433. Minimum Genetic Mutation

Minimum Genetic Mutation

Hello, devs 👋! Today we will discuss a medium problem which is a very popular problem in coding interviews.

Problem Description

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'].

Approach) Brute Force

var minMutation = function(start, end, bank) {
    let banks = new Set(bank);
    if( !banks.has(end) ) return -1;
    let chars = ['A', 'T', 'C', 'G'];        
    var Q = [[start,0]];
    var item, dist, i, j;
    
    
    while( Q.length > 0 ) {
        [item,dist] = Q.shift();
        if( item == end ) return dist;
        
        for( i=0; i<8; i++ ) {
            for( j=0; j<4; j++ ) {
                if( item[i] == chars[j] ) continue;
                var node = item.slice(0,i) + chars[j] + item.slice(i+1);
                if( banks.has(node) ) {
                    Q.push([node,dist+1]);
                    banks.delete(node);
                }
            }
        }
    }
    
    return -1;
};

Approach) Breadth-First Search (BFS)

/**
 * @param {string} start
 * @param {string} end
 * @param {string[]} bank
 * @return {number}
 */
var minMutation = function(start, end, bank) {
    var set = new Set(bank);
    if(!set.has(end))   return -1;
    var visited = new Set();
    visited.add(start);
    
    var q = [];
    q.push(start);
    var count = 1;
    while(q.length!==0){
        var size = q.length;
        
        for(var i = 0;i<size;i++){
            var a = q.shift();
            if(oneM(a,end)) return count;
            set.forEach((b)=>{
                if(!visited.has(b) && oneM(a,b)){
                    q.push(b);
                    visited.add(b);
                }
            });
        }
        
        count++;
    }
    return -1;
    
    
};


var oneM = function(a,b){
    var count = 0;
    for(var i =0;i<a.length;i++){
        if(a[i]!==b[i]){
            count++;
        }
    }
    return count===1;
};

Approach) Depth-First Search (DFS)

/**
 * @param {string} start
 * @param {string} end
 * @param {string[]} bank
 * @return {number}
 */
var minMutation = function(start, end, bank) {
    var set = new Set(bank);
    var visited = new Set();
    visited.add(start);
    if(!set.has(end))   return -1;
    var count = dfs(start);
    
    if(count === Number.MAX_VALUE)  return -1;
    return count;
    
    //@ return min_path
    
    function dfs(a){
        if(oneM(a,end)) return 1;
        
        var min = Number.MAX_VALUE;
        
        set.forEach((b)=>{
            if(oneM(a,b) && !visited.has(b)){
                visited.add(b);
                var c = dfs(b);
                if(c!==Number.MAX_VALUE){
                    min = Math.min(min,c+1);
                }
                visited.delete(b);
            }
        });
        
        return min;
    }
};


var oneM = function(a,b){
    var count = 0;
    for(var i =0;i<a.length;i++){
        if(a[i]!==b[i]){
            count++;
        }
    }
    return count===1;
};

Conclusion

That’s all folks! In this post, we solved LeetCode problem #433. Minimum Genetic Mutation

I hope you have enjoyed this post. Feel free to share your thoughts on this.

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Happy coding!