letcood brush questions - edit once

There are three editing operations for strings: insert an English character, delete an English character, or replace an English character. Given two strings, write a function to determine whether they require only one (or zero) edits.

Example 1:
Input:
first = "pales"
second = "ple"
Output: True

Example 2:
Input:
first = "pales"
second = "pal"
Output: False

The diagram is as follows:

var oneEditAway = function(first, second) {
    // 检查长度
    if(Math.abs(first.length - second.length) > 1) return false
    // 零编辑
    if (first === second) return true
    // 
    if (first.length > second.length) {
        [second, first] = [first, second]
    }
    for (let i = 0, len = first.length; i < len; i++) {
        if (first[i] !== second[i]) {
            return first.slice(i + 1) === second.slice(i + 1) // 字符串长度相等
                || first.slice(i) === second.slice(i + 1) // 字符串不等
        }
    }
    return true
};
// 解法2 ，这个比较蛮力，逻辑没有合并起来
var oneEditAway = function(first, second) {
   if(first.length == second.length) {// 替换
       return oneEditReplace(first,second)
   }else if(first.length+1 == second.length) { // 删除
       return oneEditInsert(first,second)
   }else if(first.length-1 == second.length) { // 插入
       return oneEditInsert(second,first)
   }
   return false
};
var oneEditReplace = function(first,second) {
   var findDif = false 
   for(var i=0;i<first.length;i++) {
       if(first.charAt(i)!=second.charAt(i)) {
          if(findDif) return false
          findDif = true
       } 
   }
   return true
};
var oneEditInsert = function(first,second) {
  var index1 = 0
  var index2 = 0
  while(index2 < second.length && index1 < first.length) {
      if(first.charAt(index1) != second.charAt(index2)) {
          if(index1 != index2) {
              return false
          }
      } else {
          index1++
      }
      index2++
  }
  return true
}

// 执行结果：通过显示详情
// 执行用时：72 ms, 在所有 TypeScript 提交中击败了78.95%的用户
// 内存消耗：43.8 MB, 在所有 TypeScript 提交中击败了76.32%的用户
// 通过测试用例：1146 / 1146

function oneEditAway(first: string, second: string): boolean {
    if(first.length > second.length + 1 || first.length < second.length - 1) return false
    if(first.length === second.length) {
        if(first === second) return true
        let cut = 0
        for(let i = 0; i < first.length; i++){
            if(first[i] !== second[i]) cut++
            if(cut > 1) return false
        }
        return true
    }else{
        const flag = first.length > second.length
        let long = flag ? first : second,
            short = flag ? second : first,
            temp = ''
        for(let i = 0; i < long.length; i++){
            if(long[i] !== short[i]){
                if(temp + long[i] + short.substring(i) === long || temp + long.substring(i + 1) === short) return true
                else return false
            }
            temp += long[i]
        }
    }
};