Hello everyone lately I encountered quite an easy algorithm problem which at my great surprise I encountered a difficulty. The task was to return true if a particular array contained a duplicate element else false
Actually I don't understand why my program produces an unexpected output . I know many other solutions to solve that problem but I want to understand why this solutions doesn't not work . And it returns False from input [1,2,3,1]
In facts it's always returns false nor matter the input. It never passes the condition in the try statement
dic = {}
for i in range(0,len(nums)):
try:
if dic[nums[i]]:
return True
except:
dic[nums[i]] = i
return False
It wouldn't always return False. It would return True if your duplicated value is NOT at the start of nums (which I assume is a list of numbers).
Notice that on line dic[nums[i]] = i on the first iteration you are updating your dictionary with i equals to 0. That is why later on you can't catch the case in which your duplicated value is at the start of the list.
Related
I'm practicing leetcode and checking other people's solutions for the same exercise on my terminal gives a totally different output than expected. If I run it on leetcode it says the answer is correct...
What might be causing this?
reversing the array of chars - ["h","e","l","l","o"] - like so:
def reverse_string(s)
len = s.length
(0...len/2).each { |i|
temp = s[i]
swap_index = len-i-1
s[i] = s[swap_index]
s[swap_index] = temp
}
end
p reverse_string([["h","e","l","l","o"]])
output in my terminal -- 0...2
output in leetcode is as expected - ["o","l","l","e","h"]
Why does this happen?
thanks
Ruby's p function will print the value you pass to it.
The value you are choosing to pass to it is the return value of reverse_string.
You should add a line to the end of reverse_string that specifies what you want to return from it. In this case the line can simply consist of s. You don't even need an explict return in this case.
def large_prime(n)
return [] if n==1
factor = (2..n).find {|x| n % x == 0}
[factor] + large_prime(n/factor)
end
I got this solution from somewhere else. I don't understand the 4th line of code where large_prime is called recursively and appended onto factor.
When I change the first line "return []" and leave out the '[]' after the return, I get an error message for on line 4, that says '+':no implicit conversion of nil into Array.
So why does this code work? Thanks
P.S. I'm obviously a noob and everything is very new to me.
The 3rd line finds the first divisor of n between 2 and n. This line itself does not involve recursion.
I don't really get the code you modified, but it seems to return nil in some case, while the original method always return an Array.
You must return an empty array when passed 1 to terminate the recursion. Any positive argument other than one will result in another call to large_prime, but an argument of 1 results in large_prime simply returning an empty array.
At each level of recursion, the program adds an array with the single factor it found to an array consisting of all factors found for the value n/factor. When the last factor (other than 1) is found, the final call to large_prime is made with an argument of 1, large_prime returns an empty array which is then added to the array containing the last factor, giving an array containing just the last factor. This array is then returned and you have
[next-to-last-factor] + [last-factor], giving a return array of [next-to-last-factor, last-factor] which is added to [next-to-next-to-last-factor] giving [next-to-next-to-last-factor, next-to-last-factor, last-factor]. This is then added to an array [next-to-next-to-next-to-last-factor], giving... lather, rinse, repeat until we reach the largest factor and add it in.
You must return an empty array because you can't add nil to an array in Ruby.
This question already has answers here:
Better way to write "matching balanced parenthesis" program in Ruby
(8 answers)
Closed 8 years ago.
I just went through those interview coding quizzes for the first time and I'm somewhere between submerging myself in a tub of dran-o and investing in No Tears bubble bath products along with a bunch of toasters.
The problem was as follows:
If you're given a string like "zx(c)abcde[z{x]}", write a function that returns true if the syntax is correct and false if the syntax is incorrect: for example, in that string the brackets and braces are messed up. In other words "{hello}mot[o]" will pass but "{hello}mo{[t}" would not.
My throught process went like: keep a list of opening and closing bracket/brace/parens positions, then see if there is overlap. But that wasn't an optimal solution so I bombed it.
I'd like some help understanding how to solve this problem.
Thanks in advance.
[Edit: I've incorporated both of #sawa's excellent suggestions.]
One way you can do this is with a stack.
MATCH = { '['=>']', '('=>')', '{'=>'}' }
OPENING = MATCH.keys
CLOSING = MATCH.values
def check_for_match(str)
str.chars.each_with_object([]) do |c, arr|
case c
when *OPENING
arr << c
when *CLOSING
return false unless c.eql?(MATCH[arr.pop])
end
end.empty?
end
check_for_match("zx(c)abcde[z{x]}") #=> false
check_for_match("zx(c)abcde[z{x}]") #=> true
[Edit: I thought this question seemed familiar. I and several others answered it a while ago.]
Another way to do this is to first strip out the irrelevant characters, then sequentially remove adjacent matching pairs until either the string is empty (return true) or the string is not empty and there are no more matching adjacent pairs (return false).
def check_for_match(str)
str = str.gsub(/[^\(\)\[\]\{\}]/, '')
while str.gsub!(/\(\)|\[\]|\{\}/, ''); end
str.empty?
end
check_for_match("zx(c)abcde[z{x]}") #=> false
check_for_match("zx(c)abcde[z{x}]") #=> true
Reader challenge: provide a proof that the syntax is incorrect when false is returned.
