【算法】Vampire Numbers

Vampire Numbers

recursion permutation brute force

A Vampire Number is a positive integer greater than 99, that rearranged in all of its possible digits permutations, with every permutation being split into two parts, is equal to the product of at least one of its permutations.

  • If the number has an even quantity of digits, left and right parts will have the same length in every permutation;
  • If the number has an odd quantity of digits and at least three digits, the left and right parts will present different lengths for every possible permutation, alternating between them in the range +1 and -1.

Given a positive integer n, implement a function that returns the type of n as a string:

  • 'Normal Number' if n is lower than 100 or if no permutations return a product of their parts equal to n.
  • 'Pseudovampire' if n it is a Vampire with an odd quantity of digits.
  • 'True Vampire' if n it is a Vampire with an even quantity of digits.
Examples
javascript 复制代码
isVampire(1260) // "True Vampire"
// Has an even number of digits and is greater than 99)
// Permutations:
// 12 * 60 = 720
// 16 * 20 = 320
// 10 * 26 = 260
// 21 * 60 = 1260

isVampire(126) // "Pseudovampire"
// Has an odd number of digits and is greater than 99
// Permutations:
// 12 * 6 = 72
// 1 * 26 = 26
// 21 * 6 = 126

isVampire(67) // "Normal Number"
// Is lower than 100
// Permutations:
// 6 * 7 = 7 * 6 = 42
Notes
  • Trivially, a number from 1 to 99 is a Normal Number by the definitions: a single-digit number can't be split into two parts, and the product of the permutated two digits of a number will always be lower than the number itself.
Solutions
javascript 复制代码
// solution1: recursion + permutation
const isVampire = (n) => {
    if(n < 100){
        return 'Normal Number'
    }
    let t = n,d=[];
    while(t){
        d.unshift(t%10)
        t = Math.floor(t/10)
    }
    let o = d.length&0b1;
    let res = compute(d,0,n,o,d.length)
    return res?(!o?'True Vampire':'Pseudovampire'):'Normal Number';
}
const compute = (d,k,n,o,l)=>{
    if(k==l){
        k = l>>>1
        let pair = toIntPair(d,k);
        let res = pair[0]*pair[1] === n;
        if(o && !res){
            pair = toIntPair(d,k+1);
            res = pair[0]*pair[1] === n;
        }
        return res;
    }
    for(let i=k;i<l;i++){
        [d[i],d[k]] = [d[k],d[i]]
        let res = compute(d,k+1,n,o,l)
        if(res){
            return true;
        }
        [d[k],d[i]] = [d[i],d[k]]
    }
    return false;
};
const toIntPair=(d,k)=>{
    let res = [0,0];
    for(let i=0;i<k;i++){
        res[0] = res[0]*10 + d[i]
    }
    for(let i=k;i<d.length;i++){
        res[1] = res[1]*10 + d[i]
    }
    return res
}
javascript 复制代码
// solution2: brute force
const isVampire = (n) => {
    if(n < 100){
        return 'Normal Number'
    }
    let t = n,c=0;
    while(t){
        c++;
        t = Math.floor(t/10);
    }
    let o = c&0b1;
    c >>>= 1
    let max = 1 ;
    while(c-->0){
        max *= 10
    }
    let d=Array(10).fill(0);
    for(let i = Math.floor(max/10); i < max;i++){
            let j = Math.floor(n/i);
            let m = i*j;
            let [ii,jj,mm] = [i,j,m];
            while(mm){
                d[mm%10]++;
                mm = Math.floor(mm/10);
            }
            while(ii){
                d[ii%10]--;
                ii = Math.floor(ii/10);
            }
            while(jj){
                d[jj%10]--;
                jj = Math.floor(jj/10);
            }
            let is = true;
            for(let i=0;i<10;i++){
                if(d[i]){
                    is = false;
                    d[i] = 0;
                }
            }
            if(is && m === n){
                return !o?'True Vampire':'Pseudovampire';
            }
    }
    return 'Normal Number';
}
TestCases
javascript 复制代码
let Test = (function(){
    return {
        assertEquals:function(actual,expected){
            if(actual !== expected){
                let errorMsg = `actual is ${actual},${expected} is expected`;
                throw new Error(errorMsg);
            }
        }
    }
})();

Test.assertEquals(isVampire(1260), "True Vampire", "Example #1")
Test.assertEquals(isVampire(126), "Pseudovampire", "Example #2")
Test.assertEquals(isVampire(67), "Normal Number", "Example #3")
Test.assertEquals(isVampire(1), "Normal Number")
Test.assertEquals(isVampire(645), "Normal Number")
Test.assertEquals(isVampire(688), "Pseudovampire")
Test.assertEquals(isVampire(1345), "Normal Number")
Test.assertEquals(isVampire(1395), "True Vampire")
Test.assertEquals(isVampire(12964), "Pseudovampire")
Test.assertEquals(isVampire(98765), "Normal Number")
Test.assertEquals(isVampire(124421), "Normal Number")
Test.assertEquals(isVampire(125460), "True Vampire")
相关推荐
xnglan9 分钟前
蓝桥杯手算题和杂题简易做法
数据结构·数据库·c++·python·算法·职场和发展·蓝桥杯
凛冬君主11 分钟前
插入排序专栏
java·算法·排序算法
stbomei36 分钟前
C 语言判断一个数是否是素数的三种方法文章提纲
c语言·开发语言·算法
小牛壮士38 分钟前
Tokenizer(切词器)的不同实现算法
开发语言·算法·c#
汤永红44 分钟前
week1-[顺序结构]大海
c++·算法·信睡奥赛
博哥爱吃肉1 小时前
第2篇_Go语言基础语法_变量常量与数据类型
开发语言·算法·golang
Xの哲學4 小时前
TCP 连接管理:深入分析四次握手与三次挥手
网络·网络协议·算法
华清远见成都中心7 小时前
基于深度学习的异常检测算法在时间序列数据中的应用
人工智能·深度学习·算法
এ᭄画画的北北8 小时前
力扣-347.前K个高频元素
算法·leetcode
月殇_木言11 小时前
算法基础 第3章 数据结构
数据结构·算法