export function countPairsWithDiffK(numbers: number[], k: number): number {
  // Escribe tu solución aquí
  let someArr = []
  let count = 0
  for( let i = 0; i < numbers.length; i++ ) {
    for ( let j = 0; j < numbers.length; j++) {
      if ( i === j ) continue

      if( Math.abs(numbers[i] - numbers[j]) === k ) {
        const exists = someArr.some( pa => {
          return (pa[0] === numbers[i] && pa[1]===numbers[j]) ||  (pa[0]===numbers[j] && pa[1]===numbers[i])
        })
        if( exists ) continue

        someArr.push([ numbers[i], numbers[j] ])
        count++
      } 
    }
  }
  return count;
}