diff --git a/best-time-to-buy-and-sell-stock/Jeehay28.js b/best-time-to-buy-and-sell-stock/Jeehay28.js
new file mode 100644
index 000000000..1964dd1e5
--- /dev/null
+++ b/best-time-to-buy-and-sell-stock/Jeehay28.js
@@ -0,0 +1,41 @@
+/**
+ * @param {number[]} prices
+ * @return {number}
+ */
+
+// TC : O(n)
+// SC : O(1)
+
+var maxProfit = function (prices) {
+  if (prices.length === 1) {
+    return 0;
+  }
+
+  // Two variables (profitMax and priceMin) are used to store the maximum profit and minimum price seen, which require O(1) space.
+  let profitMax = 0;
+  let priceMin = prices[0];
+
+  for (const price of prices) {
+    const profit = price - priceMin;
+    profitMax = Math.max(profit, profitMax);
+    priceMin = Math.min(price, priceMin);
+  }
+
+  return profitMax;
+};
+
+// Why Constants Are Ignored in Big-O
+// In Big-O notation, O(2) is simplified to O(1) because constants are irrelevant in asymptotic analysis.
+// Big-O focuses on how resource usage scales with input size, not fixed values.
+
+// Using 2 variables: O(1)
+// Using 10 variables: O(1)
+// Using 100 variables: O(1)
+
+// What Space Complexity Looks Like for Larger Growth
+// O(n): Memory grows linearly with the input size (e.g., storing an array of n elements).
+// O(n^2): Memory grows quadratically (e.g., a 2D matrix with n*n elements).
+// 𝑂(log 𝑛): Memory grows logarithmically (e.g., recursive calls in binary search).
+// O(1): Fixed memory usage, regardless of input size (e.g., using a fixed number of variables).
+
+
diff --git a/group-anagrams/Jeehay28.js b/group-anagrams/Jeehay28.js
new file mode 100644
index 000000000..40310ddee
--- /dev/null
+++ b/group-anagrams/Jeehay28.js
@@ -0,0 +1,88 @@
+// Guided approach
+// TC : O(n*k), where n is the number of strings, and k is the average length of each string.
+// SC : O(n*k)
+// overal time complexity improved : from O(n * klogk) to O(n * k)
+
+/**
+ * Time Complexity Breakdown:
+ *
+ * Step                                    | Time Complexity     | Explanation
+ * --------------------------------------- | ------------------- | ----------------------------------------
+ * Outer loop over strings (`for` loop)    | O(n)                | Iterate over each string in the input array `strs`.
+ * Create key (`createKey`)                | O(k) per string     | For each string, count character frequencies, with k being the length of the string.
+ * Map operations (`set` and `get`)        | O(1) per string     | Inserting and retrieving values from a Map.
+ * Result array                            | O(n * k)            | Storing grouped anagrams in the result array.
+ *
+ * Overall Time Complexity:                | O(n * k)            | Total time complexity considering all steps.
+ *
+ * Space Complexity Breakdown:
+ *
+ * Step                                    | Space Complexity    | Explanation
+ * --------------------------------------- | ------------------- | -----------------------------------------
+ * Map to store grouped anagrams           | O(n * k)            | Map stores n groups with each group having at most k characters.
+ * Auxiliary space for `createKey`         | O(1)                | The frequency array used to count characters (constant size of 26).
+ * Space for the result array              | O(n * k)            | Result array storing n groups of up to k elements.
+ *
+ * Overall Space Complexity:               | O(n * k)            | Total space complexity considering all storage.
+ */
+
+/**
+ * @param {string[]} strs
+ * @return {string[][]}
+ */
+
+var groupAnagrams = function (strs) {
+  const createKey = (str) => {
+    const arr = new Array(26).fill(0);
+
+    for (const ch of str) {
+      const idx = ch.charCodeAt() - "a".charCodeAt();
+      arr[idx] += 1;
+    }
+
+    return arr.join("#");
+  };
+
+  let map = new Map();
+
+  for (const str of strs) {
+    const key = createKey(str);
+    map.set(key, [...(map.get(key) || []), str]);
+  }
+
+  return Array.from(map.values(map));
+};
+
+// *My own approach
+
+// Time Complexity
+// 1. Sorting Each String:
+// Sorting a string takes O(k*logk), where k is the length of the string.
