Closest pair sum in two sorted arrays

后端 未结 3 463
感动是毒
感动是毒 2021-01-03 04:40

Given two sorted arrays of integers, a and b, and an integer c, I have to find i,j such that:

a[i] + b[j]         


        
相关标签:
3条回答
  • 2021-01-03 04:55

    I think you dont have to search the whole array b[] next time.......u have to search in between starting of array b and the lowest bound u found till now....for the next element in a[].It would definitely reduce your time complexity...and when u find the given target 'c' you must stop your search.

    0 讨论(0)
  • 2021-01-03 05:05

    The below solution is in linear Time Complexity O(n), Space Complexity O(1)

    public class ClosestPair {
    
        public static void main(String[] args)
        {
            int ar2[] = {4, 5, 7};
            int ar1[] = {10, 20, 30, 40};
            int x = 10 ;
            closest(ar1,ar2,x);
        }
        public static void closest(int[] ar1, int[] ar2, int x)
        {   int diff=Integer.MAX_VALUE;
            int first_num=0;
            int second_num=0;
            int second_diff=Integer.MAX_VALUE;
            for(int i=0; i<ar1.length; i++)
            {
                if(x==ar1[i] )
                {
                    System.out.println("no pair possible");
                    return;
                }
            }
            for(int i=0; i<ar2.length; i++)
            {
                if(x==ar2[i])
                {
                    System.out.println("no pair possible");
                    return;
                }
            }
            for(int i=0; i<ar2.length; i++)
            {
    
                if(Math.abs(x-ar2[i])<=diff)
                {
                    diff=Math.abs(x-ar2[i]);
                    first_num=ar2[i];
                }
            }
           diff=x-first_num;
           for(int i=0; i<ar1.length; i++)
           {
               if(Math.abs(diff-ar1[i])<=second_diff)
               {
                   second_diff= Math.abs(diff-ar1[i]);
                   second_num= ar1[i];
               }
           }
           System.out.println(first_num + " "+ second_num);
        }
    }
    
    0 讨论(0)
  • 2021-01-03 05:21

    Thinking a bit about it, then you could ask yourself:
    "Is it necessary, each time, to search in the sorted b-array for successive values from a[]?"

    0 讨论(0)
提交回复
热议问题