From 00c361363af6a5c6092eb30daa31ea8748e243af Mon Sep 17 00:00:00 2001
From: sainathek <erukullasainath@gmail.com>
Date: Mon, 12 Jan 2026 02:18:01 -0500
Subject: [PATCH] Done Hashing-2

---
 problem1.java | 35 +++++++++++++++++++++++++++++++++++
 problem2.java | 37 +++++++++++++++++++++++++++++++++++++
 problem3.java | 33 +++++++++++++++++++++++++++++++++
 3 files changed, 105 insertions(+)
 create mode 100644 problem1.java
 create mode 100644 problem2.java
 create mode 100644 problem3.java

diff --git a/problem1.java b/problem1.java
new file mode 100644
index 00000000..5c390329
--- /dev/null
+++ b/problem1.java
@@ -0,0 +1,35 @@
+// Time Complexity : O(n)
+// Space Complexity : O(n)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No, just needed to understand prefix sum logic.
+
+
+// Your code here along with comments explaining your approach in three sentences only
+// We use prefix sum and a hashmap to store how many times each prefix sum has appeared.
+// At each index, we check if (currentSum - k) exists in the map and add its count to answer.
+// Then we store the currentSum into the map and continue.
+
+import java.util.*;
+
+class Solution {
+    public int subarraySum(int[] nums, int k) {
+
+        HashMap<Integer, Integer> map = new HashMap<>();
+        map.put(0, 1);
+
+        int sum = 0;
+        int count = 0;
+
+        for (int n : nums) {
+            sum += n;
+
+            if (map.containsKey(sum - k)) {
+                count += map.get(sum - k);
+            }
+
+            map.put(sum, map.getOrDefault(sum, 0) + 1);
+        }
+
+        return count;
+    }
+}
diff --git a/problem2.java b/problem2.java
new file mode 100644
index 00000000..1ba1fd19
--- /dev/null
+++ b/problem2.java
@@ -0,0 +1,37 @@
+// Time Complexity : O(n)
+// Space Complexity : O(n)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No, just had to convert 0 to -1.
+
+
+// Your code here along with comments explaining your approach in three sentences only
+// We treat 0 as -1 and 1 as +1 and keep a running sum.
+// If the same sum appears again, the subarray between them has equal number of 0s and 1s.
+// We store the first index of each sum in a hashmap to get the maximum length.
+
+import java.util.*;
+
+class Solution {
+    public int findMaxLength(int[] nums) {
+
+        HashMap<Integer, Integer> map = new HashMap<>();
+        map.put(0, -1);
+
+        int sum = 0;
+        int maxLen = 0;
+
+        for (int i = 0; i < nums.length; i++) {
+
+            if (nums[i] == 0) sum -= 1;
+            else sum += 1;
+
+            if (map.containsKey(sum)) {
+                maxLen = Math.max(maxLen, i - map.get(sum));
+            } else {
+                map.put(sum, i);
+            }
+        }
+
+        return maxLen;
+    }
+}
diff --git a/problem3.java b/problem3.java
new file mode 100644
index 00000000..7f3281ae
--- /dev/null
+++ b/problem3.java
@@ -0,0 +1,33 @@
+// Time Complexity : O(n)
+// Space Complexity : O(1)  (since alphabet size is fixed)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No, just had to count odd frequency characters.
+
+
+// Your code here along with comments explaining your approach in three sentences only
+// We count frequency of each character using a hash set.
+// If a character appears even times, it can fully be used in palindrome, and if odd, one can be placed in center.
+// Finally, we add all usable pairs and add one odd if any exists.
+
+import java.util.*;
+
+class Solution {
+    public int longestPalindrome(String s) {
+
+        HashSet<Character> set = new HashSet<>();
+        int count = 0;
+
+        for (char c : s.toCharArray()) {
+            if (set.contains(c)) {
+                set.remove(c);
+                count += 2;
+            } else {
+                set.add(c);
+            }
+        }
+
+        if (!set.isEmpty()) count += 1;
+
+        return count;
+    }
+}