Integer divisions of a number

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How do I split a number into equal parts or as close to equal as possible. See example below:

If I have a value of 61 that i want to distribute between two groups, it would be 30.5 and 30.5. Doubles (decimals) are no good, so in this regard the closest split is 30 and 31.

Similarly 42 / 5 = 8.4, however I need the system to return (8, 8, 8, 9, 9) which is the closest split with whole numbers.

Solved it guys:

 if(sum % numberOfTeams != 0) 
 al.add(0, sNOT);
 for(int i = 0; i < numberOfTeams - 1; i++) 
 int remover = sum - sNOT;
 if(remover % (sNOT + 1) == 0) 
 al.add(i+1, sNOT + 1);
 else 
 al.add(i + 1, sNOT);

}

edited Aug 22 at 18:59

asked Aug 19 at 10:41

M.Shiz

224

5

Not taking away from an interesting problem, but curious as to why a question with no effort gets several upvotes?
â€“Â achAmhÃ¡in
Aug 19 at 11:13

3

I agree entirely with @Dukeling.. I had a little squabble about this on the meta site yesterday. Interesting questions are more likely to receive upvotes (regardless of shown effort) than uninteresting ones.
â€“Â Taslim
Aug 19 at 11:58

@achAmhÃ¡in That is typical of the algorithm tag.
â€“Â m69
Aug 19 at 13:20

add a commentÂ |Â

up vote
5
down vote

favorite

How do I split a number into equal parts or as close to equal as possible. See example below:

If I have a value of 61 that i want to distribute between two groups, it would be 30.5 and 30.5. Doubles (decimals) are no good, so in this regard the closest split is 30 and 31.

Similarly 42 / 5 = 8.4, however I need the system to return (8, 8, 8, 9, 9) which is the closest split with whole numbers.

Solved it guys:

 if(sum % numberOfTeams != 0) 
 al.add(0, sNOT);
 for(int i = 0; i < numberOfTeams - 1; i++) 
 int remover = sum - sNOT;
 if(remover % (sNOT + 1) == 0) 
 al.add(i+1, sNOT + 1);
 else 
 al.add(i + 1, sNOT);

}

edited Aug 22 at 18:59

asked Aug 19 at 10:41

M.Shiz

224

5

Not taking away from an interesting problem, but curious as to why a question with no effort gets several upvotes?
â€“Â achAmhÃ¡in
Aug 19 at 11:13

3

I agree entirely with @Dukeling.. I had a little squabble about this on the meta site yesterday. Interesting questions are more likely to receive upvotes (regardless of shown effort) than uninteresting ones.
â€“Â Taslim
Aug 19 at 11:58

@achAmhÃ¡in That is typical of the algorithm tag.
â€“Â m69
Aug 19 at 13:20

add a commentÂ |Â

up vote
5
down vote

favorite

How do I split a number into equal parts or as close to equal as possible. See example below:

If I have a value of 61 that i want to distribute between two groups, it would be 30.5 and 30.5. Doubles (decimals) are no good, so in this regard the closest split is 30 and 31.

Similarly 42 / 5 = 8.4, however I need the system to return (8, 8, 8, 9, 9) which is the closest split with whole numbers.

Solved it guys:

 if(sum % numberOfTeams != 0) 
 al.add(0, sNOT);
 for(int i = 0; i < numberOfTeams - 1; i++) 
 int remover = sum - sNOT;
 if(remover % (sNOT + 1) == 0) 
 al.add(i+1, sNOT + 1);
 else 
 al.add(i + 1, sNOT);

}

edited Aug 22 at 18:59

asked Aug 19 at 10:41

M.Shiz

224

How do I split a number into equal parts or as close to equal as possible. See example below:

If I have a value of 61 that i want to distribute between two groups, it would be 30.5 and 30.5. Doubles (decimals) are no good, so in this regard the closest split is 30 and 31.

Similarly 42 / 5 = 8.4, however I need the system to return (8, 8, 8, 9, 9) which is the closest split with whole numbers.