+// Since we sort each string in the input array of size n, the total cost for sorting is O(n*klogk).
+
+// 2. Hash Map Operations:
+// Inserting into the hash map is O(1) on average. Over n strings, the cost remains O(n).
+
+// Overall Time Complexity:
+// O(n*klogk), where n is the number of strings and k is the average length of a string.
+
+// /**
+//  * @param {string[]} strs
+//  * @return {string[][]}
+//  */
+
+// var groupAnagrams = function (strs) {
+//   // helper function
+//   const sorted = (str) => {
+//     return str.split("").sort().join("");
+//   };
+
+//   let obj = {};
+
+//   for (const str of strs) {
+//     const key = sorted(str);
+//     obj[key] = [...(obj[key] || []), str];
+//   }
+
+//   return Object.values(obj);
+// };
diff --git a/implement-trie-prefix-tree/Jeehay28.js b/implement-trie-prefix-tree/Jeehay28.js
new file mode 100644
index 000000000..d1fa20af7
--- /dev/null
+++ b/implement-trie-prefix-tree/Jeehay28.js
@@ -0,0 +1,69 @@
+// Space complexity: O(n * m), where n is the number of words and m is the average length of the words stored in the trie.
+var Trie = function () {
+  this.root = {}; // Initialize the trie with a root node
+};
+
+/**
+ * @param {string} word
+ * @return {void}
+ */
+
+// Time Complexity: O(m), where m is the length of the word being inserted
+Trie.prototype.insert = function (word) {
+  let currentNode = this.root;
+  for (any of word) {
+    // If the character doesn't exist, create a new node
+    if (!currentNode[any]) {
+      currentNode[any] = {};
+    }
+    currentNode = currentNode[any]; // Move to the next node
+  }
+  currentNode.end = true; // Mark the end of the word
+};
+
+/**
+ * @param {string} word
+ * @return {boolean}
+ */
+// Time Complexity: O(m), where m is the length of the word being searched
+Trie.prototype.search = function (word) {
+  let currentNode = this.root;
+  for (any of word) {
+    // If the character doesn't exist in the trie, return false
+    if (!currentNode[any]) {
+      return false;
+    }
+    currentNode = currentNode[any]; // Move to the next node
+  }
+
+  return currentNode.end === true;
+};
+
+/**
+ * @param {string} prefix
+ * @return {boolean}
+ */
+// Time Complexity: O(m), where m is the length of the prefix
+Trie.prototype.startsWith = function (prefix) {
+  let currentNode = this.root;
+
+  for (any of prefix) {
+    // If the character doesn't exist, return false
+    if (!currentNode[any]) {
+      return false;
+    }
+    currentNode = currentNode[any]; // Move to the next node
+  }
+
+  return true; // Return true if the prefix exists
+};
+
+/**
+ * Your Trie object will be instantiated and called as such:
+ * var obj = new Trie()
+ * obj.insert(word)
+ * var param_2 = obj.search(word)
+ * var param_3 = obj.startsWith(prefix)
+ */
+
+
diff --git a/word-break/Jeehay28.js b/word-break/Jeehay28.js
new file mode 100644
index 000000000..a8a8ebddc
--- /dev/null
+++ b/word-break/Jeehay28.js
@@ -0,0 +1,36 @@
+/**
+ * @param {string} s
+ * @param {string[]} wordDict
+ * @return {boolean}
+ */
+
+// Time Complexity: O(n * w * m)
+// - n is the length of the string s.
+// - w is the number of words in the dictionary wordDict.
+// - m is the average length of words in wordDict.
+
+// Space Complexity: O(n)
+// - The dp array of size n + 1 is the primary contributor to space usage, where n is the length of the string s.
+var wordBreak = function (s, wordDict) {
+  dp = new Array(s.length + 1).fill(false);
+  dp[0] = true;
+
+  // O(n)
+  for (let i = 1; i <= s.length; i++) {
+    // O(w)
+    for (word of wordDict) {
+      if (i >= word.length && s.slice(i - word.length, i) === word) {
+        // s.slice(i - word.length, i), the slicing operation takes O(m), where m is the length of the word being checked
+        dp[i] = dp[i - word.length];
+      }
+
+      if (dp[i]) {
+        break;
+      }
+    }
+  }
+
+  return dp[s.length];
+};
+
+