Java > Array-2 > matchUp (CodingBat Solution)

Problem:

Given arrays nums1 and nums2 of the same length, for every element in nums1, consider the corresponding element in nums2 (at the same index). Return the count of the number of times that the two elements differ by 2 or less, but are not equal.

matchUp({1, 2, 3}, {2, 3, 10}) → 2
matchUp({1, 2, 3}, {2, 3, 5}) → 3
matchUp({1, 2, 3}, {2, 3, 3}) → 2


Solution:

public int matchUp(int[] nums1, int[] nums2) {
  int count =0;
  for (int i =0; i <nums1.length ;i++)
  if (Math.abs(nums1[i] - nums2[i]) == 1 || 
               Math.abs(nums1[i] -nums2[i]) == 2 )
  count++;
  return count;
  
}

3 comments:

  1. UnknownSeptember 14, 2018 at 3:25 AM

    public int matchUp(int[] nums1, int[] nums2) {
    int count = 0;

    for(int i = 0; i < nums1.length; i++){
    if( (Math.abs(nums1[i] - nums2[i] ) <= 2) && nums1[i] != nums2[i] ){
    count ++;
    }
    }
    return count;
    }

    ReplyDelete
  2. SinuserOctober 28, 2021 at 8:38 AM

    int listA = 0;

    for(int i = 0; i < nums1.length; i++)
    {
    if(Math.abs(nums1[i] - nums2[i]) <= 2 && nums1[i] != nums2[i]) listA += 1;
    }

    return listA;
    }

    ReplyDelete
  3. ADecember 22, 2022 at 10:57 PM

    public int matchUp(int[] nums1, int[] nums2) {
    int sum=0;
    for(int i=0;i0)
    sum++;
    }

    return sum;
    }

    ReplyDelete