[Jeehay28] WEEK 09

find-minimum-in-rotated-sorted-array/Jeehay28.js

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+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+
+// Binary Search
+// Time Complexity: O(log n) (Binary search halving the search space each step)
+// Space Complexity: O(1) (No extra space except for a few variables)
+var findMin = function (nums) {
+  // Example Rotations:
+  // nums = [0,1,2,4,5,6,7]
+  // [4,5,6,7,0,1,2] -> Rotated 4 times
+  // [0,1,2,4,5,6,7] -> 7 times rotated
+
+  // Initial State:
+  // [4, 5, 6, 7, 0, 1, 2]
+  //  ↑                 ↑
+  // (left)           (right)
+  //
+  // Find mid:
+  // [4, 5, 6, 7, 0, 1, 2]
+  //  ↑       🎯        ↑
+  // (left)   (mid)   (right)
+  //
+  // If nums[mid] > nums[right], move left to search in the right half:
+  // [4, 5, 6, 7, 0, 1, 2]
+  //              ↑     ↑
+  //          (left)   (right)
+
+  let left = 0;
+  let right = nums.length - 1;
+
+  while (left < right) {
+    let mid = Math.floor((left + right) / 2);
+
+    if (nums[mid] > nums[right]) {
+      // Minimum must be in the right half
+      // Need to update left to search in the right half
+      left = mid + 1;
+    } else {
+      // Minimum is in the left half (including mid)
+      // Need to update right to search in the left half
+      right = mid;
+    }
+  }
+
+  return nums[left];
+};
+

linked-list-cycle/Jeehay28.js

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+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
+ */
+
+/**
+ * @param {ListNode} head
+ * @return {boolean}
+ */
+
+// Time Complexity: O(n)
+// Space Complexity: O(1)
+
+// - In the worst case, we might traverse the entire linked list, but because the fast pointer moves at twice the speed of the slow pointer, they will meet within O(N) steps if a cycle exists.
+// - If there is no cycle, the fast pointer reaches the end in O(N) steps.
+// - Only two pointers (slow and fast) are used, which require O(1) extra space.
+// - No additional data structures (like arrays or hash sets) are used.
+
+var hasCycle = function (head) {
+  // If there is a cycle in the linked list, Floyd's Tortoise and Hare algorithm guarantees
+  // that the fast and slow pointers will eventually meet.
+  let fast = head;
+  let slow = head;
+
+  while (fast && fast.next) {
+    slow = slow.next; // Move slow pointer one step.
+    fast = fast.next.next; // Move fast pointer two steps.
+
+    if (slow === fast) {
+      return true; // Cycle detected.
+    }
+  }
+
+  return false;
+};
+

maximum-product-subarray/Jeehay28.js

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+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+
+// ✅ DP approach
+// Time Complexity: O(n)
+// Space Complexity: O(1)
+var maxProduct = function (nums) {
+  // Track the minimum and maximum product ending at the current position
+  // Consider three possibilities for each element:
+  // 1. Current number
+  // 2. Previous min product * current number (to handle negative numbers)
+  // 3. Previous max product * current number (to handle negative numbers)
+
+  let maxProduct = nums[0];
+  let previousMinProduct = 1;
+  let previousMaxProduct = 1;
+
+  for (const current of nums) {
+    const temp1 = previousMinProduct * current;
+    const temp2 = previousMaxProduct * current;
+
+    // Update min and max product
+    previousMinProduct = Math.min(current, temp1, temp2);
+    previousMaxProduct = Math.max(current, temp1, temp2);
+
+    // Update maxProduct
+    maxProduct = Math.max(maxProduct, previousMaxProduct);
+  }
+
+  return maxProduct;
+};
+
+// 🤔 more efficient than the previous one, but there might be a further optimization
+// using dynamic programming approach to reduce the complexity to O(n)
+// Time Complexity: O(n^2)
+// Space Complexity: O(1)
+
+// var maxProduct = function (nums) {
+//   let max = nums[0];
+
+//   for (let s = 0; s < nums.length; s++) {
+//     let temp = 1;
+//     for (let e = s; e < nums.length; e++) {
+//       temp *= nums[e];
+//       max = Math.max(max, temp);
+//     }
+//   }
+//   return max;
+// };
+
+// 😱 Time Limit Exceeded!
+// Time Complexity: O(n^3)
+// Space Complexity: O(1)
+// var maxProduct = function (nums) {
+//   let max = nums[0];
+
+//   for (let s = 0; s < nums.length; s++) {
+//     for (let e = s; e < nums.length; e++) {
+//       let temp = 1;
+
+//       for (let i = s; i <= e; i++) {
+//         temp *= nums[i];
+//       }
+
+//       max = Math.max(max, temp);
+//     }
+//   }
+
+//   return max;
+// };
+
+

