient fields?Java & Merge SortWhy is it faster to process a sorted array than an unsorted array?Why does this code using random strings print “hello world”?merge sort not working when copying the result of merge operation directly to auxiliary arrayMerge Sort : need of copying elements from temporary arrayhow to make bottom up merge sort workImplementing Merge SortMergesort algorithm implementation






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0















I am setting up merge sort to sort my array. The goal is to sort an array with any length.



I have tried looking at the mergesort function, but i do not see anything wrong with it. the sort works for some array length be it odd or even but for array length such as length of 10, i get an over bound exception.



import java.util.Arrays;

class MergeSort

// Merge two sorted sub-arrays A[from .. mid] and A[mid + 1 .. to]
public static void merge(int[] A, int[] temp, int from, int mid, int to)

int k = from, i = from, j = mid + 1;

// loop till there are elements in the left and right runs
while (i <= mid && j <= to)
if (A[i] < A[j])
temp[k++] = A[i++];
else
temp[k++] = A[j++];



// Copy remaining elements
while (i <= mid)
temp[k++] = A[i++];


// Don't need to copy second half

// copy back to the original array to reflect sorted order
for (i = from; i <= to; i++)
A[i] = temp[i];



// Iteratively sort array A[low..high] using temporary array
public static void mergesort(int[] A)

int low = 0;
int high = A.length - 1;

// sort array A[] using temporary array temp
int[] temp = Arrays.copyOf(A, A.length);

// divide the array into blocks of size m
// m = [1, 2, 4, 8, 16...]
for (int m = 1; m <= high - low; m = 2*m)

// for m = 1, i = 0, 2, 4, 6, 8...
// for m = 2, i = 0, 4, 8, 12...
// for m = 4, i = 0, 8, 16...
// ...
for (int i = low; i < high; i += 2*m)

int from = i;
int mid = i + m - 1;
int to = Integer.min(i + 2 * m - 1, high);

merge(A, temp, from, mid, to);




// Iterative Implementation of Mergesort algorithm
public static void main(String[] args)

int[] A = 5, 7, -9, 3, -4, 2, 8, 8, 10, 11 ;

System.out.println("Original Array : " + Arrays.toString(A));
mergesort(A);
System.out.println("Modified Array : " + Arrays.toString(A));











share|improve this question






















  • It is caused by A[i++] at line 19

    – Kars
    Mar 22 at 19:27











  • Thank you for the response, I have already found that problem. how to I fix it tho.

    – chrisuye
    Mar 22 at 19:46











  • This looks like a perfect time to use the debugger

    – Joakim Danielson
    Mar 22 at 19:55

















0















I am setting up merge sort to sort my array. The goal is to sort an array with any length.



I have tried looking at the mergesort function, but i do not see anything wrong with it. the sort works for some array length be it odd or even but for array length such as length of 10, i get an over bound exception.



import java.util.Arrays;

class MergeSort

// Merge two sorted sub-arrays A[from .. mid] and A[mid + 1 .. to]
public static void merge(int[] A, int[] temp, int from, int mid, int to)

int k = from, i = from, j = mid + 1;

// loop till there are elements in the left and right runs
while (i <= mid && j <= to)
if (A[i] < A[j])
temp[k++] = A[i++];
else
temp[k++] = A[j++];



// Copy remaining elements
while (i <= mid)
temp[k++] = A[i++];


// Don't need to copy second half

// copy back to the original array to reflect sorted order
for (i = from; i <= to; i++)
A[i] = temp[i];



// Iteratively sort array A[low..high] using temporary array
public static void mergesort(int[] A)

int low = 0;
int high = A.length - 1;

// sort array A[] using temporary array temp
int[] temp = Arrays.copyOf(A, A.length);

// divide the array into blocks of size m
// m = [1, 2, 4, 8, 16...]
for (int m = 1; m <= high - low; m = 2*m)

// for m = 1, i = 0, 2, 4, 6, 8...
// for m = 2, i = 0, 4, 8, 12...
// for m = 4, i = 0, 8, 16...
// ...
for (int i = low; i < high; i += 2*m)

int from = i;
int mid = i + m - 1;
int to = Integer.min(i + 2 * m - 1, high);

merge(A, temp, from, mid, to);




// Iterative Implementation of Mergesort algorithm
public static void main(String[] args)

int[] A = 5, 7, -9, 3, -4, 2, 8, 8, 10, 11 ;

System.out.println("Original Array : " + Arrays.toString(A));
mergesort(A);
System.out.println("Modified Array : " + Arrays.toString(A));











share|improve this question






















  • It is caused by A[i++] at line 19

    – Kars
    Mar 22 at 19:27











  • Thank you for the response, I have already found that problem. how to I fix it tho.

