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Q: 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.
A: BFS (Breadth-First Search)124Please respect copyright.ＰＥＮＡＮＡ3gaLW0c1u7
// better than use DFS as it just need to find out the shortest path.
class Solution {124Please respect copyright.ＰＥＮＡＮＡVn8YZab9A6
public int minMutation(String start, String end, String[] bank) {124Please respect copyright.ＰＥＮＡＮＡRmcK2df4Cj
// Initialize a queue queue and a set seen. The queue will be used for BFS and the set will be used to prevent visiting a node more than once. Initially, the queue and set should hold start.124Please respect copyright.ＰＥＮＡＮＡRoHoElz6CB
Queue<String> queue = new LinkedList<>();124Please respect copyright.ＰＥＮＡＮＡSdEhcRxEtI
Set<String> seen = new HashSet<>();124Please respect copyright.ＰＥＮＡＮＡkmZvtZTgi4
queue.add(start);124Please respect copyright.ＰＥＮＡＮＡMH75nto1hw
seen.add(start);124Please respect copyright.ＰＥＮＡＮＡURmc6pODar
124Please respect copyright.ＰＥＮＡＮＡrfUY0zkqcU
int steps = 0;124Please respect copyright.ＰＥＮＡＮＡiZvsrDpxtF
124Please respect copyright.ＰＥＮＡＮＡEGn97vRQJy
while (!queue.isEmpty()) {124Please respect copyright.ＰＥＮＡＮＡK3kQaMhQ4m
int nodesInQueue = queue.size();124Please respect copyright.ＰＥＮＡＮＡZVnxfNjQT6
for (int j = 0; j < nodesInQueue; j++) {124Please respect copyright.ＰＥＮＡＮＡDIva9nzDv9
String node = queue.remove();124Please respect copyright.ＰＥＮＡＮＡ8PztaIoQBo
// Perform a BFS. At each node, if node == end, return the number of steps so far. Otherwise, iterate over all the neighbors. For each neighbor, if neighbor is not in seen and neighbor is in bank, add it to queue and seen.
if (node.equals(end)) {124Please respect copyright.ＰＥＮＡＮＡU5HXdXI55z
return steps;124Please respect copyright.ＰＥＮＡＮＡcWFEJ53L3M
}
for (char c: new char[] {'A', 'C', 'G', 'T'}) {124Please respect copyright.ＰＥＮＡＮＡDJ4YMPTUs5
for (int i = 0; i < node.length(); i++) {124Please respect copyright.ＰＥＮＡＮＡ65noKPxuUC
String neighbor = node.substring(0, i) + c + node.substring(i + 1);124Please respect copyright.ＰＥＮＡＮＡy6aObbZbPS
if (!seen.contains(neighbor) && Arrays.asList(bank).contains(neighbor)) {124Please respect copyright.ＰＥＮＡＮＡbozfa01pYx
queue.add(neighbor);124Please respect copyright.ＰＥＮＡＮＡX1S990pSC3
seen.add(neighbor);124Please respect copyright.ＰＥＮＡＮＡiPAjiaODbb
}124Please respect copyright.ＰＥＮＡＮＡPA5GEUtrXC
}124Please respect copyright.ＰＥＮＡＮＡ26cwtaQOV3
}124Please respect copyright.ＰＥＮＡＮＡzmlNRX7wHg
}124Please respect copyright.ＰＥＮＡＮＡLaau16x8l0
124Please respect copyright.ＰＥＮＡＮＡCO8PMtV3p3
steps++;124Please respect copyright.ＰＥＮＡＮＡJwgyiw41fL
}124Please respect copyright.ＰＥＮＡＮＡAJvpPiNmoe
// If we finish the BFS and did not find end, return -1.124Please respect copyright.ＰＥＮＡＮＡ6gVKRy143U
return -1;124Please respect copyright.ＰＥＮＡＮＡ0O54ddtipH
}124Please respect copyright.ＰＥＮＡＮＡe7LLF5pTXJ
}
