625. Minimum Factorization

最新推荐文章于 2021-03-02 00:10:41 发布

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原文链接：http://www.cnblogs.com/wdw828/p/7058217.html

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#java #开发工具

本文介绍了一个LeetCode题目：给定一个正整数a，找出最小的正整数b，使得b的每一位数字相乘等于a。若不存在这样的b或者b不在32位有符号整数范围内，则返回0。文章提供了一种高效解决方案，并分析了其时间复杂度。

Problem statement

Given a positive integer a, find the smallest positive integer b whose multiplication of each digit equals to a.

If there is no answer or the answer is not fit in 32-bit signed integer, then return 0.

Example 1
Input:
48 
Output:
68
Example 2
Input:
15
Output:
35

Solution

This is the third problem for weekly contest 37. Initially, I thought it is a DFS problem. Actually, it is much easier. Although it is tagged with recursion in leetcode, I prefer it is a pure math.

Since the input number should only contains the factors which are less than 10. We loop from 9 to 2, to divide the input number, until it is not divisible by any number in [2, 9].

Suppose the input number is a, general idea is as following:

Loop i from 9 to 2.
In each loop, if a is divisible by i, update the min factorization value and a. Goes to next loop until a is not divisible by i
After exiting the loop,
1. if a < 2, means all it`s factors are single digit number, return the value(if it is in the range),
2. Otherwise, it means there is at least one factor, that is greater than 9, return 0.

Time complexity is O(8loga), space complexity is O(1).

class Solution {
public:
    int smallestFactorization(int a) {
        if(a < 2){
            return a;
        }
        long min_factorization = 0;
        long mul = 1;
        for(int i = 9; i >= 2; i--){
            while(a % i == 0){
                min_factorization += mul * i;
                mul *= 10;
                a /= i;
            }
        }
        return (a < 2 && min_factorization < INT_MAX) ? min_factorization : 0;
    }
};