Find the busiest tray run

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The pickup counter records how many drinks leave during each minute. A tray runner will cover exactly width consecutive minutes. Return [start, drinks], where start is the zero-based first minute of the run serving the most drinks and drinks is its total. If several runs tie, choose the earliest starting minute. Every minute in the chosen run must be included.

Example 1

INPUT[2,5,1,6,3], 3OUTPUT[1,12]

The three totals are 8, 12, and 10. The best run starts at minute 1.

Example 2

INPUT[4,0,4,0], 2OUTPUT[0,4]

All three runs serve four drinks, so choose the earliest start, 0.

Keep in mind

1 <= orders.length <= 20,000
Each orders entry is an integer from 0 to 1,000,000.
1 <= width <= orders.length
Totals fit exactly in a JavaScript Number.

/** @param {number[]} orders
 *  @param {number} width
 *  @returns {number[]} */
function solve(orders, width) {
  // Return the earliest busiest run and its drink total.
}

/** @param {number[]} orders
 *  @param {number} width
 *  @returns {number[]} */
function solve(orders, width) {
  // Return the earliest busiest run and its drink total.
}

INPUT[2,5,1,6,3], 3

EXPECTED OUTPUT[1,12]

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Find the busiest tray run

Example 1

Example 2

Keep in mind

How the solution works.

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