    – chrisuye
    Mar 22 at 19:46











  • This looks like a perfect time to use the debugger

    – Joakim Danielson
    Mar 22 at 19:55













0












0








0


0






I am setting up merge sort to sort my array. The goal is to sort an array with any length.



I have tried looking at the mergesort function, but i do not see anything wrong with it. the sort works for some array length be it odd or even but for array length such as length of 10, i get an over bound exception.



import java.util.Arrays;

class MergeSort

// Merge two sorted sub-arrays A[from .. mid] and A[mid + 1 .. to]
public static void merge(int[] A, int[] temp, int from, int mid, int to)

int k = from, i = from, j = mid + 1;

// loop till there are elements in the left and right runs
while (i <= mid && j <= to)
if (A[i] < A[j])
temp[k++] = A[i++];
else
temp[k++] = A[j++];



// Copy remaining elements
while (i <= mid)
temp[k++] = A[i++];


// Don't need to copy second half

// copy back to the original array to reflect sorted order
for (i = from; i <= to; i++)
A[i] = temp[i];



// Iteratively sort array A[low..high] using temporary array
public static void mergesort(int[] A)

int low = 0;
int high = A.length - 1;

// sort array A[] using temporary array temp
int[] temp = Arrays.copyOf(A, A.length);

// divide the array into blocks of size m
// m = [1, 2, 4, 8, 16...]
for (int m = 1; m <= high - low; m = 2*m)

// for m = 1, i = 0, 2, 4, 6, 8...
// for m = 2, i = 0, 4, 8, 12...
// for m = 4, i = 0, 8, 16...
// ...
for (int i = low; i < high; i += 2*m)

int from = i;
int mid = i + m - 1;
int to = Integer.min(i + 2 * m - 1, high);

merge(A, temp, from, mid, to);




// Iterative Implementation of Mergesort algorithm
public static void main(String[] args)

int[] A = 5, 7, -9, 3, -4, 2, 8, 8, 10, 11 ;

System.out.println("Original Array : " + Arrays.toString(A));
mergesort(A);
System.out.println("Modified Array : " + Arrays.toString(A));











share|improve this question














I am setting up merge sort to sort my array. The goal is to sort an array with any length.



I have tried looking at the mergesort function, but i do not see anything wrong with it. the sort works for some array length be it odd or even but for array length such as length of 10, i get an over bound exception.



import java.util.Arrays;

class MergeSort

// Merge two sorted sub-arrays A[from .. mid] and A[mid + 1 .. to]
public static void merge(int[] A, int[] temp, int from, int mid, int to)

int k = from, i = from, j = mid + 1;

// loop till there are elements in the left and right runs
while (i <= mid && j <= to)
if (A[i] < A[j])
temp[k++] = A[i++];
else
temp[k++] = A[j++];



// Copy remaining elements
while (i <= mid)
temp[k++] = A[i++];


// Don't need to copy second half

// copy back to the original array to reflect sorted order
for (i = from; i <= to; i++)
A[i] = temp[i];



// Iteratively sort array A[low..high] using temporary array
public static void mergesort(int[] A)

int low = 0;
int high = A.length - 1;

// sort array A[] using temporary array temp
int[] temp = Arrays.copyOf(A, A.length);

// divide the array into blocks of size m
// m = [1, 2, 4, 8, 16...]
for (int m = 1; m <= high - low; m = 2*m)

// for m = 1, i = 0, 2, 4, 6, 8...
// for m = 2, i = 0, 4, 8, 12...
// for m = 4, i = 0, 8, 16...
// ...
for (int i = low; i < high; i += 2*m)

int from = i;
int mid = i + m - 1;
int to = Integer.min(i + 2 * m - 1, high);

merge(A, temp, from, mid, to);




// Iterative Implementation of Mergesort algorithm
public static void main(String[] args)

int[] A = 5, 7, -9, 3, -4, 2, 8, 8, 10, 11 ;

System.out.println("Original Array : " + Arrays.toString(A));
mergesort(A);
System.out.println("Modified Array : " + Arrays.toString(A));








java mergesort






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Mar 22 at 19:18









chrisuyechrisuye

11




11












  • It is caused by A[i++] at line 19

    – Kars
    Mar 22 at 19:27











  • Thank you for the response, I have already found that problem. how to I fix it tho.

