g0401_0500.s0433_minimum_genetic_mutation.Solution Maven / Gradle / Ivy

Go to download
Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-java21 Show documentation
Java-based LeetCode algorithm problem solutions, regularly updated
There is a newer version: 1.38
package g0401_0500.s0433_minimum_genetic_mutation;

// #Medium #String #Hash_Table #Breadth_First_Search #Graph_Theory_I_Day_12_Breadth_First_Search
// #2022_07_16_Time_1_ms_(90.95%)_Space_41.9_MB_(56.72%)

import java.util.ArrayList;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
import java.util.Set;

/**
 * 433 - Minimum Genetic Mutation\.
 *
 * Medium
 *
 * A gene string can be represented by an 8-character long string, with choices from `'A'`, `'C'`, `'G'`, and `'T'`.
 *
 * Suppose we need to investigate a mutation from a gene string `start` to a gene string `end` where one mutation is defined as one single character changed in the gene string.
 *
 * *   For example, `"AACCGGTT" --> "AACCGGTA"` is one mutation.
 *
 * There is also a gene bank `bank` that records all the valid gene mutations. A gene must be in `bank` to make it a valid gene string.
 *
 * Given the two gene strings `start` and `end` and the gene bank `bank`, return _the minimum number of mutations needed to mutate from_ `start` _to_ `end`. If there is no such a mutation, return `-1`.
 *
 * Note that the starting point is assumed to be valid, so it might not be included in the bank.
 *
 * **Example 1:**
 *
 * **Input:** start = "AACCGGTT", end = "AACCGGTA", bank = ["AACCGGTA"]
 *
 * **Output:** 1 
 *
 * **Example 2:**
 *
 * **Input:** start = "AACCGGTT", end = "AAACGGTA", bank = ["AACCGGTA","AACCGCTA","AAACGGTA"]
 *
 * **Output:** 2 
 *
 * **Example 3:**
 *
 * **Input:** start = "AAAAACCC", end = "AACCCCCC", bank = ["AAAACCCC","AAACCCCC","AACCCCCC"]
 *
 * **Output:** 3 
 *
 * **Constraints:**
 *
 * *   `start.length == 8`
 * *   `end.length == 8`
 * *   `0 <= bank.length <= 10`
 * *   `bank[i].length == 8`
 * *   `start`, `end`, and `bank[i]` consist of only the characters `['A', 'C', 'G', 'T']`.
**/
@SuppressWarnings("java:S3012")
public class Solution {
    private List isInBank(Set set, String cur) {
        List res = new ArrayList<>();
        for (String each : set) {
            int diff = 0;
            for (int i = 0; i < each.length(); i++) {
                if (each.charAt(i) != cur.charAt(i)) {
                    diff++;
                    if (diff > 1) {
                        break;
                    }
                }
            }
            if (diff == 1) {
                res.add(each);
            }
        }
        return res;
    }

    public int minMutation(String start, String end, String[] bank) {
        Set set = new HashSet<>();
        for (String s : bank) {
            set.add(s);
        }
        Queue queue = new LinkedList<>();
        queue.offer(start);
        int step = 0;
        while (!queue.isEmpty()) {
            int curSize = queue.size();
            while (curSize-- > 0) {
                String cur = queue.poll();
                if (cur.equals(end)) {
                    return step;
                }
                for (String next : isInBank(set, cur)) {
                    queue.offer(next);
                    set.remove(next);
                }
            }
            step++;
        }
        return -1;
    }
}