diff --git a/2023_10_02/maximum-path-sum/Cargo.toml b/2023_10_02/maximum-path-sum/Cargo.toml
new file mode 100644
index 0000000..ef62e33
--- /dev/null
+++ b/2023_10_02/maximum-path-sum/Cargo.toml
@@ -0,0 +1,8 @@
+[package]
+name = "maximum-path-sum"
+version = "0.1.0"
+edition = "2021"
+
+# See more keys and their definitions at https://doc.rust-lang.org/cargo/reference/manifest.html
+
+[dependencies]
diff --git a/2023_10_02/maximum-path-sum/src/main.rs b/2023_10_02/maximum-path-sum/src/main.rs
new file mode 100644
index 0000000..e7a11a9
--- /dev/null
+++ b/2023_10_02/maximum-path-sum/src/main.rs
@@ -0,0 +1,3 @@
+fn main() {
+    println!("Hello, world!");
+}
diff --git a/2023_10_05/longest-k-good-segment/Cargo.toml b/2023_10_05/longest-k-good-segment/Cargo.toml
new file mode 100644
index 0000000..f5e5734
--- /dev/null
+++ b/2023_10_05/longest-k-good-segment/Cargo.toml
@@ -0,0 +1,8 @@
+[package]
+name = "longest-k-good-segment"
+version = "0.1.0"
+edition = "2021"
+
+# See more keys and their definitions at https://doc.rust-lang.org/cargo/reference/manifest.html
+
+[dependencies]
diff --git a/2023_10_05/longest-k-good-segment/README.md b/2023_10_05/longest-k-good-segment/README.md
new file mode 100644
index 0000000..256a768
--- /dev/null
+++ b/2023_10_05/longest-k-good-segment/README.md
@@ -0,0 +1,84 @@
+# Comments on the solution
+
+The problem requires finding the longest segment of an array that contains no more than `k` different values. We can use a sliding window approach. We maintain a hashmap to store the count of each value in the current segment. We start with the left and right indices at 0 and keep incrementing the right index until the number of different values in the segment exceeds `k`. At this point, we increment the left index until the number of different values in the segment is less than or equal to `k`. We keep track of the maximum length of the segment seen so far and return the left and right indices of this segment.
+
+### Complexity analysis
+
+* **Time complexity**: $O(n)$ since we only iterate through the array once and perform constant time operations on each element.
+
+* **Space complexity**: $O(k)$ since we store the count of each value in the hashmap.
+
+---
+
+### Rust Solution
+
+This version reads the input from stdin and writes the output to stdout (as required by the problem statement)
+
+```rust
+fn main() {
+    let mut input = String::new();
+    std::io::stdin().read_line(&mut input).unwrap();
+    let mut iter = input.split_whitespace();
+    let n: usize = iter.next().unwrap().parse().unwrap();
+    let k: usize = iter.next().unwrap().parse().unwrap();
+
+    let mut input = String::new();
+    std::io::stdin().read_line(&mut input).unwrap();
+    let mut iter = input.split_whitespace();
+    let mut a: Vec<usize> = Vec::with_capacity(n);
+    for _ in 0..n {
+        a.push(iter.next().unwrap().parse().unwrap());
+    }
+
+    let mut l = 0;
+    let mut r = 0;
+    let mut max = 0;
+    let mut map: std::collections::HashMap<usize, usize> = std::collections::HashMap::new();
+    while r < n {
+        let count = map.entry(a[r]).or_insert(0);
+        *count += 1;
+        while map.len() > k {
+            let count = map.entry(a[l]).or_insert(0);
+            *count -= 1;
+            if *count == 0 {
+                map.remove(&a[l]);
+            }
+            l += 1;
+        }
+        if r - l + 1 > max {
+            max = r - l + 1;
+        }
+        r += 1;
+    }
+    println!("{} {}", l + 1, r);
+}
+```
+
+Here is a version that takes the input as arguments and returns the output as a tuple.
+
+
+```rust
+fn find_k_good_segment(n: usize, k: usize, a: &[usize]) -> (usize, usize) {
+    let mut l = 0;
+    let mut r = 0;
+    let mut max = 0;
+    let mut map: std::collections::HashMap<usize, usize> = std::collections::HashMap::new();
+    while r < n {
+        let count = map.entry(a[r]).or_insert(0);
+        *count += 1;
+        while map.len() > k {
+            let count = map.entry(a[l]).or_insert(0);
+            *count -= 1;
+            if *count == 0 {
+                map.remove(&a[l]);
+            }
+            l += 1;
+        }
+        if r - l + 1 > max {
+            max = r - l + 1;
+        }
+        r += 1;
+    }
+    (l + 1, r)
+}
+```
diff --git a/2023_10_05/longest-k-good-segment/src/main.rs b/2023_10_05/longest-k-good-segment/src/main.rs
new file mode 100644
index 0000000..46a6bcd
--- /dev/null
+++ b/2023_10_05/longest-k-good-segment/src/main.rs
@@ -0,0 +1,26 @@
+fn main() {
+    // reads two integers n and k from the standard input.
+    let mut input = String::new();
+    std::io::stdin().read_line(&mut input).unwrap(); // read line from stdin
+    let mut iter = input.split_whitespace(); // split input string by whitespace
+    let n: usize = iter.next().unwrap().parse().unwrap(); // parse first element to usize
+    let k: usize = iter.next().unwrap().parse().unwrap(); // parse second element to usize
+
+    // reads n integers from the standard input and stores them in an array a.
+    let mut input = String::new(); // input string
+    std::io::stdin().read_line(&mut input).unwrap(); // read line from stdin
+    let mut iter = input.split_whitespace(); // split input string by whitespace
+    let mut a: Vec<usize> = Vec::with_capacity(n); // create vector with capacity n
+    for _ in 0..n {
+        a.push(iter.next().unwrap().parse().unwrap()); // parse each element to usize and push to vector
+    }
+
+    let mut l = 0; // left index
+    let mut r = 0; // right index
+    let mut max = 0; // max length
+    let mut count = 0; // count of different elements
+
+    // We maintain a hashmap to store the count of each value in the current segment. We start with the left and right indices at 0 and keep incrementing the right index until the number of different values in the segment exceeds `k`. At this point, we increment the left index until the number of different values in the segment is less than or equal to `k`. We keep track of the maximum length of the segment seen so far and return the left and right indices of this segment.
+
+
+}