    – chrisuye
    Mar 22 at 19:46











  • This looks like a perfect time to use the debugger

    – Joakim Danielson
    Mar 22 at 19:55

















  • It is caused by A[i++] at line 19

    – Kars
    Mar 22 at 19:27











  • Thank you for the response, I have already found that problem. how to I fix it tho.

    – chrisuye
    Mar 22 at 19:46











  • This looks like a perfect time to use the debugger

    – Joakim Danielson
    Mar 22 at 19:55
















It is caused by A[i++] at line 19

– Kars
Mar 22 at 19:27





It is caused by A[i++] at line 19

– Kars
Mar 22 at 19:27













Thank you for the response, I have already found that problem. how to I fix it tho.

– chrisuye
Mar 22 at 19:46





Thank you for the response, I have already found that problem. how to I fix it tho.

– chrisuye
Mar 22 at 19:46













This looks like a perfect time to use the debugger

– Joakim Danielson
Mar 22 at 19:55





This looks like a perfect time to use the debugger

– Joakim Danielson
Mar 22 at 19:55












2 Answers
2






active

oldest

votes


















1














Your mid calculation is incorrect. You are sometimes setting it outside the range of the area.



This below change fixes the algorithm by preventing mid from going out of bounds, similar to how you did with preventing to from going out of bounds.



Change int mid = i + m - 1; to int mid = Math.min(i + m - 1, A.length - 1);



Explanation: As mentioned in your comment, the slices of area you are examining are increasing in size. So here is how your array is sorted, and when the out of bounds error occurs, and why it did not occur on sizes that are powers of 2:



 [ -9, 5, 7, 3, -4, 2, 8, 8, 10, 11 ] Array size
First pass: [] [] [] [] [] [] [] [] [] [] 1
Second: [ ] [ ] [ ] [ ] [ ] 2
Third: [ ] [ ] [ ERROR] 4





share|improve this answer

























  • Thank you, It fixed the problem. if you have any additional input I will gladly accept. and I will work towards finding any other problem I find.

    – chrisuye
    Mar 22 at 20:11











  • Updated my answer. I believe the approach is actually the best way to solve the problem and cap the number. The alternatives involve rewriting more code and in some cases slower anyways.

    – Daryll Chandler
    Mar 22 at 20:14











  • This approach does not reduce the number of calls made to the merge function, for a 10 number array the calls made are 11. Where as if you add the if conditional statement the number of calls gets reduces to 9. Check below solution for the same. Please do correct me if I am wrong

    – VSB
    Mar 22 at 20:20











  • You're correct, this implementation of merge sort could be optimized. A break or conditional would stop it from unnecessarily re-sorting that section.

    – Daryll Chandler
    Mar 22 at 20:37



















0














Here is your fix:



Added an if statement for your mid not to cross the array bound



if(mid<high) 
merge(A, temp, from, mid, to);



Complete code:



// Merge two sorted sub-arrays A[from .. mid] and A[mid + 1 .. to]
public static void merge(int[] A, int[] temp, int from, int mid, int to)

int k = from, i = from, j = mid + 1;

// loop till there are elements in the left and right runs
while (i <= mid && j <= to)
if (A[i] < A[j])
temp[k++] = A[i++];
else
temp[k++] = A[j++];



// Copy remaining elements
while (i <= mid)
temp[k++] = A[i++];


// Don't need to copy second half

// copy back to the original array to reflect sorted order
for (i = from; i <= to; i++)
A[i] = temp[i];



// Iteratively sort array A[low..high] using temporary array
public static void mergesort(int[] A)

int low = 0;
int high = A.length - 1;

// sort array A[] using temporary array temp
int[] temp = Arrays.copyOf(A, A.length);
//System.out.println("temp Array : " + Arrays.toString(temp));

// divide the array into blocks of size m
// m = [1, 2, 4, 8, 16...]
for (int m = 1; m <= high - low; m = 2*m)

