diff --git a/leetcode-409.py b/leetcode-409.py
new file mode 100644
index 00000000..f1fbe817
--- /dev/null
+++ b/leetcode-409.py
@@ -0,0 +1,23 @@
+'''
+https://leetcode.com/problems/longest-palindrome/
+
+Time Complexity: O(n)
+Space comlexity: O(n)
+TIP: Count the total number of even occurences and add it to the totalSum. For odd occurences add `occurence - 1`, and set the oddPresent to true - in the end add 1 to res if oddPresent == true
+Refer to optimized way in the video solution s30 to use hashset for adding count
+'''
+from collections import Counter
+class Solution:
+    def longestPalindrome(self, s: str) -> int:
+        count = Counter(s)
+        res = 0
+        oddPresent = False
+        for el in count.values():
+            if el%2 == 0:
+                res += el
+            else:
+                res += (el - 1)
+                oddPresent = True            
+        if oddPresent:
+            res += 1
+        return res
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diff --git a/leetcode-525.py b/leetcode-525.py
new file mode 100644
index 00000000..375e0e86
--- /dev/null
+++ b/leetcode-525.py
@@ -0,0 +1,31 @@
+'''
+https://leetcode.com/problems/contiguous-array/description/
+
+Time Complexity: O(n)
+Space complexity: O(n)
+
+TIP: brute force: n*n loop to generate subarrays -> and finding the longest canditate. starting from each index in the array.
+How do you eliminate the n*n loop? use running sum to remove the iteration..
+since its 0-indexed - you dont need to add an extra 1 to find the diff in the length. this is wrong - i - minIndex[currSum] + 1
+'''
+class Solution:
+    def findMaxLength(self, nums: List[int]) -> int:
+        #when you encounter 0 -> add -1 to running sum - handles edge case, which says when we start the iteration, sum is initially 0 at index -1
+        runningSum = []
+        minIndex = {}
+        minIndex[0] = -1 #handle edge case for first occurence of 0 sum.
+        currSum = 0
+        longestLength = 0
+        for i, el in enumerate(nums):
+            if el == 0:
+                currSum -= 1
+            else:
+                currSum += 1
+
+            # runningSum.append(currSum)
+            if currSum not in minIndex:
+                minIndex[currSum] = i
+            else:
+                longestLength = max(longestLength, i - minIndex[currSum] )
+        
+        return longestLength
diff --git a/leetcode-560.py b/leetcode-560.py
new file mode 100644
index 00000000..b70ea459
--- /dev/null
+++ b/leetcode-560.py
@@ -0,0 +1,31 @@
+'''
+https://leetcode.com/problems/subarray-sum-equals-k/
+
+
+Time Complexity: O(n)
+Space Complexity: O(n)
+TIP:
+You need to store dictionary as {runningSum: count} cuz you are tyring to count the occurence of the sum. which will give the total count
+'''
+class Solution:
+    def subarraySum(self, nums: List[int], target: int) -> int:
+        runningSumDict = {}
+        runningSumDict[0] = 1 #edge case - refer to the video to understand the edge case
+        runningSum = 0
+        totalCount = 0
+        
+        for el in nums:
+            runningSum += el
+            complement = runningSum - target
+
+            if  complement in runningSumDict:
+                totalCount += runningSumDict[complement]
+
+            if runningSum not in runningSumDict:
+                runningSumDict[runningSum] = 1  
+            else:
+                runningSumDict[runningSum] += 1  
+        
+        return totalCount
+            
+        
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