hashing-2 problems Done

Problem1_SubarraySumEqualsK.py

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+# Time Complexity : O(n) where n is the length of the array
+# Space Complexity : O(n) for storing cumulative sums in hashmap
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+
+# Approach: Use a hashmap to store cumulative sums and their frequencies as we iterate through the array.
+# For each element, calculate the cumulative sum and check if (cumulative_sum - k) exists in the hashmap.
+# If it exists, it means there are subarrays ending at current index with sum k, so add their count to result.
+
+class Solution:
+    def subarraySum(self, nums: List[int], k: int) -> int:
+        count = 0
+        cumulative_sum = 0
+        sum_map = {0: 1}  # Initialize with sum 0 having count 1
+
+        for num in nums:
+            cumulative_sum += num
+
+            # Check if (cumulative_sum - k) exists in map
+            if (cumulative_sum - k) in sum_map:
+                count += sum_map[cumulative_sum - k]
+
+            # Update the frequency of current cumulative sum
+            sum_map[cumulative_sum] = sum_map.get(cumulative_sum, 0) + 1
+
+        return count
+

Problem2_ContiguousArray.py

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+# Time Complexity : O(n) where n is the length of the array
+# Space Complexity : O(n) for storing cumulative sum indices in hashmap
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+
+# Approach: Transform the problem by treating 0s as -1 and 1s as +1, making it equivalent to finding longest subarray with sum 0.
+# Use a hashmap to store the first occurrence index of each cumulative sum encountered.
+# When the same cumulative sum appears again, calculate the length between current and stored index to find equal 0s and 1s.
+
+class Solution:
+    def findMaxLength(self, nums: List[int]) -> int:
+        sum_map = {0: -1}  # Initialize with sum 0 at index -1
+        max_length = 0
+        cumulative_sum = 0
+
+        for i, num in enumerate(nums):
+            # Treat 0 as -1 and 1 as +1
+            cumulative_sum += 1 if num == 1 else -1
+
+            # If this cumulative sum was seen before, calculate length
+            if cumulative_sum in sum_map:
+                max_length = max(max_length, i - sum_map[cumulative_sum])
+            else:
+                # Store first occurrence of this cumulative sum
+                sum_map[cumulative_sum] = i
+
+        return max_length
+

Problem3_LongestPalindrome.py

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+# Time Complexity : O(n) where n is the length of the string
+# Space Complexity : O(1) since hashmap has at most 52 entries (uppercase and lowercase letters)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+
+# Approach: Count frequency of each character using a hashmap to track occurrences.
+# For each character, add the largest even number of occurrences (count // 2 * 2) to the palindrome length.
+# If any character has an odd count, add one more character to the center of the palindrome.
+
+class Solution:
+    def longestPalindrome(self, s: str) -> int:
+        char_count = {}
+
+        # Count frequency of each character
+        for char in s:
+            char_count[char] = char_count.get(char, 0) + 1
+
+        length = 0
+        odd_found = False
+
+        # Calculate palindrome length
+        for count in char_count.values():
+            # Add largest even number of characters
+            length += count // 2 * 2
+
+            # Check if there's an odd count character for center
+            if count % 2 == 1:
+                odd_found = True
+
+        # Add one center character if any odd count exists
+        if odd_found:
+            length += 1
+
+        return length
+