I recently came up across a problem/solution that used Loop Do. I seldom have seen this so far in my learning Ruby Programming (I am a beginner with no CS experience).
# Write a function, `nearest_larger(arr, i)` which takes an array and an
# index. The function should return another index, `j`: this should
# satisfy:
#
# (a) `arr[i] < arr[j]`, AND
# (b) there is no `j2` closer to `i` than `j` where `arr[i] < arr[j]`.
#
# In case of ties (see example beow), choose the earliest (left-most)
# of the two indices. If no number in `arr` is largr than `arr[i]`,
# return `nil`.
#
# Difficulty: 2/5
describe "#nearest_larger" do
it "handles a simple case to the right" do
nearest_larger([2,3,4,8], 2).should == 3
end
it "handles a simple case to the left" do
nearest_larger([2,8,4,3], 2).should == 1
end
it "treats any two larger numbers like a tie" do
nearest_larger([2,6,4,8], 2).should == 1
end
it "should choose the left case in a tie" do
nearest_larger([2,6,4,6], 2).should == 1
end
it "handles a case with an answer > 1 distance to the left" do
nearest_larger([8,2,4,3], 2).should == 0
end
it "handles a case with an answer > 1 distance to the right" do
nearest_larger([2,4,3,8], 1).should == 3
end
it "should return nil if no larger number is found" do
nearest_larger( [2, 6, 4, 8], 3).should == nil
end
end
SOLUTION
def nearest_larger(arr, idx)
diff = 1
loop do
left = idx - diff
right = idx + diff
if (left >= 0) && (arr[left] > arr[idx])
return left
elsif (right < arr.length) && (arr[right] > arr[idx])
return right
elsif (left < 0) && (right >= arr.length)
return nil
end
diff += 1
end
end
nearest_larger([2,4,3,8], 1)
Can someone please explain to me when is the best time to use a "loop do" construct instead of the usual "while" or "unless" or "each" construct?