diff --git a/007-MaximumSubArray.py b/007-MaximumSubArray.py
new file mode 100644
index 0000000..320af94
--- /dev/null
+++ b/007-MaximumSubArray.py
@@ -0,0 +1,46 @@
+# Brute Force Approach:
+
+def maxSubArray1(nums):
+    max_sum = 0
+    n = len(nums)
+
+    for i in range(n):
+        for j in range(i, n):
+            curr_sum = sum(nums[i:j+1])
+            max_sum = max(max_sum, curr_sum)                        # Time Complexity = O(n^3)
+                                                                    # Space Complexity= O(n)
+    return max_sum
+
+nums = [-2, 1, -3, 4, -1, 2, 1, -5, 4]
+print(maxSubArray1(nums))
+
+
+# Better Approach:
+
+def maxSubArray2(nums):
+    maxSub = nums[0]
+    curr_sum = 0
+
+    for n in nums:
+        if curr_sum < 0:
+            curr_sum = 0
+        curr_sum += n
+        maxSub = max(curr_sum, maxSub)
+    return maxSub
+
+nums = [-2, 1, -3, 4, -1, 2, 1, -5, 4]                               # Time Complexity = O(n)
+print(maxSubArray2(nums))                                            # Space Complexity= O(1)
+
+
+# Kadane's Algorithm:
+
+def maxSubArray3(nums):
+    max_sum, curr_sum = 0, 0
+
+    for i in nums:
+        curr_sum = max(i, curr_sum + i)
+        max_sum = max(max_sum, curr_sum)
+    return max_sum
+nums = [-2, 1, -3, 4, -1, 2, 1, -5, 4]                               # Time Complexity = O(n)
+print(maxSubArray3(nums))                                            # Space Complexity= O(1)
+