Javascript solutions

javaScript/easy/twoNumberSum.js

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+// SOLUTION 1:
+// O(n) time | O(n) space
+function twoNumberSum(array, targetSum) {
+  let obj = {};
+
+  for (let x of array) {
+    if (obj[targetSum - x]) {
+      return [targetSum - x, x];
+    } else {
+      obj[x] = true;
+    }
+  }
+
+  return [];
+}
+
+// SOLUTION 2:
+// O(nlog(n)) | O(1) space
+function twoNumberSum(array, targetSum) {
+  let obj = {};
+
+  for (let x of array) {
+    if (obj[targetSum - x]) {
+      return [targetSum - x, x];
+    } else {
+      obj[x] = true;
+    }
+  }
+
+  return [];
+}

javaScript/hard/patternMatcher.js

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+// pattern = "xxyxxy"
+// string = "gogopowerrangergogopowerranger"
+// output: ["go", "powerranger"]
+
+function patternMatcher(pattern, string) {
+  if (pattern.length > string.length) {
+    return [];
+  }
+
+  const newPattern = getNewPattern(pattern);
+  const hasPatternSwitched = pattern[0] !== newPattern[0];
+
+  const counts = { x: 0, y: 0 };
+
+  const firstYIndex = getCountsAndFirstYIndex(newPattern, counts);
+
+  if (counts.y !== 0) {
+    for (let lengOfX = 1; lengOfX <= string.length; lengOfX++) {
+      const lengOfY = (string.length - lengOfX * counts.x) / counts.y;
+      if (lengOfY <= 0 || lengOfY % 1 !== 0) {
+        continue;
+      }
+
+      const x = string.slice(0, lengOfX);
+      const yPosition = firstYIndex * lengOfX;
+      const y = string.slice(yPosition, yPosition + lengOfY);
+
+      const potentialMatch = newPattern
+        .map((char) => (char === 'x' ? x : y))
+        .join('');
+      if (potentialMatch === string) {
+        return hasPatternSwitched ? [y, x] : [x, y];
+      }
+    }
+  } else {
+    const lengOfX = string.length / counts.x;
+    if (lengOfX % 1 !== 0) {
+      return [];
+    }
+
+    const x = string.slice(0, lengOfX);
+
+    const potentialMatch = newPattern.map(() => x).join('');
+
+    if (potentialMatch === string) {
+      return hasPatternSwitched ? ['', x] : [x, ''];
+    }
+  }
+
+  return [];
+}
+
+function getNewPattern(pattern) {
+  let patternArray = pattern.split('');
+  if (patternArray[0] === 'x') {
+    return patternArray;
+  }
+
+  return patternArray.map((item) => (item === 'x' ? 'y' : 'x'));
+}
+
+function getCountsAndFirstYIndex(newPattern, counts) {
+  let firstYIndex = null;
+  newPattern.map((item, i) => {
+    if (item === 'y' && firstYIndex === null) {
+      firstYIndex = i;
+    }
+
+    counts[item]++;
+  });
+
+  return firstYIndex;
+}
+
+const result = patternMatcher('xxyxxy', 'gogopowerrangergogopowerranger');
+console.log(result);

javaScript/hard/solveSudoku.js

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+function solveSudoku(board) {
+  // Write your code here.
+  solveSudokuPartial(board, 0, 0);
+  return board;
+}
+
+function solveSudokuPartial(board, row, col) {
+  let currentRow = row;
+  let currentCol = col;
+
+  if (currentCol === board[currentRow].length) {
+    currentRow++;
+    currentCol = 0;
+
+    if (currentRow === board.length) return true;
+  }
+
+  if (board[currentRow][currentCol] === 0) {
+    return trialAllDigits(board, currentRow, currentCol);
+  }
+
+  return solveSudokuPartial(board, currentRow, currentCol + 1);
+}
+
+function trialAllDigits(board, row, col) {
+  for (let digit = 1; digit < 10; digit++) {
+    if (isNumberValidAtPosition(board, row, col, digit)) {
+      board[row][col] = digit;
+      if (solveSudokuPartial(board, row, col + 1)) return true;
+    }
+  }
+
+  board[row][col] = 0;
+  return false;
+}
+
+function isNumberValidAtPosition(board, row, col, value) {
+  return !(
+    hasNumberInRow(board, row, value) ||
+    hasNumberInCol(board, col, value) ||
+    hasNumberInBox(board, row, col, value)
+  );
+}
+
+function hasNumberInRow(board, row, value) {
+  return board[row].includes(value);
+}
+
+function hasNumberInCol(board, col, value) {
+  return board.map((r) => r[col]).includes(value);
+}
+
+function hasNumberInBox(board, row, col, value) {
+  const rowStart = Math.floor(row / 3) * 3;
+  const colStart = Math.floor(col / 3) * 3;
+
+  for (let rowIdx = 0; rowIdx < 3; rowIdx++) {
+    for (let colIdx = 0; colIdx < 3; colIdx++) {
+      const rowToCheck = rowStart + rowIdx;
+      const colToCheck = colStart + colIdx;
+      const currentValue = board[rowToCheck][colToCheck];
+      if (currentValue === value) return true;
+    }
+  }
+
+  return false;
+}
+// Do not edit the line below.
+exports.solveSudoku = solveSudoku;

