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1665. Minimum Initial Energy to Finish Tasks

Description

You are given an array tasks where tasks[i] = [actuali, minimumi]:

  • actuali is the actual amount of energy you spend to finish the ith task.
  • minimumi is the minimum amount of energy you require to begin the ith task.

For example, if the task is [10, 12] and your current energy is 11, you cannot start this task. However, if your current energy is 13, you can complete this task, and your energy will be 3 after finishing it.

You can finish the tasks in any order you like.

Return the minimum initial amount of energy you will need to finish all the tasks.

 

Example 1:

Input: tasks = [[1,2],[2,4],[4,8]]
Output: 8
Explanation:
Starting with 8 energy, we finish the tasks in the following order:
    - 3rd task. Now energy = 8 - 4 = 4.
    - 2nd task. Now energy = 4 - 2 = 2.
    - 1st task. Now energy = 2 - 1 = 1.
Notice that even though we have leftover energy, starting with 7 energy does not work because we cannot do the 3rd task.

Example 2:

Input: tasks = [[1,3],[2,4],[10,11],[10,12],[8,9]]
Output: 32
Explanation:
Starting with 32 energy, we finish the tasks in the following order:
    - 1st task. Now energy = 32 - 1 = 31.
    - 2nd task. Now energy = 31 - 2 = 29.
    - 3rd task. Now energy = 29 - 10 = 19.
    - 4th task. Now energy = 19 - 10 = 9.
    - 5th task. Now energy = 9 - 8 = 1.

Example 3:

Input: tasks = [[1,7],[2,8],[3,9],[4,10],[5,11],[6,12]]
Output: 27
Explanation:
Starting with 27 energy, we finish the tasks in the following order:
    - 5th task. Now energy = 27 - 5 = 22.
    - 2nd task. Now energy = 22 - 2 = 20.
    - 3rd task. Now energy = 20 - 3 = 17.
    - 1st task. Now energy = 17 - 1 = 16.
    - 4th task. Now energy = 16 - 4 = 12.
    - 6th task. Now energy = 12 - 6 = 6.

 

Constraints:

  • 1 <= tasks.length <= 105
  • 1 <= actual​i <= minimumi <= 104

 

Solutions

Solution: Greedy

  • Time complexity: O(nlogn)
  • Space complexity: O(1)

 

JavaScript

js
/**
 * @param {number[][]} tasks
 * @return {number}
 */
const minimumEffort = function (tasks) {
  let current = 0;
  let result = 0;

  tasks.sort((a, b) => {
    const diffA = a[1] - a[0];
    const diffB = b[1] - b[0];

    return diffB - diffA;
  });

  for (const [actual, minimum] of tasks) {
    const energy = Math.max(current, minimum);

    result += energy - current;
    current = energy - actual;
  }

  return result;
};

Released under the MIT license