**Problem**

An array is increasing if each element is less than its succeeding element except for the last element.

Given an array A of n elements return the beginning and ending indices of a longest increasing subarray of A.

**Solution**

Let S[i] be the longest increasing subarray between the indexes (0,i – 1).

Then

S[i] = a,b if A[i] > A[i – 1],

i,i otherwise

where a,b = S[i – 1]

In haskell this would be:

longest_contig_inc_subarray :: (Ord a) => [a] -> (Int, Int)
longest_contig_inc_subarray [] = (-1, -1)
longest_contig_inc_subarray (x:xs) = longest_contig_inc_subarray' (0, x, 0, x) xs
where
longest_contig_inc_subarray' (i,ai,j,aj) [] = (i,j)
longest_contig_inc_subarray' (i,ai,j,aj) (x:xs)
| x >= aj = longest_contig_inc_subarray' (i,ai,j + 1,x) xs
| otherwise = longest_contig_inc_subarray' (j + 1,x,j + 1,x) xs

A heuristic to improve the best case complexity (but does nothing in the worst case) is to realise that if the length of the longest subarray till i is L (and A[i + 1] < A[i] – indicating an end of the longest subarray), then a larger increasing subarray must contain *atleast* L elements. So we only need to start with L items in front and check backwards.

The code for this is (here the start index and the length of the subarray are returned instead):

-- Returns the size of the largest increasing "prefix" in an array
largest_inc_prefix [] = 0
largest_inc_prefix (x:[]) = 1
largest_inc_prefix (x:y:xs)
| x = y = 1 + largest_dec_prefix (y:xs)
| otherwise = 1
-- Returns the size of the largest decreasing "prefix" in an array
largest_dec_prefix [] = 0
largest_dec_prefix (x:[]) = 1
largest_dec_prefix (x:y:xs)
| x >= y = 1 + largest_dec_prefix (y:xs)
| otherwise = 1
lcisa :: (Ord a) => [a] -> (Int, Int)
lcisa [] = (-1,-1)
lcisa xs = lcisa' (0,1) 0 xs
where
lcisa' (start,maxlen) i [] = (start,maxlen)
lcisa' (start,maxlen) i xs
| nextlen > maxlen = lcisa' nextbest
(i + maxlen + inc_prefix)
(drop inc_prefix rest)
| otherwise = lcisa' (start,maxlen) (i + maxlen) rest
where
first_l = take maxlen xs
rest = drop maxlen xs
dec_prefix = largest_dec_prefix (reverse first_l)
inc_prefix = largest_inc_prefix rest
nextlen = inc_prefix + dec_prefix
nextbest = (i + maxlen - dec_prefix, nextlen)

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EPI 6.6 – Longest contiguous increasing subarray”