Similar String Groups

// Accepted solution for LeetCode #839: Similar String Groups
class UnionFind {
    private final int[] p;
    private final int[] size;

    public UnionFind(int n) {
        p = new int[n];
        size = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
            size[i] = 1;
        }
    }

    public int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }

    public boolean union(int a, int b) {
        int pa = find(a), pb = find(b);
        if (pa == pb) {
            return false;
        }
        if (size[pa] > size[pb]) {
            p[pb] = pa;
            size[pa] += size[pb];
        } else {
            p[pa] = pb;
            size[pb] += size[pa];
        }
        return true;
    }
}

class Solution {
    public int numSimilarGroups(String[] strs) {
        int n = strs.length, m = strs[0].length();
        UnionFind uf = new UnionFind(n);
        int cnt = n;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                int diff = 0;
                for (int k = 0; k < m; ++k) {
                    if (strs[i].charAt(k) != strs[j].charAt(k)) {
                        ++diff;
                    }
                }
                if (diff <= 2 && uf.union(i, j)) {
                    --cnt;
                }
            }
        }
        return cnt;
    }
}

// Accepted solution for LeetCode #839: Similar String Groups
type unionFind struct {
	p, size []int
}

func newUnionFind(n int) *unionFind {
	p := make([]int, n)
	size := make([]int, n)
	for i := range p {
		p[i] = i
		size[i] = 1
	}
	return &unionFind{p, size}
}

func (uf *unionFind) find(x int) int {
	if uf.p[x] != x {
		uf.p[x] = uf.find(uf.p[x])
	}
	return uf.p[x]
}

func (uf *unionFind) union(a, b int) bool {
	pa, pb := uf.find(a), uf.find(b)
	if pa == pb {
		return false
	}
	if uf.size[pa] > uf.size[pb] {
		uf.p[pb] = pa
		uf.size[pa] += uf.size[pb]
	} else {
		uf.p[pa] = pb
		uf.size[pb] += uf.size[pa]
	}
	return true
}

func numSimilarGroups(strs []string) int {
	n := len(strs)
	uf := newUnionFind(n)
	for i, s := range strs {
		for j, t := range strs[:i] {
			diff := 0
			for k := range s {
				if s[k] != t[k] {
					diff++
				}
			}
			if diff <= 2 && uf.union(i, j) {
				n--
			}
		}
	}
	return n
}

# Accepted solution for LeetCode #839: Similar String Groups
class UnionFind:
    def __init__(self, n):
        self.p = list(range(n))
        self.size = [1] * n

    def find(self, x):
        if self.p[x] != x:
            self.p[x] = self.find(self.p[x])
        return self.p[x]

    def union(self, a, b):
        pa, pb = self.find(a), self.find(b)
        if pa == pb:
            return False
        if self.size[pa] > self.size[pb]:
            self.p[pb] = pa
            self.size[pa] += self.size[pb]
        else:
            self.p[pa] = pb
            self.size[pb] += self.size[pa]
        return True


class Solution:
    def numSimilarGroups(self, strs: List[str]) -> int:
        n, m = len(strs), len(strs[0])
        uf = UnionFind(n)
        for i, s in enumerate(strs):
            for j, t in enumerate(strs[:i]):
                if sum(s[k] != t[k] for k in range(m)) <= 2 and uf.union(i, j):
                    n -= 1
        return n

// Accepted solution for LeetCode #839: Similar String Groups
/**
 * [0839] Similar String Groups
 *
 * Two strings X and Y are similar if we can swap two letters (in different positions) of X, so that it equals Y. Also two strings X and Y are similar if they are equal.
 * For example, "tars" and "rats" are similar (swapping at positions 0 and 2), and "rats" and "arts" are similar, but "star" is not similar to "tars", "rats", or "arts".
 * Together, these form two connected groups by similarity: {"tars", "rats", "arts"} and {"star"}.  Notice that "tars" and "arts" are in the same group even though they are not similar.  Formally, each group is such that a word is in the group if and only if it is similar to at least one other word in the group.
 * We are given a list strs of strings where every string in strs is an anagram of every other string in strs. How many groups are there?
 *  
 * Example 1:
 *
 * Input: strs = ["tars","rats","arts","star"]
 * Output: 2
 *
 * Example 2:
 *
 * Input: strs = ["omv","ovm"]
 * Output: 1
 *
 *  
 * Constraints:
 *
 * 	1 <= strs.length <= 300
 * 	1 <= strs[i].length <= 300
 * 	strs[i] consists of lowercase letters only.
 * 	All words in strs have the same length and are anagrams of each other.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/similar-string-groups/
// discuss: https://leetcode.com/problems/similar-string-groups/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

