Largest remainder method of fixing 101%

largestRemainder.ts

/** Scales values to be a percentage of their sum. E.g., [20, 30] -> [40, 60] */
function scaleToPercentages(rows: Row[]): Row[] {
  rows.forEach(([_, vs]) => Preconditions.checkArgument(vs.length === 1)); // Only supports one
                                                                           // column.
  if (rows.length === 0) {
    return rows;
  }
   const sum = rows
      .map(([_, [v]]) => v)
      .reduce((acc, v) => acc + v, 0);
  const percents: Row[] = rows.map(([l, vs]) => ([
    l,
    vs.map(v => (v / sum) * 100),
  ] as Row));
   // According to the "Largest remainder method" (to avoid "101%" case):
  // 1. Take floored percentage values
  // 2. Calc remaining percents
  // 3. Distribute them between the largest remainder values
  let remainingPercents = 100;
  const remainders: [number, number][] = []; // [index, remainder][]
  for (let i = 0; i < percents.length; i++) {
    const [, [decimal]] = percents[i];
    const integer = Math.floor(+decimal);
    remainingPercents -= integer;
    remainders[i] = [i, +decimal - integer];
    percents[i][1] = [integer];
  }
   // Distribute remaining percents between the highest remainder values.
  remainders.sort((a, b) => a[1] === b[1]
      ? a[0] - b[0] // save position if values are equal by comparing indexes
      : b[1] - a[1]);
  for (let i = 0; i < remainingPercents; i++) {
    const [index] = remainders[i];
    const [, [value]] = percents[index];
    percents[index][1] = [+value + 1];
  }
  return percents;
}