LeetCode 625: Minimum Factorization Solution

Master LeetCode problem 625 (Minimum Factorization), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

625. Minimum Factorization

Medium

Math Greedy

Problem Explanation

Explanation:

This problem asks us to find the smallest number that can be obtained by multiplying some given numbers. To solve this problem, we can use a backtracking approach to find all possible factorizations of the given number and then choose the minimum one. We start by finding the smallest prime factor of the given number and then recursively find the next smallest prime factor until the number becomes 1.

Here are the detailed steps for the algorithm:

Start with the smallest prime factor, which is 2.
Try to divide the number by 2 until it is no longer divisible by 2.
If the number is divisible by 2, continue the division process with the next smallest prime factor.
If the number is not divisible by 2, move on to the next smallest prime factor.
Repeat steps 2-4 until the number becomes 1.
If we find a factorization that gives a smaller number than the current minimum, update the minimum value.

Time complexity: The time complexity of this algorithm is O(log n), where n is the given number. This is because the number of divisions needed to factorize the number is logarithmic in the value of n.

Space complexity: The space complexity of this algorithm is O(log n) due to the recursive calls and the stack space used.

Solution Code

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

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