diff --git a/PartitionLabels.java b/PartitionLabels.java
new file mode 100644
index 0000000..22fea99
--- /dev/null
+++ b/PartitionLabels.java
@@ -0,0 +1,34 @@
+/*
+* Approach: Greedy Approach
+ - Idea is to be greedy and find the last index of the seen element and make the partition till that index
+ - Firstly, create a map or char array of size 26 to store the last index of the characters
+ - In the second pass, iterate over the string, get the first element and set end to last index of that element
+ - keep iterating and keep udating the end and once we have iterated to all the elements till the end, then thats our partition
+ - TC: O(n)
+ - SC: O(n)
+ */
+class Solution {
+    public List<Integer> partitionLabels(String s) {
+        HashMap<Character, Integer> map = new HashMap<>();
+        List<Integer> res = new ArrayList<>();
+        int end = 0;
+        //start and end to get the size of the partition
+        int start = 0;
+
+        for(int i = 0; i < s.length(); i++){
+            char ch = s.charAt(i);
+            map.put(ch, i);
+        }
+
+        for(int i = 0; i < s.length(); i++){
+            char curr = s.charAt(i);
+            end = Math.max(end, map.get(curr));
+            //once the i reaches the end index, that means we have found our first partition, so add it to the result and update the start pointer to point to the next element
+            if(i == end){
+                res.add(end - start + 1);
+                start = i + 1;
+            } 
+        }
+        return res;
+    }
+}
diff --git a/QueueReconstructionByHeight.java b/QueueReconstructionByHeight.java
new file mode 100644
index 0000000..a72c922
--- /dev/null
+++ b/QueueReconstructionByHeight.java
@@ -0,0 +1,28 @@
+/*
+* Approach: Greedy approach
+- Idea is to think about relative positions and think of people with the tallest height
+- Sort people array by heights first and then by number of people in the front. 
+- Iterate through the sorted array (consider example 1)
+    - first person will be of height tallest and least number of people in the front so go ahead and add them to the list (7,0)
+    - second person is either of height smaller or the same height but number of people will either be same/diff for diff height or different for same height (7,1), go ahead and add it to the list
+    - similary third is (6,1) -> so height 6 can have one person of height greater or equal to 6 in front of it so go ahead and slide person of height (7,1) 
+- TC: O(n^2logn) -> nlogn for sorting the people array and n^2 for sliding of people in the front
+- SC: O(n)
+ */
+class Solution {
+    public int[][] reconstructQueue(int[][] people) {
+        List<int[]> q = new ArrayList<>();
+        Arrays.sort(people, (a,b) ->{ //nlogn
+            if(a[0] == b[0]){ 
+                return a[1] - b[1];
+            } 
+            return b[0] - a[0];
+        });
+
+        for(int[] per: people){//n^2 - sliding of people in front that is costly
+            q.add(per[1], per);
+        }
+        
+        return q.toArray(new int[0][]);
+    }
+}
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diff --git a/TaskScheduler.java b/TaskScheduler.java
new file mode 100644
index 0000000..1dea77a
--- /dev/null
+++ b/TaskScheduler.java
@@ -0,0 +1,36 @@
+/*
+* Approach: The idea is to find how many idle cpus will be there in order to finish the given tasks
+- Firstly, we need to be greedy and look for most critical or highest freq tasks and how many highfreq tasks are there
+- Then calculate the partitions, available slots, pending tasks and idle state
+    - consider ex1: maxFreq = 3, totalmaxFreqTasks = 2 and n = 2
+    A _ _ A _ _ A -> from this, we can calculate 
+    - partitions:  2 -> maxFreq - 1;
+    - available slots: partitions * (n - (totalmaxFreqTasks - 1)) -> 4 slots -> if there are two maxFreq tasks, we need to put them together to get the local minima, resulting into global minima. 
+    - pending tasks: once all high freq tasks are done, we will have tasks remaining -> task.length - (maxFreq * totalmaxFreqTasks)
+    - idle: max of (0, (available slots - pending)), max because if the answer is negative that means we dont have any idle cpu so in that case the number of idle cpus should be 0 and not negative because that will mean we are producing extra timelol
+- Finally, return the total number of tasks + idle cpu intervals
+TC: O(n) -> iterate over tasks array to count maxFreq and iterate over keySet to get totalnumofmaxtasks
+SC: O(n) -> worst case for storing tasks and its freq
+ */
+class Solution {
+    public int leastInterval(char[] tasks, int n) {
+        HashMap<Character, Integer> map = new HashMap<>();
+        int maxFreq = 0;
+        int numOfMaxFreqTasks = 0;
+        for(char c: tasks){
+            map.put(c, map.getOrDefault(c, 0) + 1);
+            maxFreq = Math.max(maxFreq, map.get(c));
+        }
+        for(char k: map.keySet()){
+            if(map.get(k) == maxFreq)
+                numOfMaxFreqTasks++;
+        }
+
+        int partitions = maxFreq - 1;
+        int availableSlots = partitions * (n - (numOfMaxFreqTasks - 1));
+        int pendingTasks = tasks.length - (maxFreq * numOfMaxFreqTasks);
+        int idle = Math.max(0, (availableSlots - pendingTasks));
+
+        return tasks.length + idle;
+    }
+}
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