// for m = 1, i = 0, 2, 4, 6, 8...
// for m = 2, i = 0, 4, 8, 12...
// for m = 4, i = 0, 8, 16...
// ...
for (int i = low; i < high; i += 2*m)

int from = i;
int mid = i + m - 1;
int to = Integer.min(i + 2 * m - 1, high);

if(mid<high)
merge(A, temp, from, mid, to);






// Iterative Implementation of Mergesort algorithm
public static void main(String[] args)

int[] A = 5, 7, -9, 3, -4, 2, 8, 8, 10, 11 ;

System.out.println("Original Array : " + Arrays.toString(A));
mergesort(A);
System.out.println("Modified Array : " + Arrays.toString(A));






share|improve this answer























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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    Your mid calculation is incorrect. You are sometimes setting it outside the range of the area.



    This below change fixes the algorithm by preventing mid from going out of bounds, similar to how you did with preventing to from going out of bounds.



    Change int mid = i + m - 1; to int mid = Math.min(i + m - 1, A.length - 1);



    Explanation: As mentioned in your comment, the slices of area you are examining are increasing in size. So here is how your array is sorted, and when the out of bounds error occurs, and why it did not occur on sizes that are powers of 2:



     [ -9, 5, 7, 3, -4, 2, 8, 8, 10, 11 ] Array size
    First pass: [] [] [] [] [] [] [] [] [] [] 1
    Second: [ ] [ ] [ ] [ ] [ ] 2
    Third: [ ] [ ] [ ERROR] 4





    share|improve this answer

























    • Thank you, It fixed the problem. if you have any additional input I will gladly accept. and I will work towards finding any other problem I find.

      – chrisuye
      Mar 22 at 20:11











    • Updated my answer. I believe the approach is actually the best way to solve the problem and cap the number. The alternatives involve rewriting more code and in some cases slower anyways.

      – Daryll Chandler
      Mar 22 at 20:14











    • This approach does not reduce the number of calls made to the merge function, for a 10 number array the calls made are 11. Where as if you add the if conditional statement the number of calls gets reduces to 9. Check below solution for the same. Please do correct me if I am wrong

      – VSB
      Mar 22 at 20:20











    • You're correct, this implementation of merge sort could be optimized. A break or conditional would stop it from unnecessarily re-sorting that section.

      – Daryll Chandler
      Mar 22 at 20:37
















    1














    Your mid calculation is incorrect. You are sometimes setting it outside the range of the area.



    This below change fixes the algorithm by preventing mid from going out of bounds, similar to how you did with preventing to from going out of bounds.



    Change int mid = i + m - 1; to int mid = Math.min(i + m - 1, A.length - 1);



    Explanation: As mentioned in your comment, the slices of area you are examining are increasing in size. So here is how your array is sorted, and when the out of bounds error occurs, and why it did not occur on sizes that are powers of 2:



     [ -9, 5, 7, 3, -4, 2, 8, 8, 10, 11 ] Array size
    First pass: [] [] [] [] [] [] [] [] [] [] 1
    Second: [ ] [ ] [ ] [ ] [ ] 2
    Third: [ ] [ ] [ ERROR] 4





    share|improve this answer

























    • Thank you, It fixed the problem. if you have any additional input I will gladly accept. and I will work towards finding any other problem I find.

      – chrisuye
      Mar 22 at 20:11











    • Updated my answer. I believe the approach is actually the best way to solve the problem and cap the number. The alternatives involve rewriting more code and in some cases slower anyways.

      – Daryll Chandler
      Mar 22 at 20:14











    • This approach does not reduce the number of calls made to the merge function, for a 10 number array the calls made are 11. Where as if you add the if conditional statement the number of calls gets reduces to 9. Check below solution for the same. Please do correct me if I am wrong

      – VSB
      Mar 22 at 20:20











    • You're correct, this implementation of merge sort could be optimized. A break or conditional would stop it from unnecessarily re-sorting that section.

      – Daryll Chandler
      Mar 22 at 20:37














    1












    1








    1







    Your mid calculation is incorrect. You are sometimes setting it outside the range of the area.



    This below change fixes the algorithm by preventing mid from going out of bounds, similar to how you did with preventing to from going out of bounds.



