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| 1 | +/** |
| 2 | + * @param {number[]} jobs |
| 3 | + * @param {number} k |
| 4 | + * @return {number} |
| 5 | + */ |
| 6 | +const minimumTimeRequired = function (jobs, k) { |
| 7 | + if (jobs.length <= k) { |
| 8 | + return Math.max(...jobs) |
| 9 | + } |
| 10 | + |
| 11 | + // create a store to hold the number of hours each worker worked |
| 12 | + const workers = new Array(k).fill(0) |
| 13 | + |
| 14 | + let minLongestWorkingTime = Infinity |
| 15 | + const dfs = (i) => { |
| 16 | + if (i === jobs.length) { |
| 17 | + // if we assigned all the jobs, see if we have a better result |
| 18 | + minLongestWorkingTime = Math.min( |
| 19 | + minLongestWorkingTime, |
| 20 | + Math.max(...workers) |
| 21 | + ) |
| 22 | + return |
| 23 | + } |
| 24 | + const lengthOfWork = jobs[i] |
| 25 | + |
| 26 | + for (let worker = 0; worker < k; worker++) { |
| 27 | + workers[worker] += lengthOfWork |
| 28 | + |
| 29 | + // if this combination is has a chance of decreasing our |
| 30 | + // answer, try it, otherwise skip it to save on time. |
| 31 | + if (workers[worker] <= minLongestWorkingTime) { |
| 32 | + dfs(i + 1) |
| 33 | + } |
| 34 | + workers[worker] -= lengthOfWork |
| 35 | + |
| 36 | + // We want to minimize the width of the tree |
| 37 | + // so if the worker has gotten their first job |
| 38 | + // don't try any workers after this worker. |
| 39 | + // All other workers after this worker will be 0 as well |
| 40 | + // so the combination is exactly the same. |
| 41 | + if (workers[worker] === 0) break |
| 42 | + } |
| 43 | + } |
| 44 | + |
| 45 | + dfs(0) |
| 46 | + return minLongestWorkingTime |
| 47 | +} |
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