diff --git a/Level 1/Week 06/Queue/2073. Time Needed to Buy Tickets/code.cpp b/Level 1/Week 06/Queue/2073. Time Needed to Buy Tickets/code.cpp
index 2e0f3e1..a950221 100644
--- a/Level 1/Week 06/Queue/2073. Time Needed to Buy Tickets/code.cpp	
+++ b/Level 1/Week 06/Queue/2073. Time Needed to Buy Tickets/code.cpp	
@@ -1,50 +1,154 @@
-#include <queue>
-#include <vector>
-using namespace std;
-
-// 🧠 Approach:
-// This problem simulates a queue of people buying tickets with time tracking.
-// 1. Use a queue to store pairs of (position, remaining_tickets) for each person.
-// 2. Track time as each person takes 1 second to buy a ticket.
-// 3. When a person buys a ticket, decrement their remaining tickets.
-// 4. If they still have tickets, they go to the back of the queue.
-// 5. If they're done (0 tickets), they leave the queue.
-// 6. Continue until the person at position k finishes buying all their tickets.
-// 7. The key insight is to track the target person and count time for each ticket purchase.
-
-// ✅ Time Complexity: O(tickets[k] * n), where n is the number of people
-// ✅ Space Complexity: O(n), for the queue
+/*
+There are 3 approaches, but we will solve it using a queue because in this repo we are studying queues
 
+
+Understanding
+-------------
+Problem:
+- Given an array 'tickets' where tickets[i] = number of tickets the i-th person wants.
+- Index 'k' = target person.
+- Each person buys 1 ticket per second, and after buying 1 ticket, goes back to the 
+  end of the line if they have more tickets.
+- Goal: Find total time (seconds) for person at index 'k' to finish buying all tickets.
+
+Observations:
+1. Each person contributes to the total time depending on their position relative to 'k'.
+2. Target person's total time = their own tickets + contributions from others.
+---
+
+SOLUTION 1: Formula-based (O(n) time, O(1) space)
+--------------------------------------------------
+approch:
+- Identify target person at index k with tickets[k] tickets.
+- Total time = target person's own tickets + contribution of all other people.
+- Contribution rules:
+  1. If i <= k → person i will be in line until target finishes last ticket 
+     → contribution = min(tickets[i], tickets[k])
+  2. If i > k → person i may not reach the last round after target finishes 
+     → contribution = min(tickets[i], tickets[k]-1)
+- Sum all contributions → total time for target to finish buying tickets.
+- This avoids simulating the entire queue and gives O(n) time.
+
+Time Complexity: O(n) → single pass over tickets array.
+Space Complexity: O(1) → only variables used (result, loop counter).
+
+*/
 class Solution {
 public:
     int timeRequiredToBuy(vector<int>& tickets, int k) {
-        queue<pair<int, int>> queue; // (position, remaining_tickets)
-        
-        // Initialize queue with all people
+        int result=0;
+        for(int i=0;i<tickets.size();i++)
+        {
+            if(i<=k)
+                result += min(tickets[k], tickets[i]);
+            else
+                result += min(tickets[k]-1, tickets[i]);
+        }
+        return result;
+    }
+};
+
+/*
+Example:
+tickets = [2,3,2], k=2
+Target tickets = 2
+
+Index 0: min(2,2)=2
+Index 1: min(3,2)=2
+Index 2: min(2,2)=2
+Total time = 2+2+2=6
+*/
+
+/*
+---
+
+SOLUTION 2: Queue-based simulation (O(n*m) time, O(n) space)
+--------------------------------------------------------------
+approch:
+- Use a queue to store (index, remaining tickets) of each person.
+- Each second, person at front buys 1 ticket:
+    1. decrement remaining tickets
+    2. increment time
+    3. if remaining > 0 → push back to queue
+    4. if remaining == 0 and person is target → return time
+- This is a literal simulation of the queue buying process.
+
+Time Complexity: O(n * m) → worst case every person buys ticket in multiple rounds.
+Space Complexity: O(n) → queue stores up to n people at a time.
+
+*/
+class Solution2 {
+public:
+    int timeRequiredToBuy(vector<int>& tickets, int k) {
+
+        queue<pair<int, int>> q;
+
         for (int i = 0; i < tickets.size(); i++) {
-            queue.push({i, tickets[i]});
+            q.push({i, tickets[i]});
         }
-        
+
         int time = 0;
-        
-        while (!queue.empty()) {
-            auto [position, remaining] = queue.front();
-            queue.pop();
-            
-            time++; // 1 second to buy a ticket
-            
-            remaining--; // Buy one ticket
-            
-            if (remaining > 0) {
-                // Person still has tickets, goes to back of queue
-                queue.push({position, remaining});
-            } else if (position == k) {
-                // Target person finished buying all tickets
-                break;
+
+        while (!q.empty()) {
+            auto [pos, remain] = q.front();
+            q.pop();
+            remain--;
+            time++;
+            if (remain > 0) {
+                q.push({pos, remain});
             }
-            // If person is done and not the target, they just leave
+            if(pos==k && remain == 0)
+                return time;
         }
-        
         return time;
     }
-}; 
\ No newline at end of file
+};
+
+/*
+---
+
+SOLUTION 3: Circular Simulation / Direct Array Modification (O(n*m) time, O(1) space)
+----------------------------------------------------------------------------------------
+approch:
+- Use two variables: 'sec' = total seconds, 'i' = current person index.
+- Loop until target person tickets[k] = 0:
+    1. If tickets[i] > 0 → decrement tickets[i], increment sec
+    2. Move to next person: i++  
+       - If i == tickets.size() → i = 0 (start new round)
+- When tickets[k] = 0 → return sec.
+- This simulates queue circularly **without extra memory**.
+
+Time Complexity: O(n * m) → worst case loop runs tickets[k] * n times.
+Space Complexity: O(1) → only counters used, tickets array modified in-place.
+
+*/
+class Solution3 {
+public:
+    int timeRequiredToBuy(vector<int>& tickets, int k) {
+        int sec = 0;
+        int i = 0;
+
+        while (tickets[k] != 0) {
+            if (tickets[i] > 0) {  
+                tickets[i]--; 
+                sec++;
+            }
+            i++;
+            if (i == tickets.size()) i = 0; // complete a round
+        }
+        return sec;
+    }
+};
+
+/*
+---
+Time & Space Complexity:
+
+| Approach                  | Time Complexity | Space Complexity |
+|----------------------------|----------------|----------------  |
+| Formula-based              | O(n)           | O(1)             |
+| Queue simulation           | O(n*m)         | O(n)             |
+| Circular array simulation  | O(n*m)         | O(1)             |
+
+
+*/