LeetCode #433 — MEDIUM

Minimum Genetic Mutation

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

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 startGene to a gene string endGene 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 startGene and endGene and the gene bank bank, return the minimum number of mutations needed to mutate from startGene to endGene. 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: startGene = "AACCGGTT", endGene = "AACCGGTA", bank = ["AACCGGTA"]
Output: 1

Example 2:

Input: startGene = "AACCGGTT", endGene = "AAACGGTA", bank = ["AACCGGTA","AACCGCTA","AAACGGTA"]
Output: 2

Constraints:

0 <= bank.length <= 10
startGene.length == endGene.length == bank[i].length == 8
startGene, endGene, and bank[i] consist of only the characters ['A', 'C', 'G', 'T'].

Step 01

Brute Force Baseline

Problem summary: 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 startGene to a gene string endGene 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 startGene and endGene and the gene bank bank, return the minimum number of mutations needed to mutate from startGene to endGene. 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map

Example 1

"AACCGGTT"
"AACCGGTA"
["AACCGGTA"]

Example 2

"AACCGGTT"
"AAACGGTA"
["AACCGGTA","AACCGCTA","AAACGGTA"]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #433: Minimum Genetic Mutation
class Solution {
    public int minMutation(String startGene, String endGene, String[] bank) {
        Deque<String> q = new ArrayDeque<>();
        q.offer(startGene);
        Set<String> vis = new HashSet<>();
        vis.add(startGene);
        int depth = 0;
        while (!q.isEmpty()) {
            for (int m = q.size(); m > 0; --m) {
                String gene = q.poll();
                if (gene.equals(endGene)) {
                    return depth;
                }
                for (String next : bank) {
                    int c = 2;
                    for (int k = 0; k < 8 && c > 0; ++k) {
                        if (gene.charAt(k) != next.charAt(k)) {
                            --c;
                        }
                    }
                    if (c > 0 && !vis.contains(next)) {
                        q.offer(next);
                        vis.add(next);
                    }
                }
            }
            ++depth;
        }
        return -1;
    }
}

// Accepted solution for LeetCode #433: Minimum Genetic Mutation
func minMutation(startGene string, endGene string, bank []string) int {
	type pair struct {
		s     string
		depth int
	}
	q := []pair{pair{startGene, 0}}
	vis := map[string]bool{startGene: true}
	for len(q) > 0 {
		p := q[0]
		q = q[1:]
		if p.s == endGene {
			return p.depth
		}
		for _, next := range bank {
			diff := 0
			for i := 0; i < len(startGene); i++ {
				if p.s[i] != next[i] {
					diff++
				}
			}
			if diff == 1 && !vis[next] {
				vis[next] = true
				q = append(q, pair{next, p.depth + 1})
			}
		}
	}
	return -1
}

# Accepted solution for LeetCode #433: Minimum Genetic Mutation
class Solution:
    def minMutation(self, startGene: str, endGene: str, bank: List[str]) -> int:
        q = deque([(startGene, 0)])
        vis = {startGene}
        while q:
            gene, depth = q.popleft()
            if gene == endGene:
                return depth
            for nxt in bank:
                diff = sum(a != b for a, b in zip(gene, nxt))
                if diff == 1 and nxt not in vis:
                    q.append((nxt, depth + 1))
                    vis.add(nxt)
        return -1

// Accepted solution for LeetCode #433: Minimum Genetic Mutation
use std::collections::{HashSet, VecDeque};

impl Solution {
    pub fn min_mutation(start_gene: String, end_gene: String, bank: Vec<String>) -> i32 {
        let mut q = VecDeque::new();
        q.push_back((start_gene.clone(), 0));
        let mut vis = HashSet::new();
        vis.insert(start_gene);

        while let Some((gene, depth)) = q.pop_front() {
            if gene == end_gene {
                return depth;
            }
            for next in &bank {
                let mut c = 2;
                for k in 0..8 {
                    if gene.as_bytes()[k] != next.as_bytes()[k] {
                        c -= 1;
                    }
                    if c == 0 {
                        break;
                    }
                }
                if c > 0 && !vis.contains(next) {
                    vis.insert(next.clone());
                    q.push_back((next.clone(), depth + 1));
                }
            }
        }
        -1
    }
}

// Accepted solution for LeetCode #433: Minimum Genetic Mutation
function minMutation(startGene: string, endGene: string, bank: string[]): number {
    const q: [string, number][] = [[startGene, 0]];
    const vis = new Set<string>([startGene]);
    for (const [gene, depth] of q) {
        if (gene === endGene) {
            return depth;
        }
        for (const next of bank) {
            let c = 2;
            for (let k = 0; k < 8 && c > 0; ++k) {
                if (gene[k] !== next[k]) {
                    --c;
                }
            }
            if (c && !vis.has(next)) {
                q.push([next, depth + 1]);
                vis.add(next);
            }
        }
    }
    return -1;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(C × n × m)

Space

O(n × m)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.