sai rama krishna

I am working on below challenge
https://codingbat.com/prob/p108997

Given an array of ints, compute recursively if the array contains a 6. We'll use the convention of considering only the part of the array that begins at the given index. In this way, a recursive call can pass index+1 to move down the array. The initial call will pass in index as 0.


array6([1, 6, 4], 0) → true
array6([1, 4], 0) → false
array6([6], 0) → true

I was not clear on this description. Is it saying array contains 6 then return true if that is case what is purpose of index. I see few other challeges similar to this like next challenge to this as well. Can you help me understand this better so that i can move further

Thank you

I see some other interesting operators as below in that link which i never used or saw

AssignmentOperator:
(one of)
=  *=  /=  %=  +=  -=  <<=  >>=  >>>=  &=  ^=  |=

once added limit it worked with for loop also

public int[] copyEndy(int[] nums, int count) { int[] arr = new int[count]; int arrIndex = 0; // int numsIndex=0; for (int numsIndex = 0; numsIndex < nums.length && arrIndex < count; numsIndex++) { int n = nums[numsIndex]; // boolean endy=isEndy(i); if (isEndy(n)) { arr[arrIndex] = n; arrIndex++; } } return arr; } boolean isEndy(int i) { if ((i <= 10 && i >= 0) || (i <= 100 && i >= 90)) { return true; } else { return false; } }

Expected Run copyEndy([9, 11, 90, 22, 6], 2) → [9, 90] [9, 90] OK copyEndy([9, 11, 90, 22, 6], 3) → [9, 90, 6] [9, 90, 6] OK copyEndy([12, 1, 1, 13, 0, 20], 2) → [1, 1] [1, 1] OK copyEndy([12, 1, 1, 13, 0, 20], 3) → [1, 1, 0] [1, 1, 0] OK copyEndy([0], 1) → [0] [0] OK copyEndy([10, 11, 90], 2) → [10, 90] [10, 90] OK copyEndy([90, 22, 100], 2) → [90, 100] [90, 100] OK copyEndy([12, 11, 10, 89, 101, 4], 1) → [10] [10] OK copyEndy([13, 2, 2, 0], 2) → [2, 2] [2, 2] OK copyEndy([13, 2, 2, 0], 3) → [2, 2, 0] [2, 2, 0] OK copyEndy([13, 2, 13, 2, 0, 30], 2) → [2, 2] [2, 2] OK copyEndy([13, 2, 13, 2, 0, 30], 3) → [2, 2, 0] [2, 2, 0] OK other tests OK

i am not that used using more than one indexes. can we do this with just one index instead of 2 indexes(one to iterate given input index to check each element, other to put elements into result array to display finally)?

public int[] copyEndy(int[] nums, int count) { int[] arr = new int[count]; int arrIndex = 0; int numsIndex = 0; while (numsIndex < nums.length && arrIndex < count) { int n = nums[numsIndex]; if (n >= 0 && n <= 10 || n <= 100 && n >= 90) { arr[arrIndex] = n; arrIndex++; } numsIndex++; } return arr; }

Above is working with while loop



Expected Run copyEndy([9, 11, 90, 22, 6], 2) → [9, 90] [9, 90] OK copyEndy([9, 11, 90, 22, 6], 3) → [9, 90, 6] [9, 90, 6] OK copyEndy([12, 1, 1, 13, 0, 20], 2) → [1, 1] [1, 1] OK copyEndy([12, 1, 1, 13, 0, 20], 3) → [1, 1, 0] [1, 1, 0] OK copyEndy([0], 1) → [0] [0] OK copyEndy([10, 11, 90], 2) → [10, 90] [10, 90] OK copyEndy([90, 22, 100], 2) → [90, 100] [90, 100] OK copyEndy([12, 11, 10, 89, 101, 4], 1) → [10] [10] OK copyEndy([13, 2, 2, 0], 2) → [2, 2] [2, 2] OK copyEndy([13, 2, 2, 0], 3) → [2, 2, 0] [2, 2, 0] OK copyEndy([13, 2, 13, 2, 0, 30], 2) → [2, 2] [2, 2] OK copyEndy([13, 2, 13, 2, 0, 30], 3) → [2, 2, 0] [2, 2, 0] OK other tests OK

I could not make my original for loop approach to work yet though. I am checking on that limit as well in for loop

Thank you Paul.

"a &= b" is the same as "a = a & b" right not a && b? This is first time I ever saw this expression. Is it in new edition of java or i just never got chance to look earlier i saw others like you mention more often a+=b and a-=b  and a*=b and a/=b

Gaurd clause is same as edge case? I hear people calling edge case for similar check at the beginning of method.

Regarding other question if number of lines of code in a method more than 10 lines say 15 lines if i use multiple returns what is the downside or disadvantage of it? As most of my challenges I used returns liberally without thinking about this type rule.

Thank you Ritchie, Lacar

I have some follow up questions after reading above 2 comments

divides &= j != 0 && n % j == 0;
I never saw &= operator. What this represent?

If number of lines more if we put more return statements visibility is lost right?

What is meaning of guard clauses? Can you please give some simple example for that?

Looks interesting.  I will take some time to digest above code.

Is it bad thing to have multiple returns?

