See the original problem on HackerRank.
You are given a sequence of integers S
, your task is to find a pair
i,j
of indices with i<j
that maximizes the difference
S[j]S[i]
.
Note that what makes this challenge more interesting is the requirement of order, that i < j (without it you simply need to find the maximum and minimum element of S).
You have to output the value of the maximum difference.
Input Format
An integer N
followed by N
space separated integers S[i]
on a
new line.
Constraints
1 < N <= 10^6
0 <= S[i] <= 10^4
Output Format
The maximum difference as an integer.
Solutions
This problem has a naive quadratic solution that involves considering every pairs \(i,j\) with \(j>i\) and tracking the maximum. What makes this naive is that it is a bruce force solution that does not consider the input domain or the structure of the problem.
To see the solution, consider what pairs we should consider for the index \(j\). Let’s suppose that \(j\) is the right hand index so we need only look at elements of \(S\) that precede \(j\). The difference \(s[j]s[i]\) is maximized for the index \(i\) that contains the minimum element of \(s[1],…s[j1]\).
Therefore, instead of looking over pairs, we need only to keep track of the maximum element preceding any other index. This is linear.
Here is a possible C++ implementation of the described idea:


And a similare one in Rust:


The solution above hides two emerging patterns:
 we calculate a running minimum (
min
is updated independently at every step)  we reduce to the max value on the difference between the current element and
 the current running minimum
In Rust we can express it with scan
(to calculate the running minimum) and
max
to reduce to the maximum value.


In C++ we can express it using two functions of the standard library:
partial_sum
(where the sum
operation is overriden by min
) and
inner_product
, which accumulates inner products (where the multiply
operation is overridden by minus
) using the function max
instead of sum
.


From C++20, ranges’s view::partial_sum
can help remove the support array:


It should be clear that mins
is a lazy range and not a vector
. The entire
chain is lazy so the algorithm is onepass.
The same concept can be expressed by Rust and iterators:


Haskell can be very terse when taking advantage of standard library higher
order functions, like scanl1
. The solution below is a complete program.
It uses scanl1
(a particular form of scan
where the first element is used as
init) and zipWith
to find differences.


Another solution  less efficent and for some people preparatory for the previous solution  consists in using two prefix sums:


Just for fun, an alternative solution is based on the Maximum Subarray Sum problem. We first calculate the differences between adjacent elements and then solve the maximum subarray problem on the resulting array.
In C++ we can combine several standard algorithms:

