Minimum Initial Energy to Finish Tasks LeetCode solution in JavaScript

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var minimumEffort = function (tasks) {
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  let minimumStartingEnergy = 0;
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  let currentEnergy = 0;
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  tasks.sort((a, b) => b[1] - b[0] - (a[1] - a[0]));
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  for (let task of tasks) {
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    if (task[1] > currentEnergy) {
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      minimumStartingEnergy += task[1] - currentEnergy;
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      currentEnergy = task[1];
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    }
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    currentEnergy -= task[0];
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  }
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  return minimumStartingEnergy;
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};