Largest Forward Difference
January 20, 2015
The obvious brute-force solution computes all differences and takes the largest:
(define (largest-forward-difference xs)
(let ((lfd 0))
(do ((xs xs (cdr xs))) ((null? xs) lfd)
(do ((ys xs (cdr ys))) ((null? ys))
(set! lfd (max lfd (- (car ys) (car xs))))))))
This takes O(n2) time and O(1) space. Here are some examples:
> (largest-forward-difference '(1 5 7 2 9))
8
> (largest-forward-difference '(4 3 2 1))
0
A better solution scans the array from left to right. After initializing with the first two integers in the array, at each step during a left-to-right scan of the remainder of the array the largest forward difference is either the largest forward difference at the end of the previous step, or the difference between the current integer and the minimum integer to its left:
(define (largest-forward-difference xs)
(let loop ((min-to-left (min (car xs) (cadr xs)))
(max-so-far (- (cadr xs) (car xs)))
(xs (cddr xs)))
(if (null? xs) max-so-far
(let* ((min-to-left (min min-to-left (car xs)))
(diff-ending-here (- (car xs) min-to-left)))
(loop min-to-left (max max-so-far diff-ending-here) (cdr xs))))))
This takes O(n) time and O(1) space. Here are some examples:
> (largest-forward-difference '(1 5 7 2 9))
8
> (largest-forward-difference '(4 3 2 1))
0
You can run the program at http://programmingpraxis.codepad.org/qOdxb2Yu.
Programming Praxis 
Haskell:
var ar = new[] { 1,5,7,2,9 }; int largestForwardDiff = 0; for (int i = ar.Length - 1; i >= 1; i--) { for (int ii = 0; ii < i; ii++) { var forwardDiff = ar[i] - ar[ii]; if (forwardDiff > largestForwardDiff) { largestForwardDiff = forwardDiff; } } }O(n) solution in Python
def max_forward_diff(seq): max_diff = 0 min_n = None for n in seq: if min_n is None or n < min_n: min_n = n max_diff = max(max_diff, n - min_n) return max_diffOr as a one-liner:
import itertools as it import operator as op def max_fwd_diff(seq): return max(map(op.sub, seq, it.accumulate(seq, min)))I see that there are O(n^2) solutions.
Is a O(n) solution possible?
1. in array A find the maximum value x at position i.
2. find the minimum value y at position j where 0 <= j <= i.
3. largest forward difference is then max(x – y, 0).
That doesn’t work. Consider the array (9,1,8). The largest forward difference is 8 – 1 = 7, but you will find the maximum value 9, find nothing to the left of it, and report a largest forward difference of 0.
;; O(n) solution (defun l-f-d (arr) (loop for i below (length arr) with mins = (make-array (length arr) :initial-element (aref arr 0)) with maxs = (make-array (length arr) :initial-element (aref arr (1- (length arr)))) when (> i 0) do (setf (aref mins i) (min (aref arr i) (aref mins (1- i)))) when (< i (1- (length arr))) do (setf (aref maxs i) (max (aref arr i) (aref maxs (1+ i)))) finally (return (max 0 (loop for i below (length arr) maximize (- (aref maxs i) (aref mins i)))))))Another solution in C# more powerful than the brut force one:
int largestForwardDiff = 0; int right = ar[ar.Length - 1]; int left = ar[0]; int currentDiff = right - left; int tempRight = 0; for (int i = ar.Length - 2; i >= 0; i--) { if (ar[i] > right && ar[i] > tempRight) { tempRight = ar[i]; } else if (ar[i] < left || (tempRight - ar[i] > currentDiff)) { left = ar[i]; if (tempRight != 0) { right = tempRight; tempRight = 0; } currentDiff = right - left; } } if (right > left) { largestForwardDiff = right - left; }Here’s a nice compact C++ solution, assumes at least 1 element, don’t think arithmetic overflow is a problem:
#include <algorithm> int maxfd(int *a, int len) { int min = a[0], fd = 0; for (int i = 1; i < len; i++) { if (a[i] < min) min = a[i]; else fd = std::max(fd, a[i]-min); } return fd; }A simple brute-force solution with some minor optimizations, namely going from the rightmost element comparing that with the numbers to its left, and than going from the second right-most element etc. until you reach the left-most element, avoiding duplicate comparisons.
In Rust: http://xojoc.pw/programming-praxis/lfd.html
O(n) version in Haskell:
Simple Solution is: [1,5,7,2,9].inject {|a,b| (a-b).abs } = 8
[4, 3, 2, 1].inject {|a,b| (a-b).abs } = 0
max(x-cummin(x)) in R, I think.