I would replace each bracket with an XML tag, and just run it through an XML validator. It'll pick out weird stuff like this:
<bracket>stuff<curly>morestuff</bracket></curly>
This will fail XML validation, so you can just return that.
I am trying to understand how range.cover? works and following seems confusing -
("as".."at").cover?("ass") # true and ("as".."at").cover?("ate") # false
This example in isolation is not confusing as it appears to be evaluated dictionary style where ass comes before at followed by ate.
("1".."z").cover?(":") # true
This truth seems to be based on ASCII values rather than dictionary style, because in a dictionary I'd expect all special characters to precede even digits and the confusion starts here. If what I think is true then how does cover? decide which comparison method to employ i.e. use ASCII values or dictionary based approach.
And how does range work with arrays. For example -
([1]..[10]).cover?([9,11,335]) # true
This example I expected to be false. But on the face of it looks like that when dealing with arrays, boundary values as well as cover?'s argument are converted to string and a simple dictionary style comparison yields true. Is that correct interpretation?
What kind of objects is Range equipped to handle? I know it can take numbers (except complex ones), handle strings, able to mystically work with arrays while boolean, nil and hash values among others cause it to raise ArgumentError: bad value for range
Why does ([1]..[10]).cover?([9,11,335]) return true
Let's take a look at the source. In Ruby 1.9.3 we can see a following definition.
static VALUE
range_cover(VALUE range, VALUE val)
{
VALUE beg, end;
beg = RANGE_BEG(range);
end = RANGE_END(range);
if (r_le(beg, val)) {
if (EXCL(range)) {
if (r_lt(val, end))
return Qtrue;
}
else {
if (r_le(val, end))
return Qtrue;
}
}
return Qfalse;
}
If the beginning of the range isn't lesser or equal to the given value cover? returns false. Here lesser or equal to is determined in terms of the r_lt function, which uses the <=> operator for comparison. Let's see how does it behave in case of arrays
[1] <=> [9,11,335] # => -1
So apparently [1] is indeed lesser than [9,11,335]. As a result we go into the body of the first if. Inside we check whether the range excludes its end and do a second comparison, once again using the <=> operator.
[10] <=> [9,11,335] # => 1
Therefore [10] is greater than [9,11,335]. The method returns true.
Why do you see ArgumentError: bad value for range
The function responsible for raising this error is range_failed. It's called only when range_check returns a nil. When does it happen? When the beginning and the end of the range are uncomparable (yes, once again in terms of our dear friend, the <=> operator).
true <=> false # => nil
true and false are uncomparable. The range cannot be created and the ArgumentError is raised.
On a closing note, Range.cover?'s dependence on <=> is in fact an expected and documented behaviour. See RubySpec's specification of cover?.
What is the "best" way to determine the number of elements in an array in VBScript?
UBound() tells you how many slots have been allocated for the array, but not how many are filled--depending on the situation, those may or may not be the same numbers.
First off there is no predefined identifier called vbUndefined as the currently accepted answer appears to imply. That code only works when there is not an Option Explicit at the top of the script. If you are not yet using Option Explicit then start doing so, it will save you all manner of grief.
The value you could use in place of vbUndefined is Empty, e.g.,:-
If arr(x) = Empty Then ...
Empty is a predefined identify and is the default value of a variable or array element that has not yet had a value assigned to it.
However there is Gotcha to watch out for. The following statements all display true:-
MsgBox 0 = Empty
MsgBox "" = Empty
MsgBox CDate("30 Dec 1899") = True
Hence if you expect any of these values to be a valid defined value of an array element then comparing to Empty doesn't cut it.
If you really want to be sure that the element is truely "undefined" that is "empty" use the IsEmpty function:-
If IsEmpty(arr(x)) Then
IsEmpty will only return true if the parameter it actually properly Empty.
There is also another issue, Null is a possible value that can be held in an array or variable. However:-
MsgBox Null = Empty
Is a runtime error, "invalid use of null" and :-
MsgBox IsEmpty(Null)
is false. So you need to decide if Null means undefined or is a valid value. If Null also means undefined you need your If statement to look like:-
If IsEmpty(arr(x)) Or IsNull(arr(x)) Then ....
You might be able to waive this if you know a Null will never be assigned into the array.
I'm not aware of a non-iterative method to do this, so here's the iterative method:
Function countEmptySlots(arr)
Dim x, c
c = 0
For x = 0 To ubound(arr)
If arr(x) = vbUndefined Then c = c + 1
Next
countEmptySlots = c
End Function
As Spencer Ruport says, it's probably better to keep track yourself to begin with.
There's nothing built in to tell you what elements are filled. The best way is to keep track of this yourself using a count variable as you add/remove elements from the array.