Task :

Given a List [] of n integers , find minimum mumber to be inserted in a list, so that sum of all elements of list should equal the closest prime number .

Notes

• List size is at least 2 .

• List‘s numbers will only positives (n > 0) .

• Repeatition of numbers in the list could occur .

• The newer list‘s sum

  should equal the closest prime number .

Input >> Output Examples

1- minimumNumber ({3,1,2}) ==> return (1)

Explanation:

• Since , the sum of the list‘s elements equal to (6) , the minimum number to be inserted to transform the sum to prime number is (1) , which will make the sum of the List
  equal the closest prime number (7) .

2-  minimumNumber ({2,12,8,4,6}) ==> return (5)

Explanation:

• Since , the sum of the list‘s elements equal to (32) , the minimum number to be inserted to transform the sum to prime number is (5) , which will make the sum of the List equal the closest prime number (37)
  .

3- minimumNumber ({50,39,49,6,17,28}) ==> return (2)

Explanation:

• Since , the sum of the list‘s elements equal to (189) , the minimum number to be inserted to transform the sum to prime number is (2) , which will make the sum of the List equal the closest prime number (191) .

public class Solution
{
    public static int minimumNumber(int[] numbers)
    {
        //首先求得現在list中的數值之和
        int sumOri = 0;
        
        for(int i=0;i<numbers.length;i++){
             sumOri += numbers[i];
        }
        //用另一個數對原始和進行遞增
        int sumNow = sumOri;
        boolean searchPrime = true;
        //do...while遞增原始數值,並判定是否為素數
        while(searchPrime)
        {
           
            int count = 0;
            for(int i=2;i<=Math.floor(sumNow/2);i++){
                if(sumNow%i==0){
                   //合數進入
                   count++;
                }
            }
           
            //count為0,這個數是質數,停止查找
            if(count == 0){
                 searchPrime = false;
            }else{
              //遞增新的和
              sumNow++;
            } 
            
        }
        return sumNow-sumOri; // Do your magic!
    }
}

尋找List之和的最近素數