题解 | #删除字符串中出现次数最少的字符#

const rl = require("readline").createInterface({ input: process.stdin });
var iter = rl[Symbol.asyncIterator]();
const readline = async () => (await iter.next()).value;

void async function () {
    // Write your code here
    while(line = await readline()){
      let str = line
      let obj = {} // 收集每个字符的出现次数
      for (let i = 0; i < str.length; i++) {
        if (obj[str[i]]) {
            obj[str[i]]++
        } else {
            obj[str[i]] = 1
        }
      }
      let arr = []
      for (let k in obj) {
        arr.push([k, obj[k]])
      }
      arr.sort((a, b) =>{
        return a[1] - b[1]
      })
      
      let index = 0
      let minStr = [arr[index][0]]
      while(arr[index++][1] === arr[index][1]) {
        minStr.push(arr[index][0])
      }
      str = str.split('').filter(v => {
        return !minStr.includes(v)
      })
      console.log(str.join(''))
    }
}()