struct UnionFind {
    count: usize,
    parent: std::cell::RefCell<Vec<usize>>,
    size: Vec<usize>,
}

impl UnionFind {
    pub fn new(size: usize) -> Self {
        Self {
            count: size,
            parent: std::cell::RefCell::new((0..size).collect()),
            size: vec![1; size],
        }
    }
    pub fn count(&self) -> usize {
        self.count
    }
    pub fn find(&self, p: usize) -> usize {
        let mut root = p;
        while root != self.parent.borrow()[root] {
            root = self.parent.borrow()[root];
        }
        let mut p = p;
        while p != root {
            let next = self.parent.borrow()[p];
            self.parent.borrow_mut()[p] = root;
            p = next;
        }
        root
    }

    pub fn is_connected(&self, p: usize, q: usize) -> bool {
        self.find(p) == self.find(q)
    }
    pub fn connect(&mut self, p: usize, q: usize) {
        let (p_root, q_root) = (self.find(p), self.find(q));
        if p_root == q_root {
            return;
        }
        if self.size[p_root] < self.size[q_root] {
            self.parent.borrow_mut()[p_root] = q_root;
            self.size[q_root] += self.size[p_root];
        } else {
            self.parent.borrow_mut()[q_root] = p_root;
            self.size[p_root] += self.size[q_root];
        }
        self.count -= 1;
    }
}

impl Solution {
    pub fn num_similar_groups(strs: Vec<String>) -> i32 {
        fn is_similar(s1: &str, s2: &str) -> bool {
            s1.chars().zip(s2.chars()).filter(|(a, b)| !a.eq(b)).count() <= 2
        }
        let mut uf = UnionFind::new(strs.len());

        for i in 0..strs.len() {
            let s1 = strs.get(i).unwrap();
            for j in i + 1..strs.len() {
                if uf.is_connected(i, j) {
                    continue;
                }

                let s2 = strs.get(j).unwrap();
                if is_similar(s1, s2) {
                    uf.connect(i, j);
                }
            }
        }

        uf.count() as i32
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_0839_example_1() {
        let strs = vec_string!["tars", "rats", "arts", "star"];
        let result = 2;

        assert_eq!(Solution::num_similar_groups(strs), result);
    }

    #[test]
    fn test_0839_example_2() {
        let strs = vec_string!["omv", "ovm"];
        let result = 1;

        assert_eq!(Solution::num_similar_groups(strs), result);
    }
}

// Accepted solution for LeetCode #839: Similar String Groups
class UnionFind {
    private p: number[];
    private size: number[];

    constructor(n: number) {
        this.p = Array.from({ length: n }, (_, i) => i);
        this.size = Array(n).fill(1);
    }

    union(a: number, b: number): boolean {
        const pa = this.find(a);
        const pb = this.find(b);
        if (pa === pb) {
            return false;
        }
        if (this.size[pa] > this.size[pb]) {
            this.p[pb] = pa;
            this.size[pa] += this.size[pb];
        } else {
            this.p[pa] = pb;
            this.size[pb] += this.size[pa];
        }
        return true;
    }

    find(x: number): number {
        if (this.p[x] !== x) {
            this.p[x] = this.find(this.p[x]);
        }
        return this.p[x];
    }
}

function numSimilarGroups(strs: string[]): number {
    const n = strs.length;
    const m = strs[0].length;
    const uf = new UnionFind(n);
    let cnt = n;
    for (let i = 0; i < n; ++i) {
        for (let j = 0; j < i; ++j) {
            let diff = 0;
            for (let k = 0; k < m; ++k) {
                if (strs[i][k] !== strs[j][k]) {
                    diff++;
                }
            }
            if (diff <= 2 && uf.union(i, j)) {
                cnt--;
            }
        }
    }
    return cnt;
}