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author | Jack Sangdahl <[email protected]> | 2024-08-28 19:03:02 -0600 |
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committer | Jack Sangdahl <[email protected]> | 2024-08-28 19:03:02 -0600 |
commit | 7dc61c9997c53be6c50726f2a618df397f9ac09f (patch) | |
tree | 2eb12e54b80ba3643977e9867b43ed4ab5798d2e | |
parent | 35132e9f44831f0cf18d329b15482744bd5548de (diff) | |
download | leetcode-7dc61c9997c53be6c50726f2a618df397f9ac09f.tar.gz leetcode-7dc61c9997c53be6c50726f2a618df397f9ac09f.zip |
947. most stones removed with same row or column
-rw-r--r-- | most_stones_remove.c | 73 | ||||
-rw-r--r-- | most_stones_remove.cpp | 18 | ||||
-rw-r--r-- | most_stones_remove.py | 34 | ||||
-rw-r--r-- | most_stones_remove.rs | 44 |
4 files changed, 154 insertions, 15 deletions
diff --git a/most_stones_remove.c b/most_stones_remove.c new file mode 100644 index 0000000..18f368c --- /dev/null +++ b/most_stones_remove.c @@ -0,0 +1,73 @@ +// 947. Most Stones Removed with Same Row or Column +#include "leetcode.h" + +/* + * on a 2d plane, we place 'n' stones at some integer coordinate points. each + * coordinate point may have at most one stone. a stone can be removed if it + * shares either the same row or the same column as another stone that has not + * been removed. given an array 'stones' of length 'n' where 'stones[i] = [x_i, + * y_i]' represents the location of the i'th stone, return the largest possible + * number of stones that can be removed + */ + +struct hash { + struct hash *parent; + UT_hash_handle set_hh; + UT_hash_handle by_row_hh; + UT_hash_handle by_col_hh; +}; + +struct hash *find_parent(struct hash *stone) { + while (stone->parent) + stone = stone->parent; + return stone; +} + +int removeStones(int **stones, int stonesSize, int *stonesColSize) { + struct hash *set = NULL, *by_row = NULL, *by_col = NULL; + struct hash *const entries = (struct hash *)malloc(sizeof(struct hash)); + for (int i = 0; i < stonesSize; i++) { + struct hash *entry = &entries[i], *row_entry, *col_entry; + HASH_FIND(by_row_hh, by_row, &stones[i][0], sizeof(stones[i][0]), + row_entry); + HASH_FIND(by_col_hh, by_col, &stones[i][1], sizeof(stones[i][1]), + col_entry); + if (row_entry && col_entry) { + struct hash *row_parent = find_parent(row_entry); + struct hash *col_parent = find_parent(col_entry); + if (row_parent != col_parent) { + col_parent->parent = row_parent; + HASH_DELETE(set_hh, set, col_parent); + } + entry->parent = row_parent; + } else if (row_entry) { + entry->parent = row_entry; + HASH_ADD_KEYPTR(by_col_hh, by_col, &stones[i][1], sizeof(stones[i][1]), + entry); + } else if (col_entry) { + entry->parent = col_entry; + HASH_ADD_KEYPTR(by_row_hh, by_row, &stones[i][0], sizeof(stones[i][0]), + entry); + } else { + entry->parent = NULL; + HASH_ADD_KEYPTR(set_hh, set, entry, sizeof(entry), entry); + HASH_ADD_KEYPTR(by_row_hh, by_row, &stones[i][0], sizeof(stones[i][0]), + entry); + HASH_ADD_KEYPTR(by_col_hh, by_col, &stones[i][1], sizeof(stones[i][1]), + entry); + } + } + int ans = stonesSize - HASH_CNT(set_hh, set); + HASH_CLEAR(set_hh, set); + HASH_CLEAR(by_row_hh, by_row); + HASH_CLEAR(by_col_hh, by_col); + free(entries); + return ans; +} + +int main() { + int s1[6][2] = {{0, 0}, {0, 1}, {1, 0}, {1, 2}, {2, 1}, {2, 2}}, + s2[5][2] = {{0, 1}, {0, 2}, {1, 1}, {2, 0}, {2, 2}}, scs1 = 6, scs2 = 5; + printf("%d\n", removeStones((int **)s1, ARRAY_SIZE(s1), &scs1)); // expect: 5 + printf("%d\n", removeStones((int **)s2, ARRAY_SIZE(s2), &scs2)); // expect: 3 +} diff --git a/most_stones_remove.cpp b/most_stones_remove.cpp index 5283383..632a7fc 100644 --- a/most_stones_remove.cpp +++ b/most_stones_remove.cpp @@ -1,8 +1,18 @@ +// 947. Most Stones Removed with Same Row or Column #include "leetcode.h" +/* + * on a 2d plane, we