Solved it guys:

 if(sum % numberOfTeams != 0) 
 al.add(0, sNOT);
 for(int i = 0; i < numberOfTeams - 1; i++) 
 int remover = sum - sNOT;
 if(remover % (sNOT + 1) == 0) 
 al.add(i+1, sNOT + 1);
 else 
 al.add(i + 1, sNOT);

}

edited Aug 22 at 18:59

asked Aug 19 at 10:41

M.Shiz

224

edited Aug 22 at 18:59

asked Aug 19 at 10:41

M.Shiz

224

asked Aug 19 at 10:41

M.Shiz

224

asked Aug 19 at 10:41

M.Shiz

224

5

Not taking away from an interesting problem, but curious as to why a question with no effort gets several upvotes?
â€“Â achAmhÃ¡in
Aug 19 at 11:13

3

I agree entirely with @Dukeling.. I had a little squabble about this on the meta site yesterday. Interesting questions are more likely to receive upvotes (regardless of shown effort) than uninteresting ones.
â€“Â Taslim
Aug 19 at 11:58

@achAmhÃ¡in That is typical of the algorithm tag.
â€“Â m69
Aug 19 at 13:20

add a commentÂ |Â

5

Not taking away from an interesting problem, but curious as to why a question with no effort gets several upvotes?
â€“Â achAmhÃ¡in
Aug 19 at 11:13

3

I agree entirely with @Dukeling.. I had a little squabble about this on the meta site yesterday. Interesting questions are more likely to receive upvotes (regardless of shown effort) than uninteresting ones.
â€“Â Taslim
Aug 19 at 11:58

@achAmhÃ¡in That is typical of the algorithm tag.
â€“Â m69
Aug 19 at 13:20

Not taking away from an interesting problem, but curious as to why a question with no effort gets several upvotes?
â€“Â achAmhÃ¡in
Aug 19 at 11:13

I agree entirely with @Dukeling.. I had a little squabble about this on the meta site yesterday. Interesting questions are more likely to receive upvotes (regardless of shown effort) than uninteresting ones.
â€“Â Taslim
Aug 19 at 11:58

@achAmhÃ¡in That is typical of the algorithm tag.
â€“Â m69
Aug 19 at 13:20

add a commentÂ |Â

4 Answers
4

active

oldest

votes

up vote
3
down vote

Here is a solution which entirely uses Arrays, a little shorter with no loops. Validity check should also be applied as suggested above

 int value = 42;
 int groups = 5;

 int residue = value % groups;
 int res = new int[groups];
 int division = value / groups;
 Arrays.fill(res, 0, residue, division +1);
 Arrays.fill(res, residue, res.length, division);

 System.out.println(Arrays.toString(res));

answered Aug 19 at 13:05

Anton.P

1,47031123

add a commentÂ |Â

up vote
2
down vote

Here's one solution:

public static int closestSplit(int intToSplit, int noOfGroups) 

 int result = new int[noOfGroups];
 Arrays.fill(result, intToSplit / noOfGroups);
 for (int i = 0; i < intToSplit % noOfGroups; i++) 
 result[i]++;
 
 return result;




// usage:
System.out.println(Arrays.toString(closestSplit(42, 5)));

Basically, it creates an array of length noOfGroups first, then fills that with the integer division of intToSplit and noOfGroups. Next, it adds one to the first intToSplit mod noOfGroups elements of the array.

If you require the result to be ordered in ascending order, you can loop from the end of the array or use Arrays.sort.

edited Aug 19 at 11:18

zlakad

9171614

answered Aug 19 at 10:50

Sweeper

54.6k960117

add a commentÂ |Â

up vote
1
down vote

The major thing you need to consider while solving this question is remainders.

From the example above, we intend to split 61 into 2, 61 / 2 gives a remainder of 1. Since we're splitting into two groups, only 1 out of these 2 groups have to be ((int) 61 / 2) + 1. The remaining (one) group can be ((int) 61 / 2).

Also, consider splitting 42 into 5, 42 / 5 gives a remainder of 2. Since, we're splitting into 5 groups, only 2 out of these 5 groups have to be ((int) 42 / 5) + 1. The remaining (three) groups can be ((int) 42 / 5).

Here's the code:

public int spitNumbers(int number, int groups)

 int result = new int[groups]; //The array we'd return at the end

 int shift = number % groups; //I used this to check the remainder.

 for(int i = 0; i < result.length; i++)

 /* The if-statement below checks for two things:
 - if the remainder is zero
 - if we've reached the shifting stage.
 */
 if((shift == 0) 
 return result;

I hope this helps.. Merry coding!

edited Aug 19 at 11:52

answered Aug 19 at 11:33

Taslim

1,02921024

add a commentÂ |Â

up vote
0
down vote

Please have a look at this solution, I little tested, it works.

