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+# Problem1 (https://leetcode.com/problems/subarray-sum-equals-k/)
+# Time Complexity : O(n)
+# Space Complexity : O(n)
+# Did this code successfully run on Leetcode :yes
+# Any problem you faced while coding this :no
+
+class Solution(object):
+    def subarraySum(self, nums, k):
+        """
+        :type nums: List[int]
+        :type k: int
+        :rtype: int
+        """
+        ## for this problem we need to look at all the sub array whose sum is k, naive way is tending towards n^3
+        ## to be able to do it in a single pass, we need some way where we can look at all the past subarray without iterating 
+        ## here the prefix sum technique comes in handy
+        ## Start by having a hashmap which will be used to store the running sum and how many times it has occured
+        ## hashmap = {0:1} ## diff,ways ## adding 0 as base case because 0 has happenend once and this will help us 
+        ## to catch the index =0 cases as well. 
+        ## iterating over the nums and adding the curSum 
+        ## now we can check if the complement of it has happened in the past, which will mean we can increase the 
+        ## subbarray count by the times it has happened +1 
+        ## calculating the sum and getting the value from hashmap
+        ## incrementing the value by 1 and putting back in hashmap
+        ## finally we exaust the search space and return res
+        # curSum =0 
+        hashmap = {} ## diff, how many times it has happened
+        hashmap[0] = 1
+        curSum =0 
+        res =0 
+        for n in nums:
+            curSum +=n
+            diff = curSum-k
+            res += hashmap.get(diff,0)
+            hashmap[curSum] = 1+ hashmap.get(curSum,0)
+            
+        return res
+
+
+
+########################################
+# Problem2 (https://leetcode.com/problems/contiguous-array/)
+# Time Complexity : O(n)
+# Space Complexity : O(n)
+# Did this code successfully run on Leetcode :yes
+# Any problem you faced while coding this :no
+
+class Solution(object):
+    def findMaxLength(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+
+        ## prefix sum pattern
+        ## continous sum when we see 1 do +1 when encounter 0 do -1
+        ## maintain a hashmap for running sum , index
+        ## if the running sum exists in the map, replace count with max of existing and index in hashmap, current - previous 
+        ## maintain the count to return at the end 
+
+        hashmap={}
+        hashmap[0]=-1 ## base case cos we dont want to miss the first index
+        count =0 
+        rSum =0 
+        for i in range(0,len(nums)):
+            if nums[i] == 1:
+                rSum +=1
+            else:
+                rSum -=1
+                
+            if rSum in hashmap:
+                count = max(count, i-hashmap.get(rSum))
+            else:
+                hashmap[rSum] = i 
+        
+        return count
+             
+
+########################################
+# Problem3 (https://leetcode.com/problems/longest-palindrome)
+# Time Complexity : 
+# Space Complexity : 
+# Did this code successfully run on Leetcode :yes
+# Any problem you faced while coding this :no
+
+
+## for checking the longest palindrome I need to check if we already have seen a similar element
+## since we need to check for mirror
+## we only have to store the elements that we have seen and search in O(1) so the ds here we can use hashset
+## maintain a count and a hashset to store the element
+## iterate over the string store in hashset if I have seen it for the first time 
+## else increment the count by 2, cos pair and remove from the hashset
+## if not in hashset and seeing for first time, just add to hashset
+## return count if hashset is empty meaning complete mirror and count +1 if hashset not empty telling ud we can choose one value in mid
+
+
+class Solution(object):
+    def longestPalindrome(self, s):
+        """
+        :type s: str
+        :rtype: int
+        """
+        count =0 
+        hashset = set()
+        for c in s:
+            if c in hashset:
+                hashset.remove(c)
+                count +=2
+            else:
+                hashset.add(c)
+        
+        return count+1 if hashset else count
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