public void UnionTest2()
{
var dsu = new DSU <int>();
dsu.MakeSet(1);
dsu.MakeSet(2);
dsu.MakeSet(3);
dsu.MakeSet(4);
dsu.MakeSet(5);
Assert.AreEqual(4, dsu.Find(4));
dsu.Unite(1, 4, isRandom: false);
dsu.Unite(3, 5, isRandom: false);
Assert.AreEqual(1, dsu.Find(4));
Assert.AreEqual(1, dsu.Find(1));
Assert.AreEqual(2, dsu.Find(2));
Assert.AreEqual(3, dsu.Find(5));
Assert.AreEqual(3, dsu.Find(3));
dsu.Unite(5, 2, isRandom: false);
Assert.AreEqual(3, dsu.Find(5));
Assert.AreEqual(3, dsu.Find(3));
Assert.AreEqual(3, dsu.Find(2));
}
public void UnionTest()
{
var dsu = new DSU <int>();
dsu.MakeSet(1);
dsu.MakeSet(2);
dsu.Unite(1, 2, isRandom: false);
Assert.That(dsu.Find(1), Is.EqualTo(dsu.Find(2)));
}
/// <summary>
/// Nov. 19, 2020
/// Union find algorithm
/// study code
/// https://leetcode.com/problems/sentence-similarity-ii/solution/
/// Time complexity:
/// Time Complexity: O(N log P + P), where N is the maximum length of words1 and words2,
/// and P is the length of pairs. If we used union-by-rank, this complexity improves to
/// O(N * a(P) + P), close to O(N + P), where α is the Inverse-Ackermann function.
/// </summary>
/// <param name="words1"></param>
/// <param name="words2"></param>
/// <param name="pairs"></param>
/// <returns></returns>
public bool AreSentencesSimilarTwo(string[] words1, string[] words2, IList <IList <string> > pairs)
{
var length1 = words1.Length;
var length2 = words2.Length;
var pLength = pairs.Count;
if (length1 != length2)
{
return(false);
}
// index -> indexMap better name?
var indexMap = new Dictionary <string, int>();
int count = 0;
var dsu = new DSU(2 * pLength);
foreach (var pair in pairs)
{
// go through two words in one pair
foreach (var p in pair)
{
if (!indexMap.ContainsKey(p))
{
indexMap.Add(p, count++);
}
}
dsu.Union(indexMap[pair[0]], indexMap[pair[1]]);
}
for (int i = 0; i < length1; ++i)
{
var w1 = words1[i];
var w2 = words2[i];
if (w1.CompareTo(w2) == 0)
{
continue;
}
if (!indexMap.ContainsKey(w1) ||
!indexMap.ContainsKey(w2) ||
dsu.Find(indexMap[w1]) != dsu.Find(indexMap[w2]))
{
return(false);
}
}
return(true);
}
public int RemoveStones(int[][] stones)
{
var dsu = new DSU(20000);
foreach (var stone in stones)
{
dsu.Union(stone[0], stone[1] + 10000);
}
var seen = new HashSet <int>();
foreach (var stone in stones)
{
seen.Add(dsu.Find(stone[0]));
}
return(stones.Length - seen.Count);
}
public IList <IList <string> > AccountsMerge(IList <IList <string> > accounts)
{
EmailToId = new Dictionary <string, int>();
EmailToName = new Dictionary <string, string>();
foreach (var account in accounts)
{
if (account.Count < 2)
{
continue;
}
var name = account[0];
for (var i = 1; i < account.Count; i++)
{
if (!EmailToId.ContainsKey(account[i]))
{
var id = EmailToId.Count;
EmailToId[account[i]] = id;
EmailToName[account[i]] = name;
}
}
}
var totalAccounts = EmailToId.Count;
DSU = new DSU(totalAccounts);
foreach (var account in accounts)
{
// var name = account[0];
for (var i = 1; i < account.Count - 1; i++)
{
var id1 = EmailToId[account[i]];
var id2 = EmailToId[account[i + 1]];
if (id1 != id2)
{
DSU.Union(id1, id2);
}
}
}
var result = new Dictionary <int, List <string> >();
foreach (var email in EmailToName.Keys)
{
int index = DSU.Find(EmailToId[email]);
if (result.TryGetValue(index, out var list))
{
list.Add(email);
}
else
{
result[index] = new List <string> {
email
};
}
}
foreach (var emails in result.Values)
{
emails.Sort((x, y) =>
{
var i = 0;
while (i < x.Length && i < y.Length)
{
if (x[i] != y[i])
{
return(x[i].CompareTo(y[i]));
}
i++;
}
return(x.Length.CompareTo(y.Length));
});
emails.Insert(0, EmailToName[emails[0]]);
}
return(result.Values.ToList <IList <string> >());
}