public static int number(int number, int divider)

p.s. little simplfied

edited Aug 19 at 11:42

answered Aug 19 at 11:34

ImZ

266

add a commentÂ |Â

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

up vote
3
down vote

Here is a solution which entirely uses Arrays, a little shorter with no loops. Validity check should also be applied as suggested above

 int value = 42;
 int groups = 5;

 int residue = value % groups;
 int res = new int[groups];
 int division = value / groups;
 Arrays.fill(res, 0, residue, division +1);
 Arrays.fill(res, residue, res.length, division);

 System.out.println(Arrays.toString(res));

answered Aug 19 at 13:05

Anton.P

1,47031123

add a commentÂ |Â

up vote
3
down vote

Here is a solution which entirely uses Arrays, a little shorter with no loops. Validity check should also be applied as suggested above

 int value = 42;
 int groups = 5;

 int residue = value % groups;
 int res = new int[groups];
 int division = value / groups;
 Arrays.fill(res, 0, residue, division +1);
 Arrays.fill(res, residue, res.length, division);

 System.out.println(Arrays.toString(res));

answered Aug 19 at 13:05

Anton.P

1,47031123

add a commentÂ |Â

up vote
3
down vote

Here is a solution which entirely uses Arrays, a little shorter with no loops. Validity check should also be applied as suggested above

 int value = 42;
 int groups = 5;

 int residue = value % groups;
 int res = new int[groups];
 int division = value / groups;
 Arrays.fill(res, 0, residue, division +1);
 Arrays.fill(res, residue, res.length, division);

 System.out.println(Arrays.toString(res));

answered Aug 19 at 13:05

Anton.P

1,47031123

Here is a solution which entirely uses Arrays, a little shorter with no loops. Validity check should also be applied as suggested above

 int value = 42;
 int groups = 5;

 int residue = value % groups;
 int res = new int[groups];
 int division = value / groups;
 Arrays.fill(res, 0, residue, division +1);
 Arrays.fill(res, residue, res.length, division);

 System.out.println(Arrays.toString(res));

answered Aug 19 at 13:05

Anton.P

1,47031123

answered Aug 19 at 13:05

Anton.P

1,47031123

answered Aug 19 at 13:05

Anton.P

1,47031123

answered Aug 19 at 13:05

Anton.P

1,47031123

add a commentÂ |Â

up vote
2
down vote

Here's one solution:

public static int closestSplit(int intToSplit, int noOfGroups) 

 int result = new int[noOfGroups];
 Arrays.fill(result, intToSplit / noOfGroups);
 for (int i = 0; i < intToSplit % noOfGroups; i++) 
 result[i]++;
 
 return result;




// usage:
System.out.println(Arrays.toString(closestSplit(42, 5)));

If you require the result to be ordered in ascending order, you can loop from the end of the array or use Arrays.sort.

edited Aug 19 at 11:18

zlakad

9171614

answered Aug 19 at 10:50

Sweeper

54.6k960117

add a commentÂ |Â

up vote
2
down vote

Here's one solution:

public static int closestSplit(int intToSplit, int noOfGroups) 

 int result = new int[noOfGroups];
 Arrays.fill(result, intToSplit / noOfGroups);
 for (int i = 0; i < intToSplit % noOfGroups; i++) 
 result[i]++;
 
 return result;




// usage:
System.out.println(Arrays.toString(closestSplit(42, 5)));

If you require the result to be ordered in ascending order, you can loop from the end of the array or use Arrays.sort.

edited Aug 19 at 11:18

zlakad

9171614

answered Aug 19 at 10:50

Sweeper

54.6k960117

add a commentÂ |Â

up vote
2
down vote

Here's one solution:

public static int closestSplit(int intToSplit, int noOfGroups) 

 int result = new int[noOfGroups];
 Arrays.fill(result, intToSplit / noOfGroups);
 for (int i = 0; i < intToSplit % noOfGroups; i++) 
 result[i]++;
 
 return result;




// usage:
System.out.println(Arrays.toString(closestSplit(42, 5)));

If you require the result to be ordered in ascending order, you can loop from the end of the array or use Arrays.sort.

edited Aug 19 at 11:18

zlakad

9171614

answered Aug 19 at 10:50

Sweeper

54.6k960117

Here's one solution:

public static int closestSplit(int intToSplit, int noOfGroups) 

 int result = new int[noOfGroups];
 Arrays.fill(result, intToSplit / noOfGroups);
 for (int i = 0; i < intToSplit % noOfGroups; i++) 
 result[i]++;
 
 return result;




