Java > Array-1 > unlucky1 (CodingBat Solution)

Problem:

We'll say that a 1 immediately followed by a 3 in an array is an "unlucky" 1. Return true if the given array contains an unlucky 1 in the first 2 or last 2 positions in the array.

unlucky1({1, 3, 4, 5}) → true
unlucky1({2, 1, 3, 4, 5}) → true
unlucky1({1, 1, 1}) → false


Solution:

public boolean unlucky1(int[] nums) {
  if (nums.length < 2) {
    return false;
  } else if ((nums[0] == 1 && nums[1] == 3) || (nums[nums.length-2] == 1 && nums[nums.length-1] == 3)) {
      return true;
    } else if (nums.length > 2 && nums[1] == 1 && nums[2] == 3) {
        return true;
    } else
      return false;
    
}

10 comments:

  1. AnonymousJanuary 9, 2016 at 11:58 AM

    public boolean unlucky1(int[] nums) {
    if (nums.length<2) return false;

    if ((nums[0]==1 && nums[1]==3)||
    (nums[1]==1 && nums[2]==3)||
    (nums[nums.length-2]==1 && nums[nums.length-1]==3))
    return true;

    return false;
    }

    ReplyDelete
  2. AnonymousSeptember 19, 2020 at 4:15 PM

    public boolean unlucky1(int[] nums) {
    return (nums.length >= 2) &&
    ( nums[0] == 1 && nums[1] == 3
    || nums[1] == 1 && nums[2] == 3
    || nums[nums.length - 2] == 1 && nums[nums.length - 1] == 3);
    }

    ReplyDelete
  3. AnonymousSeptember 19, 2020 at 4:16 PM

    public boolean unlucky1(int[] nums) {
    if (nums.length <= 1) {
    return false;
    }
    return
    ( nums[0] == 1 && nums[1] == 3
    || nums[1] == 1 && nums[2] == 3
    || nums[nums.length - 2] == 1 && nums[nums.length - 1] == 3);
    }

    ReplyDelete
  4. anonymousAugust 18, 2021 at 8:40 PM

    public boolean unlucky1(int[] nums) {
    for (int i = 0; i < nums.length-1; i++) {
    if (nums.length > 4 && i == 2) i = nums.length-2;
    if (nums[i] == 1 && nums[i+1] == 3) return true;
    }
    return false;
    }

    ReplyDelete
  5. anonymousAugust 18, 2021 at 8:47 PM

    public int[] make2(int[] a, int[] b) {
    if (a.length > 1) return new int[] {a[0], a[1]};
    if (a.length == 1) return new int[] {a[0], b[0]};
    return new int[]{b[0], b[1]};
    }

    ReplyDelete
  6. UnknownOctober 27, 2021 at 12:44 PM

    public boolean unlucky1(int[] arr) {
    if(arr.length>1){
    if(((arr[0]==1&&arr[1]==3)||(arr[1]==1&&arr[2]==3))
    ||((arr[arr.length-2]==1&&arr[arr.length-1]==3)))
    return true;
    return false;}
    return false;
    }

    ReplyDelete
  7. UnknownOctober 28, 2021 at 12:12 PM

    hi Nodir

    ReplyDelete
  8. yogeshMarch 15, 2022 at 6:15 AM

    public boolean unlucky1(int[] nums) {

    for(int i=0;i<nums.length-1;i++)
    {
    if(nums[i]==1 && nums[i+1]==3)
    {
    if(nums[0]==1 && nums[1]==1 && nums[2]==1 && nums[3]==3 && nums[4]==1)
    {
    return false;
    }
    return true;
    }

    }

    return false;
    }

    ReplyDelete
  9. Shre_Jeni❤️August 1, 2022 at 6:20 AM

    public boolean unlucky1(int[] nums) {
    int a=nums.length;
    if(a<2){
    return false;
    }
    else
    if(a>2 && (nums[0]==1&&nums[1]==3) || (nums[1]==1&&nums[2]==3)){
    return true;
    }
    else if(nums[a-2]==1 && nums[a-1]==3){
    return true;
    }
    else return false;
    }

    ReplyDelete