diff --git a/Hard/2742 Painting the Walls.md b/Hard/2742 Painting the Walls.md
new file mode 100644
index 0000000..0dad8e3
--- /dev/null
+++ b/Hard/2742 Painting the Walls.md	
@@ -0,0 +1,50 @@
+# 2742. Painting the Walls
+
+## Intuition
+
+The key insight is that while a paid painter is painting one wall, the free painter can paint other walls during that time. We can use this to optimize our solution by considering the trade-off between cost and time.
+
+## Approach
+
+1. We use dynamic programming where `dp[i]` represents the minimum cost to paint `i` walls
+2. Initialize dp array with maximum values except dp[0] = 0
+3. For each wall:
+   - We consider using the paid painter for this wall
+   - While paid painter works on current wall, free painter can paint `time[i]` walls
+   - We update dp[k] where k is min(total_walls, current_walls + 1 + time[i])
+   - The minimum cost would be either existing cost or cost of painting current wall plus previous state
+4. The final answer is stored in dp[len(cost)]
+
+## Complexity
+
+- Time complexity: O(n^2)
+- Space complexity: O(n)
+
+## Keywords
+
+- Dynamic Programming
+- Array
+- Optimization
+- Minimum Cost
+- Two Painters Problem
+
+## Code
+
+```go
+func paintWalls(cost []int, time []int) int {
+    dp := make([]int, len(cost)+1)
+    for i := range dp {
+        dp[i] = math.MaxInt32
+    }
+    dp[0] = 0
+
+    for i := range cost {
+        for j := len(cost)-1; j >= 0; j -= 1 {
+            k := min(len(cost), j+1+time[i])
+            dp[k] = min(dp[k], dp[j]+cost[i])
+        }
+    }
+
+    return dp[len(cost)]
+}
+```
diff --git a/Hard/44 Wildcard Matching.md b/Hard/44 Wildcard Matching.md
new file mode 100644
index 0000000..4b3ccf5
--- /dev/null
+++ b/Hard/44 Wildcard Matching.md	
@@ -0,0 +1,52 @@
+# 44. Wildcard Matching
+
+## Intuition
+
+This problem can be solved using a greedy approach with two pointers. The key insight is to handle the '*' character differently from other characters. When we encounter a '*', we keep track of its position and the position in the string where we matched it, so we can try different matching lengths for the '*' if needed.
+
+## Approach
+
+1. Use two pointers `ss` and `pp` to track positions in string `s` and pattern `p` respectively
+2. Keep track of the latest '*' position (`latestStar`) and the position in string where we started matching this '*' (`latestStarMatch`)
+3. For each character in string s:
+   - If current characters match (same char or '?'), move both pointers forward
+   - If we see a '*', record its position and current string position, then move pattern pointer
+   - If characters don't match but we have a previous '*', backtrack to try matching one more character with that '*'
+   - If none of above works, return false
+4. After processing string s, check if remaining pattern only contains '*'
+
+## Complexity
+
+- Time complexity: O(S*P)
+- Space complexity: O(1)
+
+## Keywords
+
+- Two Pointers
+- Greedy Algorithm
+- Pattern Matching
+
+## Code
+
+```go
+func isMatch(s string, p string) bool {
+    ss, pp, latestStar, latestStarMatch := 0, 0, -1, 0
+
+    for ss < len(s) {
+        if pp < len(p) && (p[pp] == '?' || p[pp] == s[ss]) {
+            ss, pp = ss+1, pp+1
+        } else if pp < len(p) && p[pp] == '*' {
+            latestStar, latestStarMatch, pp = pp, ss, pp+1
+        } else if latestStar != -1 {
+            pp, latestStarMatch, ss = latestStar+1, latestStarMatch+1, latestStarMatch+1
+        } else {
+            return false
+        }
+    }
+
+    for pp < len(p) && p[pp] == '*' {
+        pp += 1
+    }
+    return pp == len(p)
+}
+```
diff --git a/Medium/435 Non-overlapping Intervals.md b/Medium/435 Non-overlapping Intervals.md
new file mode 100644
index 0000000..d73cc2b
--- /dev/null
+++ b/Medium/435 Non-overlapping Intervals.md	
