**Grid Coloring - Google Top Interview Questions**

### Problem Statement :

You are given a two-dimensional list of integers matrix. Each element matrix[r][c] represents a oolor 0 or 1. In one operation you can set the color of matrix[0][0] and all the connected (vertically and horizontally) cells that are of the same color as index (0, 0). Return the minimum number of operations required to change all cells in the matrix to the same color. Constraints 0 ≤ n, m ≤ 250 where n and m are the number of rows and columns in matrix matrix[r][c] = 0 or matrix[r][c] = 1 Example 1 Input matrix = [ [0, 0, 0, 1], [1, 1, 1, 1], [0, 0, 0, 0] ] Output 2 Explanation First set the color of (0, 0) to 1 and then to 0

### Solution :

` ````
Solution in C++ :
int solve(vector<vector<int>>& matrix) {
int n = matrix.size(), m = matrix[0].size();
vector<vector<int>> f(n, vector<int>(m, 1000010000));
struct State {
int r, c, f;
bool operator<(const State& that) const {
return f > that.f;
}
};
priority_queue<State> q;
q.push({0, 0, 0});
f[0][0] = 0;
int ret = 0;
const int dr[] = {-1, 0, 1, 0}, dc[] = {0, 1, 0, -1};
while (!q.empty()) {
const auto curr = q.top();
q.pop();
if (curr.f > f[curr.r][curr.c]) continue;
// cout << curr.r << ' ' << curr.c << ' ' << curr.f << endl;
for (int d = 0; d < 4; ++d) {
State next = {curr.r + dr[d], curr.c + dc[d], curr.f};
if (next.r < 0 || next.r >= n) continue;
if (next.c < 0 || next.c >= m) continue;
if (matrix[next.r][next.c] != matrix[curr.r][curr.c]) ++next.f;
if (next.f < f[next.r][next.c]) {
f[next.r][next.c] = next.f;
q.push(next);
ret = max(ret, next.f);
}
}
}
/*
for (const auto& v : f) {
for (int x : v) cout << x << ' ';
cout << endl;
}
*/
return ret;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
final int[][] dirs = new int[][] {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
public int solve(int[][] matrix) {
if (matrix.length == 0)
return 0;
// q represents the current list of coordinates of islands we want to floodfill from
Queue<int[]> q = new LinkedList();
q.offer(new int[] {0, 0});
// curr represents the id of the island we are trying to floodfill. this value will
// alternate between 1 and 0 indefinitely until the problem is solved
int curr = matrix[0][0];
// next is the queue that we add the coordinates of "borders" into. We will then loop
// through these values in our "next round" of floodfill
Queue<int[]> next = new LinkedList();
// visited array
boolean[][] vis = new boolean[matrix.length][matrix[0].length];
vis[0][0] = true;
int ret = 0;
while (!q.isEmpty()) {
// loop through queue and add all of the borders to the other queue "next"
while (!q.isEmpty()) {
int size = q.size();
int[] temp = q.poll();
int i = temp[0];
int j = temp[1];
for (int[] d : dirs) {
int neighI = i + d[0];
int neighJ = j + d[1];
// out of bounds check
if (neighI < 0 || neighJ < 0 || neighI >= matrix.length
|| neighJ >= matrix[0].length || vis[neighI][neighJ])
continue;
vis[neighI][neighJ] = true;
if (matrix[neighI][neighJ] == curr) {
q.add(new int[] {neighI, neighJ});
} else {
next.offer(new int[] {neighI, neighJ});
}
}
}
ret++;
curr = 1 - curr;
// swapping the two queues
Queue<int[]> temp1 = new LinkedList();
Queue<int[]> temp2 = new LinkedList();
temp1.addAll(q);
temp2.addAll(next);
q = temp2;
next = temp1;
}
return ret - 1;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, matrix):
if not matrix or not matrix[0]:
return 0
m, n = len(matrix), len(matrix[0])
is_filled = [[False] * n for _ in range(m)]
is_filled[0][0] = True
operation_count = -1
layer = [(0, 0)]
while layer:
operation_count += 1
next_layer = []
while layer:
y, x = layer.pop()
for y2, x2 in [(y - 1, x), (y, x + 1), (y + 1, x), (y, x - 1)]:
if not 0 <= y2 < m or not 0 <= x2 < n:
continue
if is_filled[y2][x2]:
continue
is_filled[y2][x2] = True
if matrix[y2][x2] == matrix[y][x]:
layer.append((y2, x2))
else:
next_layer.append((y2, x2))
layer = next_layer
return operation_count
```

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