Divide and Conquer to find maximum difference in an array - c++

Question

I am trying to solve a problem where given an array I need to calculate the maximum difference such that the larger element appears after the smaller element.
Here is a better problem statement:
Given the stock prices on each day for n days, what is the maximum profit a person can make by doing exactly one transaction. One transaction means that the person can buy exactly one stock on one day and sell it on a later date.
I am trying to solve this problem using divide and conquer algo.
In my recursive function i am trying to spilt the array into two halves, but i am not sure on how to proceed with logic. Do i just get the max difference in each halves and compare?
int calculateMaxDiff(int *src, int start, int end){
if(end - start == 1) return src[start];
int middle = (start + end)/ 2;
int half1_diff;
int half2_diff;
half1_diff = calculateMaxDiff(src, start, middle);
half2_diff = calculateMaxDiff(src, middle, end);
//Do i need to have two loops here that calculate the diffs for each halves
....
return max(half1_diff, half2_diff);
}
Edit: Example output
Give an array {12, 9, 18, 3, 7, 11, 6, 15, 6, 1 ,10} should return 12 as a result of difference between 15 and 3

Answer 1

The question in your problem can be translated into a better problem statement:
Given the stock prices on each day for n days, what is the maximum profit a person can make by doing exactly one transaction. One transaction means that the person can buy exactly one stock on one day and sell it on a later date.
The divide-and-conquer solution: Let's see if we can solve this by splitting the input in half, solving the problem in each subarray, then combining the two together. Turns out we actually can do this, and can do so efficiently! The intuition is as follows. If we have a single day, the best option is to buy on that day and then sell it back on the same day for no profit. Otherwise, split the array into two halves. If we think about what the optimal answer might be, it must be in one of three places:
The correct buy/sell pair occurs completely within the first half.
The correct buy/sell pair occurs completely within the second half.
The correct buy/sell pair occurs across both halves - we buy in the first half, then sell in the second half.
We can get the values for (1) and (2) by recursively invoking our algorithm on the first and second halves. For option (3), the way to make the highest profit would be to buy at the lowest point in the first half and sell in the greatest point in the second half. We can find the minimum and maximum values in the two halves by just doing a simple linear scan over the input and finding the two values. This then gives us an algorithm with the following recurrence:
T(n) = 2T(n/2) + O(n)
T(n) = O(nlogn)
Here is a simple implementation in Python. Its very simple to understand and its also easy to convert to C++:
def DivideAndConquerSingleSellProfit(arr):
# Base case: If the array has zero or one elements in it, the maximum
# profit is 0.
if len(arr) <= 1:
return 0;
# Cut the array into two roughly equal pieces.
left = arr[ : len(arr) / 2]
right = arr[len(arr) / 2 : ]
# Find the values for buying and selling purely in the left or purely in
# the right.
leftBest = DivideAndConquerSingleSellProfit(left)
rightBest = DivideAndConquerSingleSellProfit(right)
# Compute the best profit for buying in the left and selling in the right.
crossBest = max(right) - min(left)
# Return the best of the three
return max(leftBest, rightBest, crossBest)
Edit: Here is the C++ implementation for the above algorithm
#include <iostream>
#include <algorithm>
using namespace std;
int calculateMin(int a[], int low, int high)
{
int i,mini;
mini = a[low];
for(i=low;i<=high;i++)
{
if(a[i]<mini)
{
mini = a[i];
}
}
return mini;
}
int calculateMax(int a[], int low, int high)
{
int i,maxi;
maxi = a[low];
for(i=low;i<=high;i++)
{
if(a[i]>maxi)
{
maxi = a[i];
}
}
return maxi;
}
int calculateMaxDiff(int a[], int low, int high)
{
if(low>=high)
return 0;
int mid = (low+high)/2;
int leftPartition = calculateMaxDiff(a,low,mid);
int rightPartition = calculateMaxDiff(a,mid+1,high);
int left = calculateMin(a,low,mid); // calculate the min value in the left partition
int right = calculateMax(a,mid+1,high); // calculate the max value in the right partition
return max(max(leftPartition, rightPartition), (right - left));
}
int main() {
int arr[] = {12, 9, 18, 3, 7, 11, 6, 15, 6, 1 ,10};
int len = sizeof(arr)/sizeof(arr[0]);
int ans = calculateMaxDiff(arr, 0, len-1);
cout << "Maximum Profit: " <<ans<<endl;
return 0;
}
Hope it helps!!!

Answer 2

There is no need in complicated D/C algorithm because simple cycle with checking like
maxdiff = max(current - min_so_far, maxdiff)
update min_so_far
solves the problem
If you really want to apply divide and conquer method, you may return triplet {local_min, local_max, local_max_diff} from recursive function like:
left = calculateMaxDiff(start, middle)
right = calculateMaxDiff(middle + 1, end)
return {min(left.local_min, right.local_min),
max(left.local_max, right.local_max),
max(left.local_diff, right.local_diff, right.localmax - left.local_min)

Answer 3

The key for a divide and conquer algorithm is the conquer part.
For this problem the most important condition is:
the larger element appears after the smaller element
For an array src, after dividing src to 2 halves, half1 and half2, suppose the answer would be in position i and j, there are 3 cases now:
i and j are both in half1 -> half1_diff
i and j are both in half2 -> half2_diff
i is in half1 and j is in half2
So the main part is to deal with case3. As the larger one comes after, so we just need to find the minimum value min_half1 in half1 and the maximum value max_half2 in half2, and check if it meets the condition max_half2 >= min_half1 and update the result as max(half1_diff, half2_diff, max_half2-min_half1).
In order to calculate min_half1 and max_half2 efficiently, you have to keep the record of min and max value of the array, and it takes O(1) time.
So T(n) = 2T(n/2) + O(1), T(n) = O(n).
Check the example for more details
http://ideone.com/TbIL2r