@@ -0,0 +1,53 @@
+# 435. Non-overlapping Intervals
+
+## Intuition
+
+To minimize the number of intervals that need to be removed to make the remaining intervals non-overlapping, we should:
+
+1. Sort the intervals by their end time
+2. Keep track of the current end time
+3. Remove intervals that overlap with the current end time
+
+## Approach
+
+1. Sort all intervals based on their end time in ascending order
+2. Initialize variables:
+   - `end`: tracks the end time of the last valid interval (initialized to -50001)
+   - `ret`: counts the number of intervals that need to be removed
+3. Iterate through the sorted intervals:
+   - If current interval's start time is greater than or equal to previous end time:
+     - Update end time to current interval's end time
+   - Otherwise:
+     - Increment the removal counter (ret)
+4. Return the total number of intervals that need to be removed
+
+## Complexity
+
+- Time complexity: O(nlogn)
+- Space complexity: O(1)
+
+## Keywords
+
+- Array
+- Sorting
+- Greedy
+- Interval Scheduling
+
+## Code
+
+```go
+func eraseOverlapIntervals(intervals [][]int) int {
+    sort.Slice(intervals, func(i, j int) bool {
+        return intervals[i][1] < intervals[j][1]
+    })
+    end, ret := -50001, 0
+    for _, interval := range intervals {
+        if end <= interval[0] {
+            end = interval[1]
+        } else {
+            ret += 1
+        }
+    }
+    return ret
+}
+```
diff --git a/README.md b/README.md
index fa192e2..28296f7 100644
--- a/README.md
+++ b/README.md
@@ -13,6 +13,7 @@
 | 23 | Merge k Sorted Lists | [go](/Hard/23%20Merge%20k%20Sorted%20Lists.md) | H |
 | 24 | Swap Nodes in Pairs | [go](/Medium/24%20Swap%20Nodes%20in%20Pairs.md) | M |
 | 25 | Reverse Nodes in k-Group | [go](/Hard/25%20Reverse%20Nodes%20in%20k-Group.md) | H |
+| 44 | Wildcard Matching | [go](/Hard/44%20Wildcard%20Matching.md) | H |
 | 45 | Jump Game II | [go](/Medium/45%20Jump%20Game%20II.md) | M |
 | 46 | Permutations | [go](/Medium/46%20Permutations.md) | M |
 | 50 | Pow(x, n) | [go](/Medium/50%20Pow(x,%20n).md) | M |
@@ -25,6 +26,7 @@
 | 236 | Lowest Common Ancestor of a Binary Tree | [go](/Medium/236%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Tree.md) | M |
 | 239 | Sliding Window Maximum | [go](/Hard/239%20Sliding%20Window%20Maximum.md) | H |
 | 322 | Coin Change | [go](/Medium/322%20Coin%20Change.md) | M |
+| 435 | Non-overlapping Interval | [go](/Medium/435%20Non-overlapping%20Intervals.md) | M |
 | 460 | LFU Cache | [go](/Hard/460%20LFU%20Cache.md) | H |
 | 592 | Fraction Addition and Subtraction | [go](/Medium/592%20Fraction%20Addition%20and%20Subtraction.md) | M |
 | 752 | Open the Lock | [go](/Medium/752%20Open%20the%20Lock.md) | M |
@@ -38,4 +40,5 @@
 | 948 | Bag of Tokens | [go](/Medium/948%20Bag%20Of%20Tokens.md) | M |
 | 1395 | Count Number of Trams | [go](/Medium/1395%20Count%20Number%20of%20Teams.md) | M |
 | 2366 | Mimimum Replacements to Sort the Array | [go](/Hard/2366%20Minimum%20Replacements%20to%20Sort%20the%20Array.md) | H |
+| 2742 | Paiting the Walls | [go](/Hard/2742%20Painting%20the%20Walls.md) | H |
 | 2932 | Build A Matrix with Conditions | [go](/Hard/2932%20Build%20A%20Matrix%20With%20Conditions.md) | H |
diff --git a/Week9/2742 AC.png b/Week9/2742 AC.png
new file mode 100644
index 0000000..8aa983a
Binary files /dev/null and b/Week9/2742 AC.png differ
diff --git a/Week9/2742 Painting the Walls.md b/Week9/2742 Painting the Walls.md
new file mode 100644
index 0000000..0dad8e3
--- /dev/null
+++ b/Week9/2742 Painting the Walls.md	
@@ -0,0 +1,50 @@
+# 2742. Painting the Walls
+
+## Intuition
+
+The key insight is that while a paid painter is painting one wall, the free painter can paint other walls during that time. We can use this to optimize our solution by considering the trade-off between cost and time.
