找出数组对象中重复次数最多和最少的元素

以下仅适用于数据很小的情况，数据量大的情况下可能内存不够，计算时间也会很长

 // min || max 元素长度可能都大于一
  var arr = ['a', 'a', 'a', 'b', 'b', 'a', 'b', 'a', 'c', 'a', 'b', 'a', 'b', 'd']
  var obj ={}
  for (var i =0; i < arr.length; i++) {
    obj[arr[i]] = arr[i]
  }
  Object.keys(obj).forEach(item => obj[item]= arr.filter(e => e === obj[item]).length)

  var nums = Object.keys(obj).map(e => obj[e])
  var max = Math.max.call(null, nums.reduce((prev, next) => prev > next ? prev : next))
  var min = Math.min.call(null, nums.reduce((prev, next) => prev < next ? prev : next))
  var maxItem = Object.keys(obj).filter(e => obj[e] === max)
  var minItem = Object.keys(obj).filter(e => obj[e] === min)
  console.log(minItem)
  console.log(maxItem)
// console.log(obj)
// console.log(max)
// console.log(min)

var arr = ['a', 'a', 'a', 'a', 'b', 'b', 'b', 'c', 'c', 'c', 'd', 'e', 'd', 'f']
    var obj = {}
    for(var i = 0; i < arr.length; i++){
      if(obj.hasOwnProperty(arr[i]))
        obj[arr[i]] = ++obj[arr[i]]
      else
        obj[arr[i]] = 1
    }
    var max = Math.max.call(null, Object.values(obj).reduce((prev, next) => prev > next ? prev : next))
    var min = Math.min.call(null, Object.values(obj).reduce((prev, next) => prev < next ? prev : next))
    var maxItem = Object.keys(obj).filter(item => obj[item] === max)
    var minItem = Object.keys(obj).filter(item => obj[item] === min)
    console.log(obj)
    console.log(max, min)
    console.log(maxItem, minItem)