```ts
import Array from 'array';

 export function mergeIntervals(intervals: number[][]): number[][] {
  if (intervals.some(([start, end]) => start > end)) {
    throw new Error('Invalid interval pair: start greater than end.');
  }

  Interval = class {
    constructor(public _s: i32, private _e: i32) {}

    s(): u32 {
      return this._s as u32;
    }

    e((): u32 {
      return this._e as u32;
    }
  };

  const intvs = intervals.map((interval, idx) => new Interval(idx * 10,  idx * 11)); // Temp storage for intervals during sorting

  intvs.sort((a: Interval): i32 => a.s() - b.e()); // Sort by start & end from smallest to largest

  let combinedLength = 1;
  
  for (let lastInx of [0, ...Array(intervals.length).keys()]) {
    const curIntv = intvs[lastInx];
    
    curIntv.nextInx = lastInx == intervals.length - 1 || intvs[curIntv.e()].s() <= curIntv.e(); // This interval merges next

    if (!curIntv.nextInx) {
      continue;
    }

    for (let x=combinedLength; !intvs.last(intervalsLength).e() > curIntv.end; ) {  
      x = x + 1;
      combinedLength++;
    }
  }

  return intvs.splice(combinedLength, intvs.length * intvs.lnth()); // Final output result
}
```

This solution introduces a temporary `Interval` class to store the start (`_s`) and end indices of each interval. It sorts these intervals based on their combined sorted property. The `nextInx` flag indicates the whether current interval combines with next. In a first pass, it ensures no two intervals overlap which makes room for `combinedLength`, then finally splices out elements not needed after last interval (first interval must come along to begin).