minimum-window-substring/Jeehay28.js

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+/**
+ * @param {string} s
+ * @param {string} t
+ * @return {string}
+ */
+
+// 🚀 sliding window + two pointer approach
+
+// 🕒 Time Complexity: O(n), where n is the length of s
+// The inner while loop shrinks the window from the left side but never exceeds the total number of characters in s
+
+// 🗂 Space Complexity: O(m) (or worst case O(n)), where n is the length of t
+
+var minWindow = function (s, t) {
+  // early return for the critical edge case
+  if (s.length < t.length) {
+    return "";
+  }
+
+  let windowCnt = new Map();
+  let charCnt = new Map();
+  let minStart = 0;
+  let minEnd = s.length; // set this to an out-of-bounds value initially
+
+  let formed = 0;
+  let left = 0;
+
+  // initialize charCount
+  // 🔢 t = "ABC", charCount = { A: 1, B: 1, C: 1 }
+  for (const ch of t) {
+    charCnt.set(ch, (charCnt.get(ch) || 0) + 1);
+  }
+
+  // expand the windowCnt
+  for (let right = 0; right < s.length; right++) {
+    const char = s[right];
+
+    windowCnt.set(char, (windowCnt.get(char) || 0) + 1);
+
+    if (charCnt.has(char) && charCnt.get(char) === windowCnt.get(char)) {
+      formed += 1;
+    }
+
+    // shrink the window by moving the left pointer
+    while (formed === charCnt.size) {
+      if (right - left < minEnd - minStart) {
+        minStart = left;
+        minEnd = right;
+      }
+
+      const char = s[left];
+      windowCnt.set(char, windowCnt.get(char) - 1);
+
+      if (charCnt.has(char) && windowCnt.get(char) < charCnt.get(char)) {
+        formed -= 1;
+      }
+      left += 1;
+    }
+  }
+
+  return minEnd === s.length ? "" : s.slice(minStart, minEnd + 1);
+};
+

pacific-atlantic-water-flow/Jeehay28.js

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+/**
+ * @param {number[][]} heights
+ * @return {number[][]}
+ */
+
+// ✅ Time Complexity: O(m * n)
+// ✅ Space Complexity: O(m * n)
+
+var pacificAtlantic = function (heights) {
+  const m = heights.length;
+  const n = heights[0].length;
+
+  // ❌ Array(m).fill(Array(n).fill(false)) → Incorrect (all rows share the same array)
+  const pacific = Array.from({ length: m }, () => Array(n).fill(false));
+  const atlantic = Array.from({ length: m }, () => Array(n).fill(false));
+
+  // four possible movement directions: down, up, right, left
+  const directions = [
+    [1, 0], // ⬇️
+    [-1, 0], // ⬆️
+    [0, 1], // ➡️
+    [0, -1], // ⬅️
+  ];
+
+  /**
+   * Depth-First Search (DFS) to mark reachable cells
+   * @param {number} row - current row index
+   * @param {number} col - current column index
+   * @param {boolean[][]} visited - visited cells
+   * @param {number} prevHeight - previously visited cell's height
+   */
+
+  const dfs = (row, col, visited, prevHeight) => {
+    // No search needed:
+    // 1) Out of bounds
+    // 2) already visited
+    // 3) current height < previous height
+    if (
+      row < 0 ||
+      row >= m ||
+      col < 0 ||
+      col >= n ||
+      visited[row][col] ||
+      heights[row][col] < prevHeight
+    ) {
+      return;
+    }
+
+    // mark current cell as visited
+    visited[row][col] = true;
+
+    // visit all four possible directions
+    for (const [dr, dc] of directions) {
+      dfs(row + dr, col + dc, visited, heights[row][col]);
+    }
+  };
+
+  // start dfs from each border
+  for (let i = 0; i < m; i++) {
+    dfs(i, 0, pacific, heights[i][0]); // left border(Pacific ocean)
+    dfs(i, n - 1, atlantic, heights[i][n - 1]); // right border(Atlantic ocean)
+  }
+
+  for (let j = 0; j < n; j++) {
+    dfs(0, j, pacific, heights[0][j]); // top border(Pacific ocean)
+    dfs(m - 1, j, atlantic, heights[m - 1][j]); // bottom border(Atlantic ocean)
+  }
+
+  let result = [];
+
+  for (let r = 0; r < m; r++) {
+    for (let c = 0; c < n; c++) {
+      if (pacific[r][c] && atlantic[r][c]) {
+        result.push([r, c]);
+      }
+    }
+  }
+  return result;
+};
+
+// Example test
+const heights = [
+  [1, 2, 2, 3, 5],
+  [3, 2, 3, 4, 4],
+  [2, 4, 5, 3, 1],
+  [6, 7, 1, 4, 5],
+  [5, 1, 1, 2, 4],
+];
+
+const expected = [
+  [0, 4],
+  [1, 3],
+  [1, 4],
+  [2, 2],
+  [3, 0],
+  [3, 1],
+  [4, 0],
+];
+
+const output = pacificAtlantic(heights);
+
+if (JSON.stringify(output.sort()) === JSON.stringify(expected.sort())) {
+  console.log("✅ Accepted\n");
+} else {
+  console.log("❌ Not Accepted\n");
+}
+
+console.log("Input:", heights, "\n");
+console.log("Output:", output, "\n");
+console.log("Expected:", expected);
+