// usage:
System.out.println(Arrays.toString(closestSplit(42, 5)));

If you require the result to be ordered in ascending order, you can loop from the end of the array or use Arrays.sort.

edited Aug 19 at 11:18

zlakad

9171614

answered Aug 19 at 10:50

Sweeper

54.6k960117

edited Aug 19 at 11:18

zlakad

9171614

edited Aug 19 at 11:18

zlakad

9171614

edited Aug 19 at 11:18

zlakad

9171614

answered Aug 19 at 10:50

Sweeper

54.6k960117

answered Aug 19 at 10:50

Sweeper

54.6k960117

answered Aug 19 at 10:50

Sweeper

54.6k960117

add a commentÂ |Â

up vote
1
down vote

The major thing you need to consider while solving this question is remainders.

Here's the code:

public int spitNumbers(int number, int groups)

 int result = new int[groups]; //The array we'd return at the end

 int shift = number % groups; //I used this to check the remainder.

 for(int i = 0; i < result.length; i++)

 /* The if-statement below checks for two things:
 - if the remainder is zero
 - if we've reached the shifting stage.
 */
 if((shift == 0) 
 return result;

I hope this helps.. Merry coding!

edited Aug 19 at 11:52

answered Aug 19 at 11:33

Taslim

1,02921024

add a commentÂ |Â

up vote
1
down vote

The major thing you need to consider while solving this question is remainders.

Here's the code:

public int spitNumbers(int number, int groups)

 int result = new int[groups]; //The array we'd return at the end

 int shift = number % groups; //I used this to check the remainder.

 for(int i = 0; i < result.length; i++)

 /* The if-statement below checks for two things:
 - if the remainder is zero
 - if we've reached the shifting stage.
 */
 if((shift == 0) 
 return result;

I hope this helps.. Merry coding!

edited Aug 19 at 11:52

answered Aug 19 at 11:33

Taslim

1,02921024

add a commentÂ |Â

up vote
1
down vote

The major thing you need to consider while solving this question is remainders.

Here's the code:

public int spitNumbers(int number, int groups)

 int result = new int[groups]; //The array we'd return at the end

 int shift = number % groups; //I used this to check the remainder.

 for(int i = 0; i < result.length; i++)

 /* The if-statement below checks for two things:
 - if the remainder is zero
 - if we've reached the shifting stage.
 */
 if((shift == 0) 
 return result;

I hope this helps.. Merry coding!

edited Aug 19 at 11:52

answered Aug 19 at 11:33

Taslim

1,02921024

The major thing you need to consider while solving this question is remainders.

Here's the code:

public int spitNumbers(int number, int groups)

 int result = new int[groups]; //The array we'd return at the end

 int shift = number % groups; //I used this to check the remainder.

 for(int i = 0; i < result.length; i++)

 /* The if-statement below checks for two things:
 - if the remainder is zero
 - if we've reached the shifting stage.
 */
 if((shift == 0) 
 return result;

I hope this helps.. Merry coding!

edited Aug 19 at 11:52

answered Aug 19 at 11:33

Taslim

1,02921024

edited Aug 19 at 11:52

answered Aug 19 at 11:33

Taslim

1,02921024

answered Aug 19 at 11:33

Taslim

1,02921024

answered Aug 19 at 11:33

Taslim

1,02921024

add a commentÂ |Â

up vote
0
down vote

Please have a look at this solution, I little tested, it works.

public static int number(int number, int divider)

p.s. little simplfied

edited Aug 19 at 11:42

answered Aug 19 at 11:34

ImZ

266

add a commentÂ |Â

up vote
0
down vote

Please have a look at this solution, I little tested, it works.

public static int number(int number, int divider)

p.s. little simplfied

edited Aug 19 at 11:42

answered Aug 19 at 11:34

ImZ

266

add a commentÂ |Â

up vote
0
down vote

Please have a look at this solution, I little tested, it works.

public static int number(int number, int divider)

p.s. little simplfied

edited Aug 19 at 11:42

answered Aug 19 at 11:34

ImZ

266

Please have a look at this solution, I little tested, it works.

public static int number(int number, int divider)

p.s. little simplfied

edited Aug 19 at 11:42

answered Aug 19 at 11:34

ImZ

266

edited Aug 19 at 11:42

answered Aug 19 at 11:34

ImZ

266

answered Aug 19 at 11:34

ImZ

266

answered Aug 19 at 11:34

ImZ

266

add a commentÂ |Â

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