    Change int mid = i + m - 1; to int mid = Math.min(i + m - 1, A.length - 1);



    Explanation: As mentioned in your comment, the slices of area you are examining are increasing in size. So here is how your array is sorted, and when the out of bounds error occurs, and why it did not occur on sizes that are powers of 2:



     [ -9, 5, 7, 3, -4, 2, 8, 8, 10, 11 ] Array size
    First pass: [] [] [] [] [] [] [] [] [] [] 1
    Second: [ ] [ ] [ ] [ ] [ ] 2
    Third: [ ] [ ] [ ERROR] 4





    share|improve this answer















    Your mid calculation is incorrect. You are sometimes setting it outside the range of the area.



    This below change fixes the algorithm by preventing mid from going out of bounds, similar to how you did with preventing to from going out of bounds.



    Change int mid = i + m - 1; to int mid = Math.min(i + m - 1, A.length - 1);



    Explanation: As mentioned in your comment, the slices of area you are examining are increasing in size. So here is how your array is sorted, and when the out of bounds error occurs, and why it did not occur on sizes that are powers of 2:



     [ -9, 5, 7, 3, -4, 2, 8, 8, 10, 11 ] Array size
    First pass: [] [] [] [] [] [] [] [] [] [] 1
    Second: [ ] [ ] [ ] [ ] [ ] 2
    Third: [ ] [ ] [ ERROR] 4






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Mar 22 at 20:11

























    answered Mar 22 at 19:35









    Daryll ChandlerDaryll Chandler

    1487




    1487












    • Thank you, It fixed the problem. if you have any additional input I will gladly accept. and I will work towards finding any other problem I find.

      – chrisuye
      Mar 22 at 20:11











    • Updated my answer. I believe the approach is actually the best way to solve the problem and cap the number. The alternatives involve rewriting more code and in some cases slower anyways.

      – Daryll Chandler
      Mar 22 at 20:14











    • This approach does not reduce the number of calls made to the merge function, for a 10 number array the calls made are 11. Where as if you add the if conditional statement the number of calls gets reduces to 9. Check below solution for the same. Please do correct me if I am wrong

      – VSB
      Mar 22 at 20:20











    • You're correct, this implementation of merge sort could be optimized. A break or conditional would stop it from unnecessarily re-sorting that section.

      – Daryll Chandler
      Mar 22 at 20:37


















    • Thank you, It fixed the problem. if you have any additional input I will gladly accept. and I will work towards finding any other problem I find.

      – chrisuye
      Mar 22 at 20:11











    • Updated my answer. I believe the approach is actually the best way to solve the problem and cap the number. The alternatives involve rewriting more code and in some cases slower anyways.

      – Daryll Chandler
      Mar 22 at 20:14











    • This approach does not reduce the number of calls made to the merge function, for a 10 number array the calls made are 11. Where as if you add the if conditional statement the number of calls gets reduces to 9. Check below solution for the same. Please do correct me if I am wrong

      – VSB
      Mar 22 at 20:20











    • You're correct, this implementation of merge sort could be optimized. A break or conditional would stop it from unnecessarily re-sorting that section.

      – Daryll Chandler
      Mar 22 at 20:37

















    Thank you, It fixed the problem. if you have any additional input I will gladly accept. and I will work towards finding any other problem I find.

    – chrisuye
    Mar 22 at 20:11





    Thank you, It fixed the problem. if you have any additional input I will gladly accept. and I will work towards finding any other problem I find.

    – chrisuye
    Mar 22 at 20:11













    Updated my answer. I believe the approach is actually the best way to solve the problem and cap the number. The alternatives involve rewriting more code and in some cases slower anyways.

    – Daryll Chandler
    Mar 22 at 20:14





    Updated my answer. I believe the approach is actually the best way to solve the problem and cap the number. The alternatives involve rewriting more code and in some cases slower anyways.

    – Daryll Chandler
    Mar 22 at 20:14













    This approach does not reduce the number of calls made to the merge function, for a 10 number array the calls made are 11. Where as if you add the if conditional statement the number of calls gets reduces to 9. Check below solution for the same. Please do correct me if I am wrong

    – VSB
    Mar 22 at 20:20





    This approach does not reduce the number of calls made to the merge function, for a 10 number array the calls made are 11. Where as if you add the if conditional statement the number of calls gets reduces to 9. Check below solution for the same. Please do correct me if I am wrong

    – VSB
    Mar 22 at 20:20













    You're correct, this implementation of merge sort could be optimized. A break or conditional would stop it from unnecessarily re-sorting that section.

