diff --git a/problem1.java b/problem1.java
new file mode 100644
index 0000000..73f6d3f
--- /dev/null
+++ b/problem1.java
@@ -0,0 +1,53 @@
+import java.util.*;
+
+class Solution {
+
+    /**
+    Time Complexity : O(N)
+    Explanation:
+    We count task frequencies once and then scan the frequency map.
+
+    Space Complexity : O(1)
+    Explanation:
+    HashMap stores at most 26 uppercase English letters.
+
+    Did this code successfully run on LeetCode : Yes
+
+    Any problem you faced while coding this :
+    Initially struggled to calculate idle slots correctly.
+    Fixed it by using the most frequent task as the base structure:
+        - partitions = maxfreq - 1
+        - available idle slots depend on cooldown n
+        - pending tasks fill those idle slots
+    Also handled multiple tasks having the same maximum frequency.
+    */
+
+    public int leastInterval(char[] tasks, int n) {
+
+        HashMap<Character, Integer> map = new HashMap<>();
+
+        int l = tasks.length;
+        int maxfreq = 0;
+        int noofmaxfreq = 0;
+
+        // Count task frequency
+        for (char task : tasks) {
+            map.put(task, map.getOrDefault(task, 0) + 1);
+            maxfreq = Math.max(maxfreq, map.get(task));
+        }
+
+        // Count how many tasks have max frequency
+        for (char task : map.keySet()) {
+            if (map.get(task) == maxfreq) {
+                noofmaxfreq++;
+            }
+        }
+
+        int partitions = maxfreq - 1;
+        int available = partitions * (n - (noofmaxfreq - 1));
+        int pending = l - (noofmaxfreq * maxfreq);
+        int empty = Math.max(0, available - pending);
+
+        return l + empty;
+    }
+}
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diff --git a/problem2.java b/problem2.java
new file mode 100644
index 0000000..8f70a18
--- /dev/null
+++ b/problem2.java
@@ -0,0 +1,58 @@
+import java.util.*;
+
+class Solution {
+
+    /**
+    Time Complexity : O(N^2)
+    Explanation:
+    - Sorting takes O(N log N)
+    - Inserting into ArrayList at specific index takes O(N)
+    - Total worst case becomes O(N^2)
+
+    Space Complexity : O(N)
+    Explanation:
+    Extra list is used to reconstruct the queue.
+
+    Did this code successfully run on LeetCode : Yes
+
+    Any problem you faced while coding this :
+    Initially tried placing shorter people first,
+    but later insertions disturbed the arrangement.
+    Fixed it by:
+        1) Sorting taller people first
+        2) For same height, sort by smaller k first
+    Then inserting each person at index k automatically places them
+    in the correct position.
+    */
+
+    public int[][] reconstructQueue(int[][] people) {
+
+        List<int[]> li = new ArrayList<>();
+
+        // Sort:
+        // height descending
+        // k ascending
+        Arrays.sort(people, (a, b) -> {
+
+            if (a[0] == b[0]) {
+                return a[1] - b[1];
+            }
+
+            return b[0] - a[0];
+        });
+
+        // Insert person at index k
+        for (int[] person : people) {
+            li.add(person[1], person);
+        }
+
+        // Convert list to array
+        int[][] res = new int[li.size()][2];
+
+        for (int i = 0; i < li.size(); i++) {
+            res[i] = li.get(i);
+        }
+
+        return res;
+    }
+}
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diff --git a/problem3.java b/problem3.java
new file mode 100644
index 0000000..4c3748d
--- /dev/null
+++ b/problem3.java
@@ -0,0 +1,53 @@
+import java.util.*;
+
+class Solution {
+
+    /**
+    Time Complexity : O(N)
+    Explanation:
+    We scan the string once to store last occurrence of each character,
+    then scan again to create partitions.
+
+    Space Complexity : O(1)
+    Explanation:
+    HashMap stores at most 26 lowercase English letters.
+
+    Did this code successfully run on LeetCode : Yes
+
+    Any problem you faced while coding this :
+    Initially confused about where a partition should end.
+    Fixed it by storing the last index of every character.
+    While scanning, keep updating the partition end as the farthest
+    last occurrence of characters seen so far.
+    When current index reaches end, we close the partition.
+    */
+
+    public List<Integer> partitionLabels(String s) {
+
+        List<Integer> result = new ArrayList<>();
+        HashMap<Character, Integer> map = new HashMap<>();
+
+        // Store last occurrence of each character
+        for (int i = 0; i < s.length(); i++) {
+            char c = s.charAt(i);
+            map.put(c, i);
+        }
+
+        int start = 0;
+        int end = 0;
+
+        // Build partitions
+        for (int i = 0; i < s.length(); i++) {
+            char c = s.charAt(i);
+
+            end = Math.max(end, map.get(c));
+
+            if (i == end) {
+                result.add(end - start + 1);
+                start = i + 1;
+            }
+        }
+
+        return result;
+    }
+}
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