533. Lonely Pixel II

Description

Given an m x n picture consisting of black 'B' and white 'W' pixels and an integer target, return the number of black lonely pixels.

A black lonely pixel is a character 'B' that located at a specific position (r, c) where:

Row r and column c both contain exactly target black pixels.
For all rows that have a black pixel at column c, they should be exactly the same as row r.

Example 1:

Input: picture = [["W","B","W","B","B","W"],["W","B","W","B","B","W"],["W","B","W","B","B","W"],["W","W","B","W","B","W"]], target = 3
Output: 6
Explanation: All the green 'B' are the black pixels we need (all 'B's at column 1 and 3).
Take 'B' at row r = 0 and column c = 1 as an example:
 - Rule 1, row r = 0 and column c = 1 both have exactly target = 3 black pixels. 
 - Rule 2, the rows have black pixel at column c = 1 are row 0, row 1 and row 2. They are exactly the same as row r = 0.

Example 2:

Input: picture = [["W","W","B"],["W","W","B"],["W","W","B"]], target = 1
Output: 0

Constraints:

m == picture.length
n == picture[i].length
1 <= m, n <= 200
picture[i][j] is 'W' or 'B'.
1 <= target <= min(m, n)

Solutions

class Solution {
    public int findBlackPixel(char[][] picture, int target) {
        int m = picture.length, n = picture[0].length;
        int[] rows = new int[m];
        Map<Integer, List<Integer>> cols = new HashMap<>();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (picture[i][j] == 'B') {
                    ++rows[i];
                    cols.computeIfAbsent(j, k -> new ArrayList<>()).add(i);
                }
            }
        }
        boolean[][] t = new boolean[m][m];
        for (int i = 0; i < m; ++i) {
            for (int k = i; k < m; ++k) {
                t[i][k] = i == k || all(picture[i], picture[k]);
                t[k][i] = t[i][k];
            }
        }
        int res = 0;
        for (int i = 0; i < m; ++i) {
            if (rows[i] == target) {
                for (int j = 0; j < n; ++j) {
                    List<Integer> col = cols.get(j);
                    if (col != null && col.size() == target) {
                        boolean check = true;
                        for (int k : col) {
                            check = check && t[i][k];
                        }
                        if (check) {
                            ++res;
                        }
                    }
                }
            }
        }
        return res;
    }

    private boolean all(char[] row1, char[] row2) {
        int n = row1.length;
        for (int j = 0; j < n; ++j) {
            if (row1[j] != row2[j]) {
                return false;
            }
        }
        return true;
    }
}

class Solution {
public:
    int findBlackPixel(vector<vector<char>>& picture, int target) {
        int m = picture.size(), n = picture[0].size();
        vector<int> rows(m);
        unordered_map<int, vector<int>> cols;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (picture[i][j] == 'B') {
                    ++rows[i];
                    cols[j].push_back(i);
                }
            }
        }
        vector<vector<bool>> t(m, vector<bool>(m, false));
        for (int i = 0; i < m; ++i) {
            for (int k = i; k < m; ++k) {
                t[i][k] = i == k || all(picture[i], picture[k]);
                t[k][i] = t[i][k];
            }
        }
        int res = 0;
        for (int i = 0; i < m; ++i) {
            if (rows[i] == target) {
                for (int j = 0; j < n; ++j) {
                    if (cols[j].size() == target) {
                        bool check = true;
                        for (int k : cols[j]) check = check && t[i][k];
                        if (check) ++res;
                    }
                }
            }
        }
        return res;
    }

    bool all(vector<char>& row1, vector<char>& row2) {
        int n = row1.size();
        for (int j = 0; j < n; ++j)
            if (row1[j] != row2[j]) return false;
        return true;
    }
};

class Solution:
    def findBlackPixel(self, picture: List[List[str]], target: int) -> int:
        m, n = len(picture), len(picture[0])
        rows = [0] * m
        cols = defaultdict(list)
        for i in range(m):
            for j in range(n):
                if picture[i][j] == 'B':
                    rows[i] += 1
                    cols[j].append(i)
        t = [[False] * m for _ in range(m)]
        for i in range(m):
            for k in range(i, m):
                if i == k:
                    t[i][k] = True
                else:
                    t[i][k] = all([picture[i][j] == picture[k][j] for j in range(n)])
                t[k][i] = t[i][k]
        res = 0
        for i in range(m):
            if rows[i] == target:
                for j in range(n):
                    if len(cols[j]) == target and all([t[i][k] for k in cols[j]]):
                        res += 1
        return res

func findBlackPixel(picture [][]byte, target int) int {
	m, n := len(picture), len(picture[0])
	rows := make([]int, m)
	cols := make(map[int][]int)
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if picture[i][j] == 'B' {
				rows[i]++
				cols[j] = append(cols[j], i)
			}
		}
	}
	t := make([][]bool, m)
	for i := 0; i < m; i++ {
		t[i] = make([]bool, m)
	}
	for i := 0; i < m; i++ {
		for k := i; k < m; k++ {
			if i == k {
				t[i][k] = true
			} else {
				t[i][k] = all(picture[i], picture[k])
			}
			t[k][i] = t[i][k]
		}
	}
	res := 0
	for i := 0; i < m; i++ {
		if rows[i] == target {
			for j := 0; j < n; j++ {
				col, ok := cols[j]
				if ok && len(col) == target {
					check := true
					for _, k := range col {
						check = check && t[i][k]
					}
					if check {
						res++
					}
				}
			}
		}
	}
	return res
}

func all(row1, row2 []byte) bool {
	n := len(row1)
	for i := 0; i < n; i++ {
		if row1[i] != row2[i] {
			return false
		}
	}
	return true
}

function findBlackPixel(picture: string[][], target: number): number {
    const m: number = picture.length;
    const n: number = picture[0].length;
    const g: number[][] = Array.from({ length: n }, () => []);
    const rows: number[] = Array(m).fill(0);

    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (picture[i][j] === 'B') {
                ++rows[i];
                g[j].push(i);
            }
        }
    }

    let ans: number = 0;
    for (let j = 0; j < n; ++j) {
        if (g[j].length === 0 || rows[g[j][0]] !== target) {
            continue;
        }
        const i1: number = g[j][0];
        let ok: number = 0;
        if (g[j].length === rows[i1]) {
            ok = target;
            for (const i2 of g[j]) {
                if (picture[i1].join('') !== picture[i2].join('')) {
                    ok = 0;
                    break;
                }
            }
        }
        ans += ok;
    }
    return ans;
}

533 - Lonely Pixel II

533. Lonely Pixel II

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533. Lonely Pixel II

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