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Maximizing the Difference - Medium

Problem : Given an array A of integers, maximize A[j] - A[i] such that 0 <= i < j < len(A).

A very interesting problem, as always lets begin with a naive solution.

def max_diff_naive(L):
max_diff = start = end = 0
for i in xrange(len(L)):
for j in xrange(i+1,len(L)):
if max_diff < L[j] - L[i]:
start,end = i,j
max_diff = L[j] - L[i]
return max_diff,start,end
view raw max_diff_naive hosted with ❤ by GitHub
Clearly, the solution has quadratic time complexity. Now, step back and look at the solution. Essentially what we are doing is iterating over the array, splitting the array into two parts. From the first part we always pick the lowest element, (clearly since that will give us the maximum difference). Since we already know the minimum element we encountered, we simply need to check if the difference with the new element improves upon the earlier solution. This idea can be implemented in Linear Time. Here's how.

def max_diff_linear(L):
min_index = 0
max_diff = start = end = 0
for i in xrange(len(L)):
if L[i] < L[min_index]: min_index = i
if max_diff < L[i] - L[min_index]:
max_diff = L[i] - L[min_index]
start,end = min_index,i
return max_diff,start,end
view raw max_diff_linear hosted with ❤ by GitHub

Update: So I just discovered this problem is called the Single Sell Profit problem. It's been discussed quite thoroughly by templatetypedef here. It includes 4 solutions (including this one).

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