place 'n' stones at some integer coordinate points. each + * coordinate point may have at most one stone. a stone can be removed if it + * shares either the same row or the same column as another stone that has not + * been removed. given an array 'stones' of length 'n' where 'stones[i] = [x_i, + * y_i]' represents the location of the i'th stone, return the largest possible + * number of stones that can be removed + */ + class Solution { public: - int removeStones(vvd(int) & stones) { + int removeStones(vector<vector<int>> &stones) { int n = stones.size(), c = 0; unordered_map<int, vector<int>> rows, cols; for (int i = 0; i < n; ++i) { @@ -30,6 +40,8 @@ public: int main() { Solution obj; - vvd(int) stones = {{0, 0}, {0, 1}, {1, 0}, {1, 2}, {2, 1}, {2, 2}}; - cout << obj.removeStones(stones); + vvd(int) s1 = {{0, 0}, {0, 1}, {1, 0}, {1, 2}, {2, 1}, {2, 2}}, + s2 = {{0, 1}, {0, 2}, {1, 1}, {2, 0}, {2, 2}}; + printf("%d\n", obj.removeStones(s1)); // expect: 5 + printf("%d\n", obj.removeStones(s2)); // expect: 3 } diff --git a/most_stones_remove.py b/most_stones_remove.py new file mode 100644 index 0000000..1f12d3e --- /dev/null +++ b/most_stones_remove.py @@ -0,0 +1,34 @@ +# 947. Most Stones Removed with Same Row or Column + +""" +on a 2d plane, we place 'n' stones at some integer coordinate stones. each +coordinate point may have at most one stone. a stone can be removed if it +shares either the same row or the same column as another stone that has not +been removed. given an array 'stones' of length 'n' where 'stones[i] = [x_i, +y_i]' represents the location of the i'th stone, return the largest possible +number of stones that can be removed +""" + + +class Solution(object): + def removeStones(self, stones): + """ + :type stones: List[List[int]] + :rtype: int + """ + uf = {} + + def find(x): + if x != uf.setdefault(x, x): + uf[x] = find(uf[x]) + return uf[x] + + for i, j in stones: + uf[find(i)] = find(~j) + return len(stones) - len({find(x) for x in uf}) + + +if __name__ == "__main__": + obj = Solution() + print(obj.removeStones(stones=[[0, 0], [0, 1], [1, 0], [1, 2], [2, 1], [2, 2]])) + print(obj.removeStones(stones=[[0, 0], [0, 2], [1, 1], [2, 0], [2, 2]])) diff --git a/most_stones_remove.rs b/most_stones_remove.rs index 8dc890d..d32137f 100644 --- a/most_stones_remove.rs +++ b/most_stones_remove.rs @@ -1,3 +1,14 @@ +// 947. Most Stones Removed with Same Row or Column + +/* + * on a 2d plane, we place 'n' stones at some integer coordinate points. each + * coordinate point may have at most one stone. a stone can be removed if it + * shares either the same row or the same column as another stone that has not + * been removed. given an array 'stones' of length 'n' where 'stones[i] = [x_i, + * y_i]' represents the location of the i'th stone, return the largest possible + * number of stones that can be removed + */ + use std::collections::HashMap; struct Solution; @@ -9,16 +20,24 @@ impl Solution { let mut cols: HashMap<i32, usize> = HashMap::new(); for (i, v) in stones.iter().enumerate() { match rows.get(&v[0]) { - Some(&j) => if Self::union_stone(i, j, &mut parent) { - ans += 1; + Some(&j) => { + if Self::union_stone(i, j, &mut parent) { + ans += 1; + } + } + None => { + rows.insert(v[0], i); } - None => { rows.insert(v[0], i); } } match cols.get(&v[1]) { - Some(&j) => if Self::union_stone(i, j, &mut parent) { - ans += 1; + Some(&j) => { + if Self::union_stone(i, j, &mut parent) { + ans += 1; + } + } + None => { + cols.insert(v[1], i); } - None => { cols.insert(v[1], i); } } } ans @@ -39,11 +58,12 @@ impl Solution { fn main() { let stones = vec![ - vec![0,0], - vec![0,1], - vec![1,0], - vec![1,2], - vec![2,1], - vec![2,2]]; + vec![0, 0], + vec![0, 1], + vec![1, 0], + vec![1, 2], + vec![2, 1], + vec![2, 2], + ]; print!("{}", Solution::remove_stones(stones)); } |