Eliminate Maximum Number of Monsters
MediumArrayGreedySorting
Solution
export function eliminateMaximum(dists: number[], speeds: number[]): number {
const n = dists.length;
const remainMinutes = dists.map((dist, i) => Math.ceil(dist / speeds[i])).sort((a, b) => a - b);
let answer = 0;
while (answer < n && answer < remainMinutes[answer]) {
answer += 1;
}
return answer;
}