9ms C++ prime table + DP.


  • 0
    M

    This is a true dp solution with prime table which beats most raw dp solution. The code is a little bit long though. The idea is to shortcut any prime number, and optimize the non-prime number during dp:

    class Solution {
    public:
        int minSteps(int n) {
            unordered_set<int> primes = { 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997};
            if (primes.find(n) != primes.end())
            {
                return n;
            }
            
            vector<int> shortPrimes = {2,3,5,7,11,13,17,19,23,29,31};
            
            vector<int> dp(n + 1, 0);
            for (int i = 2; i <= n; ++i)
            {
                if (primes.find(i) != primes.end())
                {
                    dp[i] = i;
                    continue;
                }
                
                int d = 0;
                for (auto iter = shortPrimes.rbegin(); iter != shortPrimes.rend(); ++iter)
                {
                    if (*iter > i)
                    {
                        continue;
                    }
                    
                    if (i % *iter == 0)
                    {
                        d = *iter;
                        break;
                    }
                }
                
                dp[i] = dp[i / d] + d;
            }
    
            return dp[n];
        }
    };
    

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