diff --git a/contiguous_array.py b/contiguous_array.py
new file mode 100644
index 00000000..b12288b9
--- /dev/null
+++ b/contiguous_array.py
@@ -0,0 +1,27 @@
+''' Time Complexity : O(n)
+    Space Complexity : O(n) 
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+
+   Approach : Calculate the Running Sum at each index.
+              Key idea is, if 2 indices have same running sum, their subarray is always balanced.
+'''
+
+class Solution:
+    def findMaxLength(self, nums: List[int]) -> int:
+        rSum = 0
+        hashmap = {}
+        hashmap[0] = -1
+        result = 0
+        for i in range(len(nums)):
+            if nums[i] == 0:
+                rSum -= 1
+            else:
+                rSum +=1 
+
+            if rSum in hashmap:
+                length = i - hashmap[rSum]
+                result = max(result, length)
+            else:
+                hashmap[rSum] = i
+        return result
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diff --git a/longest_palindrome.py b/longest_palindrome.py
new file mode 100644
index 00000000..56ec1277
--- /dev/null
+++ b/longest_palindrome.py
@@ -0,0 +1,22 @@
+''' Time Complexity : O(n)
+    Space Complexity : O(n) 
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+
+   Approach : if char already present in set, means we found 1 pair, increament count by 2,
+              else add char to set,
+              at end, it set has stil has element, we can add 1 to the count
+'''
+class Solution:
+    def longestPalindrome(self, s: str) -> int:
+        hashset=set()
+        count = 0
+        for c in s:
+            if c in hashset:
+                count += 2
+                hashset.remove(c)
+            else:
+                hashset.add(c)
+        if hashset:
+            count += 1
+        return count
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diff --git a/subarray-sum-equals-k.py b/subarray-sum-equals-k.py
new file mode 100644
index 00000000..7d639772
--- /dev/null
+++ b/subarray-sum-equals-k.py
@@ -0,0 +1,21 @@
+''' Time Complexity : O(n)
+    Space Complexity : O(n) 
+    Did this code successfully run on Leetcode : Yes
+    Any problem you faced while coding this :  No
+
+   Approach : Calculate the Running Sum at each index.
+              IF diff of Running Sum and k is already in hashmap, that means we have got subarray at that running sum
+'''
+class Solution:
+    def subarraySum(self, nums: List[int], k: int) -> int:
+        result = 0
+        rSum = 0 
+        hashmap = {}
+        hashmap[0] = 1
+        for i in range(len(nums)):
+            rSum += nums[i]
+            diff = rSum - k
+            if diff in hashmap:
+                result += hashmap[diff]
+            hashmap[rSum] = hashmap.get(rSum,0) + 1
+        return result
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