I am using hacker rank and I do not understand why my ruby code only works for one test case out of like 20. Here is the question:

John Watson knows of an operation called a right circular rotation on an array of integers. One rotation operation moves the last array element to the first position and shifts all remaining elements right one. To test Sherlock's abilities, Watson provides Sherlock with an array of integers. Sherlock is to perform the rotation operation a number of times then determine the value of the element at a given position.

For each array, perform a number of right circular rotations and return the values of the elements at the given indices.

Function Description

Complete the circularArrayRotation function in the editor below.

circularArrayRotation has the following parameter(s):

• int a[n]: the array to rotate
• int k: the rotation count
• int queries[1]: the indices to report

Returns

int[q]: the values in the rotated a as requested in m

Input Format

The first line contains 3 space-separated integers, n, k, and q, the number of elements in the integer array, the rotation count and the number of queries. The second line contains n space-separated integers, where each integer i describes array element a[i] (where 0 <= i < n). Each of the q subsequent lines contains a single integer, queries[i], an index of an element in a to return.

Constraints

Sample Input 0

3 2 3
1 2 3
0
1
2

Sample Output 0

2
3
1

Here is my code :

def circularArrayRotation(a, k, queries)
    q = []
    
    while k >= 1
        m = a.pop()
        a.unshift m
        k = k - 1
    end
    
    for i in queries do
     v = a[queries[i]]
    q.push v
    
    end 
    
    return q

end

It only works for the sample text case but I can't figure out why. Thanks for any help you can provide.

CodePudding user response：

A little help with the parsing of the input:

#!/usr/bin/env ruby
  
input = ARGF.readlines.map(&:split)
n,k,q = input.shift.map(&:to_i)
a = input.shift.map(&:to_i)
queries = input.flatten.map(&:to_i)

puts <<-EOF
n = #{n}
k = #{k}
q = #{q}
a = #{a}
queries = #{queries}
EOF

output (from given input):

n = 3
k = 2
q = 3
a = [1, 2, 3]
queries = [0, 1, 2]

CodePudding user response：

I don't see any errors in your code, but I would like to suggest a more efficient way of making the calculation.

First observe that after q rotations the element at index i will at index (i q) % n.

For example, suppose

n = 3
a = [1,2,3]
q = 5

Then after q rotations the array will be as follows.

arr = Array.new(3)
arr[(0 5) % 3] = a[0] #=> arr[2] = 1
arr[(1 5) % 3] = a[1] #=> arr[0] = 2
arr[(2 5) % 3] = a[2] #=> arr[1] = 3
arr                   #=> [2,3,1] 

We therefore can write

def doit(n,a,q,queries)      
  n.times.with_object(Array.new(n)) do |i,arr|
    arr[(i q) % n] = a[i]
  end.values_at(*queries)
end

doit(3,[1,2,3],5,[0,1,2])
  #=> [2,3,1]
doit(3,[1,2,3],5,[2,1])
  #=> [1, 3]
doit(3,[1,2,3],2,[0,1,2])
  #=> [2, 3, 1]
p doit(3,[1,2,3],0,[0,1,2])
  #=> [1,2,3]
doit(20,(0..19).to_a,25,(0..19).to_a.reverse)
  #=> [14, 13, 12, 11, 10, 9, 8, 7, 6, 5,
  #    4, 3, 2, 1, 0, 19, 18, 17, 16, 15]

Alternatively, we may observe that after q rotations the element at index j was initially at index (j n-q) % n.

For the earlier example, letting nmq = n-q = 3-5 = -2, after q rotations the array will be

[a[(0-2) % 3], a[(1-2) % 3], a[(2-2) % 3]]
  #=> [a[1], a[2], a[0]]
  #=> [2,3,1]

We therefore could instead write

def doit(n,a,q,queries)
  nmq = n-q
  n.times.map { |j| a[(j nmq) % n] }.values_at(*queries)
end