## Largest Forward Difference

### January 20, 2015

In an array of integers, the forward difference between any two elements is the rightmost integer less the leftmost integer. The largest forward difference is the greatest value of all forward differences between elements of the array. If there are no positive forward differences, the largest forward difference is taken to be zero, as if an integer is subtracted from itself.

For instance, in the array [1,5,7,2,9] the largest forward difference is 9 – 1 = 8, and in the array [4, 3, 2, 1] the largest forward difference is 0.

Your task is to write a program that finds the largest forward difference in an array. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

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### 13 Responses to “Largest Forward Difference”

1. Francesco said

```fd (i:[]) = 0
fd (i:is) = max (maximum \$ map (subtract i) is) (fd is)
```
2. klettier said
```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;
}
}
}
```
3. Mike said

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_diff
```

Or 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)))
```
4. inoakoala said

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).

5. programmingpraxis said

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.

6. Axio said
```;; 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)))))))
```
7. klettier said

Another solution in C# more powerful than the brut force one:

```int largestForwardDiff = 0;

int right = ar[ar.Length - 1];
int left = ar;
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;
}
```
8. matthew said

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, 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;
}
```
9. Brett Warren said

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.

```from itertools import islice

def max_forward_diff(num_set):
diffs = list()

for i, n in list(enumerate(num_set))[::-1]:
for m in num_set[:i][::-1]:
diffs.append(n-m)
if n-m == max(num_set)-1 and n-m > 0:
return n - m

return max(diffs) if max(diffs) > 0 else 0

if __name__ == "__main__":
print(max_forward_diff([1,3,2,4]))
```
10. Mike said

```mfd xs = maximum . zipWith (-) xs \$ scanl1 min xs