max_sub_array.py: Add Maximum Subarray

Divide and Conquer method to solve maximum subarray problem

closes NITSkmOS#215

README.md

@@ -22,6 +22,7 @@ This repository contains examples of various algorithms written on different pro
 | [Bubble Sort](https://en.wikipedia.org/wiki/Bubble_sort)                                        | [:octocat:](bubble_sort/C)            | [:octocat:](bubble_sort/Cpp)          |                                       |                                       |
 | [Shell Sort](https://en.wikipedia.org/wiki/Shellsort)                                           | [:octocat:](shell_sort/C)             |                                       |                                       | [:octocat:](shell_sort/Python)        |
 | [Heap Sort](https://en.wikipedia.org/wiki/Heapsort)                                             |                                       |                                       |                                       | [:octocat:](heap_sort/python)        |
+| [Maximum Subarray Problem](https://en.wikipedia.org/wiki/Maximum_subarray_problem)              |                                       |                                       |                                       | [:octocat:](/maximum_subarray/Python)|
 
 
 ## Implemented Data Structures

maximum_subarray/Python/max_sub_array.py

@@ -0,0 +1,102 @@
+"""
+Here Maximum SubArray problem has been implemented on the Share
+Market Data Analysis. Where prices of every day has been given.
+We have to find the buying and selling day so that the profit will
+be maximum. Here the Divide and Conquer method is used and
+time-complexity is O(nlogn).For more information visit-
+<https://en.wikipedia.org/wiki/Maximum_subarray_problem>
+"""
+
+
+def max_sub_array(arr, low, high):
+    """
+    Method to find maximum sub array.
+
+    :param low: kjndn
+    :param high: kjndknfg
+    """
+    if low == high:
+        # if there is only one day
+        return low, low + 1, arr[low]
+
+    else:
+        mid = int((low + high) / 2)
+        # ll, lh, ls are respectively left-low, left-high, left-sum
+        # rl, rh, rs are respectively right-low, right-high, right-sum
+        # cl, ch, cs are respectively cross-low, cross-high, cross-sum
+
+        ll, lh, ls = max_sub_array(arr, low, mid)
+        rl, rh, rs = max_sub_array(arr, mid + 1, high)
+        cl, ch, cs = cross_sub_array(arr, low, mid, high)
+
+        max_sum = max(ls, rs, cs)
+
+        if max_sum is ls:
+            return ll, lh, ls
+        elif max_sum is rs:
+            return rl, rh, rs
+        else:
+            return cl, ch, cs
+
+
+def cross_sub_array(arr, low, mid, high):
+    """
+    :param mid: lsum is left sum
+    """
+    lsum = -10000
+    sum = 0
+    maxl = mid
+    maxr = mid + 1
+    for i in range(mid, low - 1, -1):
+        sum = sum + arr[i]
+        if sum > lsum:
+            lsum = sum
+            maxl = i
+
+    # rsum is right sum
+    rsum = -10000
+    sum = 0
+    for i in range(mid + 1, high + 1):
+        sum = sum + arr[i]
+        if sum > rsum:
+            rsum = sum
+            maxr = i + 1
+    return maxl, maxr, (lsum + rsum)
+
+
+def main():
+    # price is array of the prices where each index of array
+    # represents the day.
+    # price_difference is array of profit(either +ve or -ve)
+
+    price = [4, 9, 5, 13, 16, 7, 8]
+
+    if len(price) <= 1:
+        print('Same day purchase and sell.So no profit.')
+        return
+
+    price_difference = []
+    for day in range(1, len(price)):
+        price_difference.append(price[day] - price[day - 1])
+
+    # l is day of purchasing
+    # h is day of selling
+    # s is profit
+
+    l, h, s = max_sub_array(price_difference, 0, (len(price_difference) - 1))
+    if s < 0:
+        # if it happens that price drops and never rises to
+        # the price when it was purchased
+
+        print('No days found to get profit')
+    else:
+        # To be more real-life oriented we start counting from day-1 instead of
+        # day-0
+
+        print('Day of Purchase:', l + 1,
+              '\nDay of Selling:', h + 1,
+              '\nProfit:', s)
+
+
+if __name__ == '__main__':
+    main()