    – Daryll Chandler
    Mar 22 at 20:37






    You're correct, this implementation of merge sort could be optimized. A break or conditional would stop it from unnecessarily re-sorting that section.

    – Daryll Chandler
    Mar 22 at 20:37














    0














    Here is your fix:



    Added an if statement for your mid not to cross the array bound



    if(mid<high) 
    merge(A, temp, from, mid, to);



    Complete code:



    // Merge two sorted sub-arrays A[from .. mid] and A[mid + 1 .. to]
    public static void merge(int[] A, int[] temp, int from, int mid, int to)

    int k = from, i = from, j = mid + 1;

    // loop till there are elements in the left and right runs
    while (i <= mid && j <= to)
    if (A[i] < A[j])
    temp[k++] = A[i++];
    else
    temp[k++] = A[j++];



    // Copy remaining elements
    while (i <= mid)
    temp[k++] = A[i++];


    // Don't need to copy second half

    // copy back to the original array to reflect sorted order
    for (i = from; i <= to; i++)
    A[i] = temp[i];



    // Iteratively sort array A[low..high] using temporary array
    public static void mergesort(int[] A)

    int low = 0;
    int high = A.length - 1;

    // sort array A[] using temporary array temp
    int[] temp = Arrays.copyOf(A, A.length);
    //System.out.println("temp Array : " + Arrays.toString(temp));

    // divide the array into blocks of size m
    // m = [1, 2, 4, 8, 16...]
    for (int m = 1; m <= high - low; m = 2*m)

    // for m = 1, i = 0, 2, 4, 6, 8...
    // for m = 2, i = 0, 4, 8, 12...
    // for m = 4, i = 0, 8, 16...
    // ...
    for (int i = low; i < high; i += 2*m)

    int from = i;
    int mid = i + m - 1;
    int to = Integer.min(i + 2 * m - 1, high);

    if(mid<high)
    merge(A, temp, from, mid, to);






    // Iterative Implementation of Mergesort algorithm
    public static void main(String[] args)

    int[] A = 5, 7, -9, 3, -4, 2, 8, 8, 10, 11 ;

    System.out.println("Original Array : " + Arrays.toString(A));
    mergesort(A);
    System.out.println("Modified Array : " + Arrays.toString(A));






    share|improve this answer



























      0














      Here is your fix:



      Added an if statement for your mid not to cross the array bound



      if(mid<high) 
      merge(A, temp, from, mid, to);



      Complete code:



      // Merge two sorted sub-arrays A[from .. mid] and A[mid + 1 .. to]
      public static void merge(int[] A, int[] temp, int from, int mid, int to)

      int k = from, i = from, j = mid + 1;

      // loop till there are elements in the left and right runs
      while (i <= mid && j <= to)
      if (A[i] < A[j])
      temp[k++] = A[i++];
      else
      temp[k++] = A[j++];



      // Copy remaining elements
      while (i <= mid)
      temp[k++] = A[i++];


      // Don't need to copy second half

      // copy back to the original array to reflect sorted order
      for (i = from; i <= to; i++)
      A[i] = temp[i];



      // Iteratively sort array A[low..high] using temporary array
      public static void mergesort(int[] A)

      int low = 0;
      int high = A.length - 1;

      // sort array A[] using temporary array temp
      int[] temp = Arrays.copyOf(A, A.length);
      //System.out.println("temp Array : " + Arrays.toString(temp));

      // divide the array into blocks of size m
      // m = [1, 2, 4, 8, 16...]
      for (int m = 1; m <= high - low; m = 2*m)

      // for m = 1, i = 0, 2, 4, 6, 8...
      // for m = 2, i = 0, 4, 8, 12...
      // for m = 4, i = 0, 8, 16...
      // ...
      for (int i = low; i < high; i += 2*m)

      int from = i;
      int mid = i + m - 1;
      int to = Integer.min(i + 2 * m - 1, high);

      if(mid<high)
      merge(A, temp, from, mid, to);






      // Iterative Implementation of Mergesort algorithm
      public static void main(String[] args)

      int[] A = 5, 7, -9, 3, -4, 2, 8, 8, 10, 11 ;