Can I use for loop instead of while loop here?

can I use for loop interchangeably wherever while is used or there are any restrictions for for loop? I got used to for loop more than while loop

Below finally worked
public boolean dividesSelf(int n) { int x = n; while (x > 0) { int digit = x % 10; x = x / 10; if (digit == 0 || n % digit != 0) { return false; } } return true; }

It passed all tests
Expected Run dividesSelf(128) → true true OK dividesSelf(12) → true true OK dividesSelf(120) → false false OK dividesSelf(122) → true true OK dividesSelf(13) → false false OK dividesSelf(32) → false false OK dividesSelf(22) → true true OK dividesSelf(42) → false false OK dividesSelf(212) → true true OK dividesSelf(213) → false false OK dividesSelf(162) → true true OK other tests OK All Correct
x is the new assigned number of n which used for chop off and while loop
digit is reminder digit
I have to check n reminder by dividing with digit and chop off that x to get new digit every time in the for loop

I agree. I debugged in eclipse and changed < condition to be less than n not the length of array and it worked fine as below

public String[] wordsFront(String[] words, int n) { String counts[] = countStrMethod(words, n); return counts; } String[] countStrMethod(String[] words, int n) { String[] countStr = new String[n]; for (int i = 0; i < n; i++) { countStr[i] = words[i]; } // words[] return countStr; }


Expected Run wordsFront(["a", "b", "c", "d"], 1) → ["a"] ["a"] OK wordsFront(["a", "b", "c", "d"], 2) → ["a", "b"] ["a", "b"] OK wordsFront(["a", "b", "c", "d"], 3) → ["a", "b", "c"] ["a", "b", "c"] OK wordsFront(["a", "b", "c", "d"], 4) → ["a", "b", "c", "d"] ["a", "b", "c", "d"] OK wordsFront(["Hi", "There"], 1) → ["Hi"] ["Hi"] OK wordsFront(["Hi", "There"], 2) → ["Hi", "There"] ["Hi", "There"] OK

I agree. Lesson learnt Never test further when found required result

If 1, 2, 3 comes in random positions my solution will not work. May be I need one other for loop. I have to try that test scenario

public boolean array123(int[] nums) { boolean res = false; for (int i = 0; i < nums.length - 2; i++) { res = (nums[i] == 1 && nums[i + 1] == 2 && nums[i + 2] == 3); } return res; }

Above refactored code failing some reason one test
Expected Run array123([1, 1, 2, 3, 1]) → true false X array123([1, 1, 2, 4, 1]) → false false OK array123([1, 1, 2, 1, 2, 3]) → true true OK array123([1, 1, 2, 1, 2, 1]) → false false OK array123([1, 2, 3, 1, 2, 3]) → true true OK array123([1, 2, 3]) → true true OK array123([1, 1, 1]) → false false OK array123([1, 2]) → false false OK array123([1]) → false false OK array123([]) → false false OK

whereas below further refactored code passing all tests

public boolean array123(int[] nums) { for (int i = 0; i < (nums.length - 2); i++) { if (nums[i] == 1 && nums[i + 1] == 2 && nums[i + 2] == 3) return true; } return false; }

Expected Run array123([1, 1, 2, 3, 1]) → true true OK array123([1, 1, 2, 4, 1]) → false false OK array123([1, 1, 2, 1, 2, 3]) → true true OK array123([1, 1, 2, 1, 2, 1]) → false false OK array123([1, 2, 3, 1, 2, 3]) → true true OK array123([1, 2, 3]) → true true OK array123([1, 1, 1]) → false false OK array123([1, 2]) → false false OK array123([1]) → false false OK array123([]) → false false OK

Looks like refactored code 1 is checking whole array which is not right approach for this challenge.
Refactored code 2 check part of array and as soon as finds 1,2, 3 next to each other returns true and program exited

Very interesting old sun link. Thank you for sharing.


I mean to say below code passed all tests
public boolean array123(int[] nums) { boolean res=false; for(int i=0;i<nums.length-2;i++){   if(nums[i]==1&&nums[i+1]==2&&nums[i+2]==3){     res= true;     break;   }   } return res; }
Expected Run array123([1, 1, 2, 3, 1]) → true true OK array123([1, 1, 2, 4, 1]) → false false OK array123([1, 1, 2, 1, 2, 3]) → true true OK array123([1, 1, 2, 1, 2, 1]) → false false OK array123([1, 2, 3, 1, 2, 3]) → true true OK array123([1, 2, 3]) → true true OK array123([1, 1, 1]) → false false OK array123([1, 2]) → false false OK array123([1]) → false false OK array123([]) → false false OK

Like you mentioned let me refactor to remove those if else statements as well

Thank you. I agree with all above points. When I am using eclipse I am using debugger and sometimes system.out.println statements

I did not think about less than 0 scenario. Also did not pay much attention to 0 case as that was given in test case. I was focusing more on description.  Whatever challenges i wrote last night in codingbat were not registered not sure some issue going on with website currently

public int bunnyEars(int bunnies) {    if (bunnies == 0) {        return 0;    }    return 2+ bunnyEars(bunnies - 1); }
Definitely. Trying to understand how recursion works instead of loops. Above worked. Looks like I cannot take base case of bunnies==1 but only bunnies==0 and then keep adding 2 with recursion call of bunnies-1