The full question
You are asked to solve the following four independent coding problems.
---
1) Block Placement Simulator (Tetris-like)
You have an empty n x m grid (rows indexed top-to-bottom, columns left-to-right). You will place a sequence of pieces, each piece being one of five types A/B/C/D/E.
Each piece type is defined by a fixed set of occupied cells in its own local coordinate system (e.g., a small boolean matrix or a list of (dr, dc) offsets). Pieces:
- cannot be rotated or flipped,
- must be placed fully within the grid,
- cannot overlap already-occupied cells.
Placement rule (scan order): For each incoming piece, find the first valid placement position by scanning candidate top-left anchors in priority order: 1) smallest row index (top to bottom), 2) within the same row, smallest column index (left to right).
Place the piece at that first valid position. If no valid position exists, stop and return the grid/state according to the output requirement.
Task: Implement the simulator that processes the piece sequence and produces the final grid (or alternatively the number of pieces placed), following the scan-order placement rule.
---
2) Longest Continuous Houses After Each Build
On an integer number line, you build houses one by one at positions given by an array queries, in order. After each build, you must output the current length of the longest contiguous segment of built houses, where “contiguous” means consecutive integer positions.
Example: if houses exist at {2,3,4,10}, the longest contiguous segment length is 3 (for 2-4).
Notes:
- Coordinates can be very large (e.g., up to billions),
- The number of builds is manageable,
- Duplicate builds (building again at an already-built coordinate) should be handled (either ignored or specified by the problem).
Task: Return an array ans where ans[i] is the longest contiguous segment length after processing queries[0..i].
---
3) Rating Calculator (Max and Final Rating)
You start with rating 1500. You are given an integer array diffs, where diffs[i] is the rating change applied after the i-th event.
Let rating_0 = 1500 and rating_{i+1} = rating_i + diffs[i].
Task: Compute: 1) the maximum rating ever reached across the whole process (including the initial rating), and 2) the final rating after all changes.
Return (max_rating, final_rating).
---
4) Array Reduction Game
Given an array of non-negative integers a:
Repeat until all elements become 0: 1) Find the leftmost index i such that a[i] > 0. Let x = a[i]. 2) Starting from index i and moving right, subtract x from each element while it is legal to do so (i.e., continue for indices j = i, i+1, ... as long as a[j] >= x at the moment you are about to subtract). Stop at the first position where the current value is < x. 3) Add x to an accumulator result.
Task: Return result.
(You may assume inputs are such that the process terminates; implement according to the exact step rules above.)
Model answer
// 1) Block Placement Simulator (Tetris-like)
function blockPlacementSimulator(n, m, pieces) {
const grid = Array.from({ length: n }, () => Array(m).fill(0));
const pieceShapes = {
A: [[1, 0], [1, 1], [1, 2]],
B: [[0, 0], [0, 1], [1, 0]],
C: [[0, 0], [1, 0], [2, 0]],
D: [[0, 0], [0, 1], [1, 1]],
E: [[0, 0], [1, 0], [1, 1], [1, 2]]
};
for (const piece of pieces) {
const shape = pieceShapes[piece];
let placed = false;
for (let r = 0; r < n; r++) {
for (let c = 0; c < m; c++) {
if (canPlacePiece(grid, shape, r, c)) {
placePiece(grid, shape, r, c);
placed = true;
break;
}
}
if (placed) break;
}
if (!placed) break; // Stop if no valid position found
}
return grid;
}
function canPlacePiece(grid, shape, row, col) {
for (const [dr, dc] of shape) {
const r = row + dr;
const c = col + dc;
if (r < 0 || r >= grid.length || c < 0 || c >= grid[0].length || grid[r][c] !== 0) {
return false;
}
}
return true;
}
function placePiece(grid, shape, row, col) {
for (const [dr, dc] of shape) {
grid[row + dr][col + dc] = 1; // Mark the cell as occupied
}
}
// 2) Longest Continuous Houses After Each Build
function longestContinuousHouses(queries) {
const builtHouses = new Set();
const ans = [];
for (const position of queries) {
builtHouses.add(position);
ans.push(getLongestSegment(builtHouses));
}
return ans;
}
function getLongestSegment(builtHouses) {
const sortedHouses = Array.from(builtHouses).sort((a, b) => a - b);
let maxLength = 0;
let currentLength = 1;
for (let i = 1; i < sortedHouses.length; i++) {
if (sortedHouses[i] === sortedHouses[i - 1] + 1) {
currentLength++;
} else {
maxLength = Math.max(maxLength, currentLength);
currentLength = 1;
}
}
return Math.max(maxLength, currentLength);
}
// 3) Rating Calculator (Max and Final Rating)
function ratingCalculator(diffs) {
let maxRating = 1500;
let finalRating = 1500;
for (const diff of diffs) {
finalRating += diff;
maxRating = Math.max(maxRating, finalRating);
}
return [maxRating, finalRating];
}
// 4) Array Reduction Game
function arrayReductionGame(a) {
let result = 0;
while (true) {
let i = a.findIndex(x => x > 0);
if (i === -1) break; // All elements are zero
let x = a[i];
for (let j = i; j < a.length && a[j] >= x; j++) {
a[j] -= x;
}
result += x;
}
return result;
}
// Approach:
// - Implemented four independent functions for each problem.
// - Each function handles its specific logic and returns the required result.
// - Used helper functions where necessary for clarity and reusability.
// Complexity:
// - Block Placement: O(n * m * p) where p is the number of pieces.
// - Longest Houses: O(k log k) where k is the number of unique house positions.
// - Rating Calculator: O(n) where n is the length of diffs.
// - Array Reduction: O(n^2) in the worst case for the reduction process.
ChannelPulse