CODEWITHSUNDEEP
javaalgorithm
membersound
membersound

Reputation: 86657

How to distribute integers topdown a set of persons?

I have a sorted set of persons, and I want to distribute a defined count topdown. Example:

Scenario:

int i = 8;
personA = 0;
personB = 0;
personC = 0;

Desired result:

personA = 3;
personB = 3;
personC = 2;

Of course I could iterate with for/while loops, but if i is very large that might be inefficient.

Can the result also be optained by devision, without loops?

Pseudocode:

distributed = 0;
while (distributed < i) {
    for (int p : persons) {
        p++;
        distributed++;
    }
}

interation steps:

1-1-1 > start over > 2-2-2 > start over > 3-3-2

Upvotes: 0

Views: 76

Answers (4)

Manil Liyanage
Manil Liyanage

Reputation: 255

The following code will run for i=8 5 times, i=10 4 times, as i increase , number of time loop executed is proportionately less

        int i = 10;

        int[] arr = new int[3] { 0, 0, 0 };

        int BaseVal = i / (arr.Length);
        int Remainder = i - (BaseVal * arr.Length);

        for (int x = 0; x < arr.Length  ;x++)
        {
            arr[x] = BaseVal;
        }
        for (int y = 0; y < Remainder; y++)
        {
            arr[y] = arr[y] + 1;
        }

Upvotes: 2

RealSkeptic
RealSkeptic

Reputation: 34618

Yes, of course it's possible. Since you didn't give any particular implementation, I'll just put the numbers in an array. This is just a demonstration, of course. Assume NUM_OF_PERSONS is the number of persons in your array, and NUM_TO_DISTRIBUTE is the number you want to distribute. In your example, 8.

int persons[] = new int[NUM_OF_PERSONS];

int basicRation = NUM_TO_DISTRIBUTE / NUM_OF_PERSONS;
int peopleGettingExtra = NUM_TO_DISTRIBUTE % NUM_OF_PERSONS;

for ( int i = 0; i < NUM_OF_PERSONS; i ++ ) {
    persons[i] = basicRation + ( i < peopleGettingExtra ? 1 : 0 );
}

Test case 1: 9 to give, 3 people. The base ration is 3. The number of people getting extra is zero. In the loop, since no i will be less than 0, everybody will get basicRation + 0 which means 3.

Test case 2: 8 to give, 3 people. The base ration is 2. The number of people getting extra is 2. In the loop, people with indexes 0 and 1 will get 2 + 1, and the last person gets 2 + 0.

Upvotes: 3

Salih Erikci
Salih Erikci

Reputation: 5087

distributed = 0;
int howManyForEach = (int)(i/persons.size())

for(int p : person){
   p = howManyForEach ;
}
distrubuted = howManyForEach * persons.size();
while (distributed < i) {
    for (int p : persons) {
        p++;
        distributed++;
    }
}

Upvotes: 2

Terry Storm
Terry Storm

Reputation: 497

from the problem it is clear, that the maximal difference between any person is 1.

Just use #persons / i = base value for every person

and #persons % i = # top persons with +1 value

Example

i = 5 , persons = 3: 
baseValue = 1
#topPersons = 2 

=> 2 2 1

Upvotes: 1

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