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A,B,C are the arrays :
A = {1,2,3,4}
B = {8,1,2,3}
C = {1,2,9,3}
Result is the uncommon values from three arrays
Result = {4,8,9}
Asking for the logic what I can implement ?
There might be some other better answers. But here is the most simple one.
On line 4 I took the Intersection of 3 sets
On line 7 I took the Union of three sets
On line 10 the difference or XOR operation, of Union and intersection
I hope it helps:-
>>> a = {1,2,3,4}
>>> b = {8,1,2,3}
>>> c = {1,2,9,3}
>>> d = a & b & c
>>> print(d)
{1, 2, 3}
>>> e = a | b | c
>>> print(e)
{1, 2, 3, 4, 8, 9}
>>> f = d^e
>>> print(f)
{4, 8, 9}
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I'm new to rails and I was wondering how I compare the first value of array "a" if it is greater than the first value of array "b"?
Example:
a = [1, 2, 3]
b = [3, 2, 1]
How do I check if a[0] is greater than b[0].
You can use the first method:
a = [1, 2, 3]
b = [3, 2, 1]
a.first > b.first #=> false
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I'm looking for an algorithm to merge/compress lists of unsorted integers in to a single list while maintaining the original lists.
Example
A = [1, 1, 3]
B = [3, 2, 9]
C = [1, 3, 2]
D = [1]
A C
| |
Result[ 1 1 3 2 9 ]
| |
D B
I found lots of results with similar questions but most of them deal with sorted lists or break them up. I'm sure something like this exists but I simply don't know the correct terminology.
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How can I remove elements from a given array of integers that are evenly divisible by 3? A new array should be returned and the original array should not be altered.
For example, if
array = [1,2,3,4,1,5,2,6,7,8,9,10,7,11,12,13,14,15]
the array
[1, 2, 4, 1, 5, 2, 7, 8, 10, 7, 11, 13, 14]
should be returned.
Just found it:
array.reject{|a| a % 3 == 0}
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I had 10 items in total. I lost all of them in 3 days: 5 items on the 1st day, 3 items on the 2nd day, and 2 items on the last day. I need to get an array [5, 2, 0] of remaining items at the end of each day. How can I get the array, given total 10 and the array of lost items [5, 3, 2]?
[5, 3, 2].each_with_object([10]){|e, a| a.push(a.last - e)}.drop(1)
# => [5, 2, 0]
Know why you need the complication drop(1)? It is because, without it, the answer makes more logical sense. Your requirement is what was complicated.
[5, 3, 2].each_with_object([10]){|e, a| a.push(a.last - e)}
# => [10, 5, 2, 0]
The initial 10 represents the initial state.
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Write a program that takes as input a list of n integers and produces as output the largest difference obtained by subtracting an integer in the list from the one following it
I can think of this using the combination method.
ar = (1..12).to_a
ar.combination(2).max_by { |a,b| b - a }
# => [1, 12]
ar = [1,23,56,11]
# => [1, 23, 56, 11]
ar.combination(2).max_by { |a,b| (b - a).abs }
# => [1, 56]
It seems I misunderstood the line as OP mentioned :
output the largest difference obtained by subtracting an integer in the list from the one following it
Here is the correct one :-
ar = [1,23,56,11]
ar.each_cons(2).max_by { |a, b| (b - a).abs }
# => [56, 11]