+
+## Approach
+
+1. We use dynamic programming where `dp[i]` represents the minimum cost to paint `i` walls
+2. Initialize dp array with maximum values except dp[0] = 0
+3. For each wall:
+   - We consider using the paid painter for this wall
+   - While paid painter works on current wall, free painter can paint `time[i]` walls
+   - We update dp[k] where k is min(total_walls, current_walls + 1 + time[i])
+   - The minimum cost would be either existing cost or cost of painting current wall plus previous state
+4. The final answer is stored in dp[len(cost)]
+
+## Complexity
+
+- Time complexity: O(n^2)
+- Space complexity: O(n)
+
+## Keywords
+
+- Dynamic Programming
+- Array
+- Optimization
+- Minimum Cost
+- Two Painters Problem
+
+## Code
+
+```go
+func paintWalls(cost []int, time []int) int {
+    dp := make([]int, len(cost)+1)
+    for i := range dp {
+        dp[i] = math.MaxInt32
+    }
+    dp[0] = 0
+
+    for i := range cost {
+        for j := len(cost)-1; j >= 0; j -= 1 {
+            k := min(len(cost), j+1+time[i])
+            dp[k] = min(dp[k], dp[j]+cost[i])
+        }
+    }
+
+    return dp[len(cost)]
+}
+```
diff --git a/Week9/435 AC.png b/Week9/435 AC.png
new file mode 100644
index 0000000..7011758
Binary files /dev/null and b/Week9/435 AC.png differ
diff --git a/Week9/435 Non-overlapping Intervals.md b/Week9/435 Non-overlapping Intervals.md
new file mode 100644
index 0000000..d73cc2b
--- /dev/null
+++ b/Week9/435 Non-overlapping Intervals.md	
@@ -0,0 +1,53 @@
+# 435. Non-overlapping Intervals
+
+## Intuition
+
+To minimize the number of intervals that need to be removed to make the remaining intervals non-overlapping, we should:
+
+1. Sort the intervals by their end time
+2. Keep track of the current end time
+3. Remove intervals that overlap with the current end time
+
+## Approach
+
+1. Sort all intervals based on their end time in ascending order
+2. Initialize variables:
+   - `end`: tracks the end time of the last valid interval (initialized to -50001)
+   - `ret`: counts the number of intervals that need to be removed
+3. Iterate through the sorted intervals:
+   - If current interval's start time is greater than or equal to previous end time:
+     - Update end time to current interval's end time
+   - Otherwise:
+     - Increment the removal counter (ret)
+4. Return the total number of intervals that need to be removed
+
+## Complexity
+
+- Time complexity: O(nlogn)
+- Space complexity: O(1)
+
+## Keywords
+
+- Array
+- Sorting
+- Greedy
+- Interval Scheduling
+
+## Code
+
+```go
+func eraseOverlapIntervals(intervals [][]int) int {
+    sort.Slice(intervals, func(i, j int) bool {
+        return intervals[i][1] < intervals[j][1]
+    })
+    end, ret := -50001, 0
+    for _, interval := range intervals {
+        if end <= interval[0] {
+            end = interval[1]
+        } else {
+            ret += 1
+        }
+    }
+    return ret
+}
+```
diff --git a/Week9/44 AC.png b/Week9/44 AC.png
new file mode 100644
index 0000000..98c04f5
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diff --git a/Week9/44 Wildcard Matching.md b/Week9/44 Wildcard Matching.md
new file mode 100644
index 0000000..4b3ccf5
--- /dev/null
+++ b/Week9/44 Wildcard Matching.md	
@@ -0,0 +1,52 @@
+# 44. Wildcard Matching
+
+## Intuition
+
+This problem can be solved using a greedy approach with two pointers. The key insight is to handle the '*' character differently from other characters. When we encounter a '*', we keep track of its position and the position in the string where we matched it, so we can try different matching lengths for the '*' if needed.
+
+## Approach
+
+1. Use two pointers `ss` and `pp` to track positions in string `s` and pattern `p` respectively
+2. Keep track of the latest '*' position (`latestStar`) and the position in string where we started matching this '*' (`latestStarMatch`)
+3. For each character in string s:
+   - If current characters match (same char or '?'), move both pointers forward
+   - If we see a '*', record its position and current string position, then move pattern pointer
+   - If characters don't match but we have a previous '*', backtrack to try matching one more character with that '*'
+   - If none of above works, return false
+4. After processing string s, check if remaining pattern only contains '*'
+
+## Complexity
+
+- Time complexity: O(S*P)
+- Space complexity: O(1)
+
+## Keywords
+
+- Two Pointers
+- Greedy Algorithm
+- Pattern Matching
+
+## Code
+
+```go
+func isMatch(s string, p string) bool {
+    ss, pp, latestStar, latestStarMatch := 0, 0, -1, 0
+
+    for ss < len(s) {
+        if pp < len(p) && (p[pp] == '?' || p[pp] == s[ss]) {
+            ss, pp = ss+1, pp+1
+        } else if pp < len(p) && p[pp] == '*' {
+            latestStar, latestStarMatch, pp = pp, ss, pp+1
+        } else if latestStar != -1 {
+            pp, latestStarMatch, ss = latestStar+1, latestStarMatch+1, latestStarMatch+1
+        } else {
+            return false
+        }
+    }
+
+    for pp < len(p) && p[pp] == '*' {
+        pp += 1
+    }
+    return pp == len(p)
+}
+```