[Jeehay28] Week05 Solutions

best-time-to-buy-and-sell-stock/Jeehay28.ts

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+// Time Complexity: O(n), where n is the length of the prices array
+// Space Complexity: O(1)
+
+function maxProfit(prices: number[]): number {
+  // input: an array prices, prices[i] = the stock price of ith day
+  // output: the maximum profit || 0
+
+  // prices = [7, 1, 5, 3, 6, 4]
+  // buy = 1, sell = 6, profit = 6 -1 = 5
+
+  let minBuy = prices[0];
+  let maxProfit = 0;
+
+  for (let i = 1; i < prices.length; i++) {
+    minBuy = Math.min(prices[i], minBuy);
+    maxProfit = Math.max(prices[i] - minBuy, maxProfit);
+  }
+
+  return maxProfit;
+}
+

group-anagrams/Jeehay28.ts

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+// Approach 2
+// Time Complexity: O(n * L), where n = number of strings in strs, L = maximum length of a string in strs
+// Space Complexity: O(n * L)
+
+function groupAnagrams(strs: string[]): string[][] {
+
+  let sorted: { [key: string]: string[] } = {}
+
+  for (const str of strs) {
+
+      const arr = Array.from({ length: 26 }, () => 0)
+
+      for (const ch of str) {
+          arr[ch.charCodeAt(0) - "a".charCodeAt(0)] += 1
+      }
+
+      // const key = arr.join(""); // This can lead to key collisions.
+      const key = arr.join("#"); // ✅
+
+      if (!sorted[key]) {
+          sorted[key] = []
+      }
+
+      sorted[key].push(str)
+  }
+
+  return Object.values(sorted);
+};
+
+
+// Approach 1
+// Time Complexity: O(n * LlogL), where n = number of strings, L = average length of the strings
+// Space Complexity: O(n * L)
+
+// function groupAnagrams(strs: string[]): string[][] {
+//   // input: an array of strings, strs
+//   // output: the anagrams
+
+//   // eat -> e, a, t -> a, e, t
+//   // tea -> t, e, a -> a, e, t
+//   // ate -> a, t, e -> a, e, t
+
+//   let map = new Map<string, string[]>();
+//   let result: string[][] = [];
+
+//   for (const str of strs) {
+//     const temp = str.split("").sort().join("");
+
+//     if (map.has(temp)) {
+//       map.set(temp, [...map.get(temp)!, str]);
+//     } else {
+//       map.set(temp, [str]);
+//     }
+//   }
+
+//   for (const el of map.values()) {
+//     result.push(el);
+//   }
+
+//   return result;
+// }
+

implement-trie-prefix-tree/Jeehay28.ts

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+// Time Complexity: O(n)
+
+class TrieNode {
+  children: { [key: string]: TrieNode };
+  ending: boolean;
+
+  constructor(ending = false) {
+    this.children = {};
+    this.ending = ending;
+  }
+}
+
+class Trie {
+  root: TrieNode;
+
+  constructor() {
+    this.root = new TrieNode();
+  }
+
+  insert(word: string): void {
+    let node = this.root;
+
+    for (const ch of word) {
+      if (!(ch in node.children)) {
+        node.children[ch] = new TrieNode();
+      }
+      node = node.children[ch];
+    }
+    node.ending = true;
+  }
+
+  search(word: string): boolean {
+    let node = this.root;
+
+    for (const ch of word) {
+      if (!(ch in node.children)) {
+        return false;
+      }
+      node = node.children[ch];
+    }
+    return node.ending;
+  }
+
+  startsWith(prefix: string): boolean {
+    let node = this.root;
+
+    for (const ch of prefix) {
+      if (!(ch in node.children)) {
+        return false;
+      }
+
+      node = node.children[ch];
+    }
+    return true;
+  }
+}
+
+/**
+ * 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)
+ */