javaScript/medium/KadanesAlgorithm.js

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+// Write a function that takes in a non-empty array of integers and returns the
+// maximum sum that can be obtained by summing up all of the integers in a
+// non-empty subarray of the input array. A subarray must only contain adjacent
+// numbers (numbers next to each other in the input array).
+
+// Sample input =  [3, 5, -9, 1, 3, -2, 3, 4, 7, 2, -9, 6, 3, 1, -5, 4]
+// Sample output = 19  // [1, 3, -2, 3, 4, 7, 2, -9, 6, 3, 1]
+
+function kadanesAlgorithm(array) {
+  let maxEndingHere = array[0];
+  let maxSoFar = array[0];
+
+  for (let i = 1; i < array.length; i++) {
+    const num = array[i];
+    maxEndingHere = Math.max(maxEndingHere + num, num);
+    maxSoFar = Math.max(maxSoFar, maxEndingHere);
+  }
+
+  return maxSoFar;
+}
+
+const input = [3, 5, -9, 1, 3, -2, 3, 4, 7, 2, -9, 6, 3, 1, -5, 4];
+const result = kadanesAlgorithm(input);
+console.log(result); // 19

javaScript/medium/binaryTreeDiameter.js

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+// Space Complexity O(h), where h is the height of the tree
+// Time Complexity O(n), where n is the number of nodes in the Binary tree
+class BinaryTree {
+  constructor(value) {
+    this.value = value;
+    this.left = null;
+    this.right = null;
+  }
+}
+
+function binaryTreeDiameter(tree) {
+  // Write your code here.
+  let diameter = 0;
+  function dfsHeightCalculation(node) {
+    if (!node) {
+      return 0;
+    }
+
+    const leftHeight = dfsHeightCalculation(node.left, diameter);
+    const rightHeight = dfsHeightCalculation(node.right, diameter);
+
+    diameter = Math.max(diameter, leftHeight + rightHeight);
+
+    return 1 + Math.max(leftHeight, rightHeight);
+  }
+
+  dfsHeightCalculation(tree);
+
+  return diameter;
+}

javaScript/medium/bstTraversal.js

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+// O(n) time | O(n) space
+
+//                      10
+//                    5     15
+//                  2   5       22
+//                1
+
+function inOrderTraverse(tree, array) {
+  // [1, 2, 5, 5, 10, 15, 22]
+  if (tree !== null) {
+    inOrderTraverse(tree.left, array);
+    array.push(tree.value);
+    inOrderTraverse(tree.right, array);
+  }
+  return array;
+}
+
+function preOrderTraverse(tree, array) {
+  // [10, 5, 2, 1, 5, 15, 22]
+  if (tree !== null) {
+    array.push(tree.value);
+    preOrderTraverse(tree.left, array);
+    preOrderTraverse(tree.right, array);
+  }
+  return array;
+}
+
+function postOrderTraverse(tree, array) {
+  // [1, 2, 5, 5, 22, 15, 10]
+  if (tree !== null) {
+    postOrderTraverse(tree.left, array);
+    postOrderTraverse(tree.right, array);
+    array.push(tree.value);
+  }
+  return array;
+}

javaScript/medium/invertBinaryTree.js

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+function invertBinaryTree(tree) {
+  // Write your code here.
+  if (!tree) {
+    return;
+  }
+
+  swapLeftAndRight(tree);
+  invertBinaryTree(tree.left);
+  invertBinaryTree(tree.right);
+
+  return tree;
+}
+
+function swapLeftAndRight(tree) {
+  let tempNode = tree.left;
+  tree.left = tree.right;
+  tree.right = tempNode;
+}
+
+// This is the class of the input binary tree.
+class BinaryTree {
+  constructor(value) {
+    this.value = value;
+    this.left = null;
+    this.right = null;
+  }
+}

javaScript/medium/kthLargestValueBST.js

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+class BST {
+  constructor(value) {
+    this.value = value;
+    this.left = null;
+    this.right = null;
+  }
+}
+
+// Solution 1:
+// O(n) time O(n) space
+function findKthLargestValueInBst(tree, k) {
+  const sortedNodeValues = [];
+  inorderTraverse(tree, sortedNodeValues);
+  return sortedNodeValues[sortedNodeValues.length - k];
+}
+
+function inorderTraverse(node, sortedNodeValues) {
+  if (node === null) return;
+
+  inorderTraverse(node.left, sortedNodeValues);
+  sortedNodeValues.push(node.value);
+  inorderTraverse(node.right, sortedNodeValues);
+}
+
+// Solution 2:
+// O(h + k) time | O(h) space - where h is the height of the tree and k is the input parameter
+
+class TreeInfo {
+  constructor(numberOfNodesVisited, latestVisitedValue) {
+    this.numberOfNodesVisited = numberOfNodesVisited;
+    this.latestVisitedValue = latestVisitedValue;
+  }
+}
+
+function findKthLargestValueInBst2(tree, k) {
+  // Write your code here.
+  const treeInfo = new TreeInfo(0, -1);
+  reverseInorderTraverse(tree, treeInfo, k);
+  return treeInfo.latestVisitedValue;
+}
+
+function reverseInorderTraverse(node, treeInfo, k) {
+  if (node === null || treeInfo.numberOfNodesVisited >= k) {
+    return;
+  }
+  reverseInorderTraverse(node.right, treeInfo, k);
+  if (treeInfo.numberOfNodesVisited < k) {
+    treeInfo.numberOfNodesVisited += 1;
+    treeInfo.latestVisitedValue = node.value;
+
+    reverseInorderTraverse(node.left, treeInfo, k);
+  }
+}
+
+// Do not edit the lines below.
+exports.BST = BST;
+exports.findKthLargestValueInBst = findKthLargestValueInBst;