diff --git a/.gitignore b/.gitignore
index a1c09e19..21b5bc82 100644
--- a/.gitignore
+++ b/.gitignore
@@ -10,3 +10,5 @@ Thumbs.db
 # binary files
 *.out
 *.class
+*.idea
+
diff --git a/algorithms/Go/README.md b/algorithms/Go/README.md
index ffa0542a..bbc1737a 100644
--- a/algorithms/Go/README.md
+++ b/algorithms/Go/README.md
@@ -1,5 +1,9 @@
 # Go
 
+## Arrays
+1. [Maximum subarray sum (Kadane's Algorithm)](arrays/maximum-subarray-sum.go)
+2. [Two Sum](arrays/two-sum.go)
+
 ## Scheduling
 1. [Interval Scheduling](scheduling/interval-scheduling.go)
 
diff --git a/algorithms/Go/arrays/maximum-subarray-sum.go b/algorithms/Go/arrays/maximum-subarray-sum.go
new file mode 100644
index 00000000..01d5b239
--- /dev/null
+++ b/algorithms/Go/arrays/maximum-subarray-sum.go
@@ -0,0 +1,38 @@
+/*
+	Package arrays maximumSubarraySum problem
+	Input: An array of integers
+	Output: Subarray with maximum sum
+	The Strategy: Keep track of the previous encounter sum and the current element.
+ 				  If the current element is largest than the previous sum + current element
+				  then throw the previous sum and start with the current element as the largest sum.
+	Algorithm Type: Kadane's Algorithm
+	Time Complexity: O(n)
+	Dry Run :
+	 		[-2,1,-3,4,-1,2,1,-5,4]
+	initial := largestSum = -2 , currentSum = -2
+	i=1     -> currentSum = max(1,  (-2 + 1 )) = 1 ,      largestSum = max(-2, 1) = 1
+	i=2		-> currentSum = max(-3, (1 + -3 )) = -2,	  largestSum = max(1, -2) = 1
+	i=3 	-> currentSum = max(4,  (-2 + 4 )) = 4 , 	  largestSum = max(1, 4)  = 4
+	i=4		-> currentSum = max(-1, (4 + -1 )) = 3 , 	  largestSum = max(4, 3) = 4
+	i=5		-> currentSum = max(2,  (3 +  2 )) = 5 , 	  largestSum = max(4, 5) = 5
+	i=6		-> currentSum = max(1,  (5 + 1  )) = 6 , 	  largestSum = max(5, 6) = 6
+	i=7		-> currentSum = max(-5, (6 + -5 )) = 1 , 	  largestSum = max(6, 1) = 6
+	i=8		-> currentSum = max(4,  (1 + 4 ))  = 5 , 	  largestSum = max(6, 5) = 6
+*/
+package arrays
+
+import (
+	"math"
+)
+
+// To find the maximum subarray sum.
+func maximumSubarraySum(nums []int) int {
+	maximumSum := nums[0]
+	currentSum := maximumSum
+	for i := 1; i < len(nums); i++ {
+		currentSum = int(math.Max(float64(nums[i]), float64(nums[i]+currentSum)))
+		maximumSum = int(math.Max(float64(maximumSum), float64(currentSum)))
+	}
+
+	return maximumSum
+}
diff --git a/algorithms/Go/arrays/maximum_subarray_sum_test.go b/algorithms/Go/arrays/maximum_subarray_sum_test.go
new file mode 100644
index 00000000..93feb57b
--- /dev/null
+++ b/algorithms/Go/arrays/maximum_subarray_sum_test.go
@@ -0,0 +1,11 @@
+package arrays
+
+import "testing"
+
+func TestMaximumSubarraySum(t *testing.T) {
+	sampleArray := []int{-2, 1, -3, 4, -1, 2, 1, -5, 4}
+	want := 6
+	if got := maximumSubarraySum(sampleArray); want != got {
+		t.Errorf("want %v , got %v", want, got)
+	}
+}
diff --git a/algorithms/Go/arrays/two-sum.go b/algorithms/Go/arrays/two-sum.go
new file mode 100644
index 00000000..f86cd2ec
--- /dev/null
+++ b/algorithms/Go/arrays/two-sum.go
@@ -0,0 +1,46 @@
+/*
+	Problem Statement : Given an array of integers nums and an integer target, return indices of
+						the two numbers such that they add up to target.
+						You may assume that each input would have exactly one solution, and you may
+						not use the same element twice.
+
+	Input: An array of integers and a target (int)
+	Output: array of indexes of len(2) with sum of element at that index equal to target or nil
+*/
+
+package arrays
+
+/*
+	Using Brute Force : For every element check for another element if it exist in the array such that sum of
+						both the element is equals to the target
+	Time Complexity   : O(n^2)
+*/
+func twoSum(arr []int, target int) []int {
+	for i := 0; i < len(arr); i++ {
+		for j := i + 1; j < len(arr); j++ {
+			if arr[i]+arr[j] == target {
+				return []int{i, j}
+			}
+		}
+	}
+	return nil
+}
+
+/*
+	Using Map := While traversing every element add the element as key and its position as its value in a map
+  				 Check the required value (i.e target - arr[i]) in the map
+  				 If the map contains the required value then we have two elements with the required sum and
+				 return the positions.
+	Time Complexity : O(n)
+*/
+func twoSumEfficient(arr []int, target int) []int {
+	m := make(map[int]int)
+	for i := 0; i < len(arr); i++ {
+		compliment := target - arr[i]
+		if _, ok := m[compliment]; ok {
+			return []int{m[compliment], i}
+		}
+		m[arr[i]] = i
+	}
+	return []int{}
+}
diff --git a/algorithms/Go/arrays/two_sum_test.go b/algorithms/Go/arrays/two_sum_test.go
new file mode 100644
index 00000000..d0b92c8c
--- /dev/null
+++ b/algorithms/Go/arrays/two_sum_test.go
@@ -0,0 +1,35 @@
+package arrays
+
+import (
+	"testing"
+)
+
+func TestTwoSum(t *testing.T) {
+
+	arr := []int{2, 7, 11, 15, 1, 13, 18}
+	elem := 26
+
+	want := []int{2, 3}
+
+	if got := twoSum(arr, elem); got == nil || got[0] != want[0] && got[1] != want[1] {
+		if got == nil {
+			t.Errorf("got no indexes")
+		}
+		t.Errorf("want indexes [%d,%d] and got indexes [%d,%d]", want[0], want[1], got[0], got[1])
+	}
+}
+
+func TestTwoSumEfficient(t *testing.T) {
+
+	arr := []int{2, 7, 11, 15, 1, 13, 18}
+	elem := 26
+
+	want := []int{2, 3}
+
+	if got := twoSumEfficient(arr, elem); got == nil || got[0] != want[0] && got[1] != want[1] {
+		if got == nil {
+			t.Errorf("got no indexes")
+		}
+		t.Errorf("want indexes [%d,%d] and got indexes [%d,%d]", want[0], want[1], got[0], got[1])
+	}
+}