diff --git a/divide_and_conquer/closest_pair_of_points.py b/divide_and_conquer/closest_pair_of_points.py
new file mode 100644
index 000000000..cc5be428d
--- /dev/null
+++ b/divide_and_conquer/closest_pair_of_points.py
@@ -0,0 +1,113 @@
+"""
+The algorithm finds distance btw closest pair of points in the given n points.
+Approach used -> Divide and conquer 
+The points are sorted based on Xco-ords 
+& by applying divide and conquer approach, 
+minimum distance is obtained recursively.
+
+>> closest points lie on different sides of partition
+This case handled by forming a strip of points 
+whose Xco-ords distance is less than closest_pair_dis
+from mid-point's Xco-ords.
+Closest pair distance is found in the strip of points. (closest_in_strip)
+
+min(closest_pair_dis, closest_in_strip) would be the final answer.
+ 
+Time complexity: O(n * (logn)^2)
+"""
+
+
+import math 
+
+
+def euclidean_distance_sqr(point1, point2):
+    return pow(point1[0] - point2[0], 2) + pow(point1[1] - point2[1], 2)
+
+
+def column_based_sort(array, column = 0):
+    return sorted(array, key = lambda x: x[column])
+    
+
+def dis_between_closest_pair(points, points_counts, min_dis = float("inf")):
+    """ brute force approach to find distance between closest pair points
+
+    Parameters : 
+    points, points_count, min_dis (list(tuple(int, int)), int, int) 
+    
+    Returns : 
+    min_dis (float):  distance between closest pair of points
+
+    """
+
+    for i in range(points_counts - 1):
+        for j in range(i+1, points_counts):
+            current_dis = euclidean_distance_sqr(points[i], points[j])
+            if current_dis < min_dis:
+                min_dis = current_dis
+    return min_dis
+
+
+def dis_between_closest_in_strip(points, points_counts, min_dis = float("inf")):
+    """ closest pair of points in strip
+
+    Parameters : 
+    points, points_count, min_dis (list(tuple(int, int)), int, int) 
+    
+    Returns : 
+    min_dis (float):  distance btw closest pair of points in the strip (< min_dis)
+
+    """
+
+    for i in range(min(6, points_counts - 1), points_counts):
+        for j in range(max(0, i-6), i):
+            current_dis = euclidean_distance_sqr(points[i], points[j])
+            if current_dis < min_dis:
+                min_dis = current_dis
+    return min_dis
+
+
+def closest_pair_of_points_sqr(points, points_counts):
+    """ divide and conquer approach
+
+    Parameters : 
+    points, points_count (list(tuple(int, int)), int) 
+    
+    Returns : 
+    (float):  distance btw closest pair of points 
+
+    """
+
+    # base case
+    if points_counts <= 3:
+        return dis_between_closest_pair(points, points_counts)
+    
+    # recursion
+    mid = points_counts//2
+    closest_in_left = closest_pair_of_points(points[:mid], mid)
+    closest_in_right = closest_pair_of_points(points[mid:], points_counts - mid)
+    closest_pair_dis = min(closest_in_left, closest_in_right)
+    
+    """ cross_strip contains the points, whose Xcoords are at a 
+    distance(< closest_pair_dis) from mid's Xcoord
+    """
+
+    cross_strip = []
+    for point in points:
+        if abs(point[0] - points[mid][0]) < closest_pair_dis:
+            cross_strip.append(point)
+
+    cross_strip = column_based_sort(cross_strip, 1)
+    closest_in_strip = dis_between_closest_in_strip(cross_strip, 
+                     len(cross_strip), closest_pair_dis)
+    return min(closest_pair_dis, closest_in_strip)
+
+    
+def closest_pair_of_points(points, points_counts):
+    return math.sqrt(closest_pair_of_points_sqr(points, points_counts))
+
+
+points = [(2, 3), (12, 30), (40, 50), (5, 1), (12, 10), (0, 2), (5, 6), (1, 2)]
+points = column_based_sort(points)
+print("Distance:", closest_pair_of_points(points, len(points)))
+
+
diff --git a/divide_and_conquer/max_subarray_sum.py b/divide_and_conquer/max_subarray_sum.py
new file mode 100644
index 000000000..0428f4e13
--- /dev/null
+++ b/divide_and_conquer/max_subarray_sum.py
@@ -0,0 +1,75 @@
+""" 
+Given a array of length n, max_subarray_sum() finds 
+the maximum of sum of contiguous sub-array using divide and conquer method.
+
+Time complexity : O(n log n)
+
+Ref : INTRODUCTION TO ALGORITHMS THIRD EDITION 
+(section : 4, sub-section : 4.1, page : 70)
+
+"""
+
+
+def max_sum_from_start(array):
+    """ This function finds the maximum contiguous sum of array from 0 index
+
+    Parameters : 
+    array (list[int]) : given array
+    
+    Returns : 
+    max_sum (int) : maximum contiguous sum of array from 0 index
+
+    """
+    array_sum = 0
+    max_sum = float("-inf")
+    for num in array:
+        array_sum += num
+        if array_sum > max_sum:
+            max_sum = array_sum
+    return max_sum
+
+
+def max_cross_array_sum(array, left, mid, right):
+    """ This function finds the maximum contiguous sum of left and right arrays
+
+    Parameters : 
+    array, left, mid, right (list[int], int, int, int) 
+    
+    Returns : 
+    (int) :  maximum of sum of contiguous sum of left and right arrays
+
+    """
+
+    max_sum_of_left = max_sum_from_start(array[left:mid+1][::-1])
+    max_sum_of_right = max_sum_from_start(array[mid+1: right+1])
+    return max_sum_of_left + max_sum_of_right
+
+
+def max_subarray_sum(array, left, right):
+    """ Maximum contiguous sub-array sum, using divide and conquer method
+
+    Parameters : 
+    array, left, right (list[int], int, int) : 
+    given array, current left index and current right index
+    
+    Returns : 
+    int :  maximum of sum of contiguous sub-array
+
+    """
+
+    # base case: array has only one element
+    if left == right:
+        return array[right]
+    
+    # Recursion
+    mid = (left + right) // 2
+    left_half_sum = max_subarray_sum(array, left, mid)
+    right_half_sum = max_subarray_sum(array, mid + 1, right)
+    cross_sum = max_cross_array_sum(array, left, mid, right)
+    return max(left_half_sum, right_half_sum, cross_sum)
+
+
+array = [-2, -5, 6, -2, -3, 1, 5, -6]
+array_length = len(array)
+print("Maximum sum of contiguous subarray:", max_subarray_sum(array, 0, array_length - 1))
+