diff --git a/PartitionLabels.java b/PartitionLabels.java
new file mode 100644
index 0000000..a9c9dc6
--- /dev/null
+++ b/PartitionLabels.java
@@ -0,0 +1,30 @@
+/**
+ * Approach: Use a HashMap to store the last index of each character to determine partition boundaries.
+ * Solution: Iterate through the string, updating the current partition end to the farthest last index seen.
+ * When the current index equals the partition end, close the partition and record its length.
+ */
+public class PartitionLabels {
+    public List<Integer> partitionLabels(String s) {
+        HashMap<Character, Integer> hmap = new HashMap<>();
+
+        for(int i = 0; i < s.length() ;i++) {
+            char c = s.charAt(i);
+            hmap.put(c,i);
+        }
+
+        List<Integer> result = new ArrayList<>();
+
+        int start = 0;
+        int end = 0; 
+
+        for(int i = 0; i < s.length() ; i++) {
+            char c = s.charAt(i);
+            end = Math.max(end, hmap.get(c));
+            if(i == end) {
+                result.add(end-start+1);
+                start = i+1;
+            }
+        }
+        return result;
+    }
+}
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diff --git a/QueueReconstructionByHeight.java b/QueueReconstructionByHeight.java
new file mode 100644
index 0000000..77ecc5f
--- /dev/null
+++ b/QueueReconstructionByHeight.java
@@ -0,0 +1,31 @@
+/**
+ * Approach: Sort people by height in descending order; if heights are equal, sort by k value ascending.
+ * Solution: Insert each person into a list at the index equal to their k value, maintaining relative order.
+ * This ensures each person has exactly k taller or equal people in front when reconstruction is complete.
+ */
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.List;
+
+public class QueueReconstructionByHeight {
+    class Solution {
+    public int[][] reconstructQueue(int[][] people) {
+        Arrays.sort(people, (a,b) ->  {
+
+            if(a[0] == b [0]) {
+                return a[1] - b[1];
+            }
+
+            return b[0] - a[0];
+        });
+
+        List<int[]> result = new ArrayList<>(); 
+
+        for(int[] p : people) {
+            result.add(p[1],p);
+        }
+
+        return result.toArray(new int[0][0]);
+    }
+}
+}
diff --git a/TaskScheduler.java b/TaskScheduler.java
new file mode 100644
index 0000000..908c142
--- /dev/null
+++ b/TaskScheduler.java
@@ -0,0 +1,41 @@
+/**
+ * Approach: Use a max-heap to schedule the most frequent tasks first while respecting cooldowns using a queue.
+ * Solution: Pop tasks from the heap, execute them, and push them into a cooldown queue with the next valid time.
+ * Repeat until both heap and queue are empty, tracking total time to execute all tasks.
+ */
+public class TaskScheduler {
+    public int leastInterval(char[] tasks, int n) {
+        int[] freqMap = new int[26];
+
+        for (char ch : tasks) {
+            freqMap[ch - 'A']++;
+        }
+
+        PriorityQueue<Integer> maxHeap = new PriorityQueue<>((a, b) -> b - a);
+        for (int i = 0; i < 26; i++) {
+            if (freqMap[i] > 0) {
+                maxHeap.offer(freqMap[i]);
+            }
+        }
+        Deque<int[]> queue = new LinkedList<>(); // count, time
+        int time = 0;
+
+        while (!maxHeap.isEmpty() || !queue.isEmpty()) {
+            time++;
+
+            if (!maxHeap.isEmpty()) {
+                int currentCount = maxHeap.poll() - 1;
+                if (currentCount > 0) {
+                    queue.add(new int[] {currentCount, time + n });
+                }
+            }
+            if (!queue.isEmpty() && queue.peekFirst()[1] == time) {
+                int[] elementEligibleForMaxHeap = queue.removeFirst();
+                maxHeap.offer(elementEligibleForMaxHeap[0]);
+            }
+        }
+
+        return time;
+
+    }
+}
\ No newline at end of file