[Jeehay28] Week 09 Solutions

linked-list-cycle/Jeehay28.ts

@@ -0,0 +1,52 @@
+class ListNode {
+  val: number;
+  next: ListNode | null;
+  constructor(val?: number, next?: ListNode | null) {
+    this.val = val === undefined ? 0 : val;
+    this.next = next === undefined ? null : next;
+  }
+}
+
+// TC: O(n)
+// SC: O(1)
+function hasCycle(head: ListNode | null): boolean {
+
+  let fast: ListNode | null = head;
+  let slow: ListNode | null = head;
+
+  while (fast && fast.next) {
+    fast = fast.next.next;
+
+    if (slow !== null) {
+      slow = slow.next;
+    } else {
+      return false;
+    }
+
+    if (fast === slow) return true;
+  }
+
+  return false;
+}
+
+
+// TC: O(n)
+// SC: O(n)
+/*
+function hasCycle(head: ListNode | null): boolean {
+
+  let dummy: ListNode | null = head;
+  const visited = new Set<ListNode>();
+
+  while (dummy) {
+    if (visited.has(dummy)) return true;
+
+    visited.add(dummy);
+
+    dummy = dummy.next;
+  }
+
+  return false;
+}
+*/
+

maximum-product-subarray/Jeehay28.ts

@@ -0,0 +1,40 @@
+// TC: O(n)
+// SC: O(1)
+function maxProduct(nums: number[]): number {
+  let prevMax = nums[0];
+  let prevMin = nums[0];
+  let maxProd = nums[0];
+
+  for (let i = 1; i < nums.length; i++) {
+    const currentNum = nums[i];
+    const tempMax = Math.max(
+      currentNum,
+      prevMax * currentNum,
+      prevMin * currentNum
+    );
+    prevMin = Math.min(currentNum, prevMax * currentNum, prevMin * currentNum);
+    prevMax = tempMax;
+    maxProd = Math.max(maxProd, prevMax);
+  }
+
+  return maxProd;
+}
+
+
+// TC: O(n^2)
+// SC: O(1)
+// function maxProduct(nums: number[]): number {
+//   let largestProd = -Infinity;
+//   let subProd = 1;
+
+//   for (let i = 0; i < nums.length; i++) {
+//     subProd = 1;
+//     for (let j = i; j < nums.length; j++) {
+//       subProd *= nums[j];
+//       largestProd = Math.max(largestProd, subProd);
+//     }
+//   }
+
+//   return largestProd;
+// }
+

pacific-atlantic-water-flow/Jeehay28.ts

@@ -0,0 +1,71 @@
+// TC: O(m * n)
+// SC: O(m * n)
+
+function pacificAtlantic(heights: number[][]): number[][] {
+  const m = heights.length;
+  const n = heights[0].length;
+
+  const pacific: boolean[][] = Array.from({ length: m }, () =>
+    Array(n).fill(false)
+  );
+  const atlantic: boolean[][] = Array.from({ length: m }, () =>
+    Array(n).fill(false)
+  );
+
+  const directions = [
+    [0, 1],
+    [0, -1],
+    [1, 0],
+    [-1, 0],
+  ];
+
+  const dfs = (row: number, col: number, visited: boolean[][]) => {
+    visited[row][col] = true;
+    // Border cells are adjacent to the ocean, so they are marked as reachable (true)
+
+    // From each ocean-border cell, we explore inward:
+    // if an adjacent cell is higher or equal, water can flow from it to the ocean
+    // → mark it as reachable (true)
+    for (const [dr, dc] of directions) {
+      const newRow = row + dr;
+      const newCol = col + dc;
+
+      if (
+        newRow >= 0 &&
+        newRow < m &&
+        newCol >= 0 &&
+        newCol < n &&
+        !visited[newRow][newCol] &&
+        heights[row][col] <= heights[newRow][newCol]
+        // heights[row][col] <= heights[newRow][newCol]
+        // → Water can flow from (newRow, newCol) to (row, col),
+        // so (row, col) is reachable from the ocean through (newRow, newCol)
+      ) {
+        dfs(newRow, newCol, visited);
+      }
+    }
+  };
+
+  for (let i = 0; i < m; i++) {
+    dfs(i, 0, pacific);
+    dfs(i, n - 1, atlantic);
+  }
+
+  for (let j = 0; j < n; j++) {
+    dfs(0, j, pacific);
+    dfs(m - 1, j, atlantic);
+  }
+
+  let result: number[][] = [];
+
+  for (let i = 0; i < m; i++) {
+    for (let j = 0; j < n; j++) {
+      if (pacific[i][j] && atlantic[i][j]) {
+        result.push([i, j]);
+      }
+    }
+  }
+
+  return result;
+}
+

sum-of-two-integers/Jeehay28.ts

@@ -0,0 +1,12 @@
+// TC: O(1)
+// SC: O(1)
+function getSum(a: number, b: number): number {
+  while (b !== 0) {
+    let carry = a & b;
+    a = a ^ b; // sum WITHOUT the carry
+    b = carry << 1; // new carry
+  }
+
+  return a;
+}
+