      System.out.println("Original Array : " + Arrays.toString(A));
      mergesort(A);
      System.out.println("Modified Array : " + Arrays.toString(A));






      share|improve this answer

























        0












        0








        0







        Here is your fix:



        Added an if statement for your mid not to cross the array bound



        if(mid<high) 
        merge(A, temp, from, mid, to);



        Complete code:



        // Merge two sorted sub-arrays A[from .. mid] and A[mid + 1 .. to]
        public static void merge(int[] A, int[] temp, int from, int mid, int to)

        int k = from, i = from, j = mid + 1;

        // loop till there are elements in the left and right runs
        while (i <= mid && j <= to)
        if (A[i] < A[j])
        temp[k++] = A[i++];
        else
        temp[k++] = A[j++];



        // Copy remaining elements
        while (i <= mid)
        temp[k++] = A[i++];


        // Don't need to copy second half

        // copy back to the original array to reflect sorted order
        for (i = from; i <= to; i++)
        A[i] = temp[i];



        // Iteratively sort array A[low..high] using temporary array
        public static void mergesort(int[] A)

        int low = 0;
        int high = A.length - 1;

        // sort array A[] using temporary array temp
        int[] temp = Arrays.copyOf(A, A.length);
        //System.out.println("temp Array : " + Arrays.toString(temp));

        // divide the array into blocks of size m
        // m = [1, 2, 4, 8, 16...]
        for (int m = 1; m <= high - low; m = 2*m)

        // for m = 1, i = 0, 2, 4, 6, 8...
        // for m = 2, i = 0, 4, 8, 12...
        // for m = 4, i = 0, 8, 16...
        // ...
        for (int i = low; i < high; i += 2*m)

        int from = i;
        int mid = i + m - 1;
        int to = Integer.min(i + 2 * m - 1, high);

        if(mid<high)
        merge(A, temp, from, mid, to);






        // Iterative Implementation of Mergesort algorithm
        public static void main(String[] args)

        int[] A = 5, 7, -9, 3, -4, 2, 8, 8, 10, 11 ;

        System.out.println("Original Array : " + Arrays.toString(A));
        mergesort(A);
        System.out.println("Modified Array : " + Arrays.toString(A));






        share|improve this answer













        Here is your fix:



        Added an if statement for your mid not to cross the array bound



        if(mid<high) 
        merge(A, temp, from, mid, to);



        Complete code:



        // Merge two sorted sub-arrays A[from .. mid] and A[mid + 1 .. to]
        public static void merge(int[] A, int[] temp, int from, int mid, int to)

        int k = from, i = from, j = mid + 1;

        // loop till there are elements in the left and right runs
        while (i <= mid && j <= to)
        if (A[i] < A[j])
        temp[k++] = A[i++];
        else
        temp[k++] = A[j++];



        // Copy remaining elements
        while (i <= mid)
        temp[k++] = A[i++];


        // Don't need to copy second half

        // copy back to the original array to reflect sorted order
        for (i = from; i <= to; i++)
        A[i] = temp[i];



        // Iteratively sort array A[low..high] using temporary array
        public static void mergesort(int[] A)

        int low = 0;
        int high = A.length - 1;

        // sort array A[] using temporary array temp
        int[] temp = Arrays.copyOf(A, A.length);
        //System.out.println("temp Array : " + Arrays.toString(temp));

        // divide the array into blocks of size m
        // m = [1, 2, 4, 8, 16...]
        for (int m = 1; m <= high - low; m = 2*m)

        // for m = 1, i = 0, 2, 4, 6, 8...
        // for m = 2, i = 0, 4, 8, 12...
        // for m = 4, i = 0, 8, 16...
        // ...
        for (int i = low; i < high; i += 2*m)

        int from = i;
        int mid = i + m - 1;
        int to = Integer.min(i + 2 * m - 1, high);

        if(mid<high)
        merge(A, temp, from, mid, to);






        // Iterative Implementation of Mergesort algorithm
        public static void main(String[] args)

        int[] A = 5, 7, -9, 3, -4, 2, 8, 8, 10, 11 ;

        System.out.println("Original Array : " + Arrays.toString(A));
        mergesort(A);
        System.out.println("Modified Array : " + Arrays.toString(A));







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Mar 22 at 20:12









        VSBVSB

        25229




        25229



























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