**Number of Moves to Capture the King - Google Top Interview Questions**

### Problem Statement :

You are given a two-dimensional integer matrix board containing 0s, 1s and 2s representing some n x n chessboard. Each 0 represents an empty cell, 1 represents the knight and 2 represents the king. There is at least one knight but exactly one king. Given that the king stays still, return the minimum number of moves it would take for some knight to land on the king. If there's no solution, return -1. A knight can't land on another knight. Constraints 2 ≤ n ≤ 500 where n is the number of rows and columns in board 1 ≤ t < n * n where t is the number of knights k = 1 where k is the number of kings. Example 1 Input board = [ [1, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 2], [1, 0, 0, 0, 0], [0, 0, 0, 0, 0] ] Output 2 Explanation The knight on top left corner can jump twice to land on the king. Example 2 Input board = [ [1, 2], [1, 1] ] Output -1 Explanation There is no way to land on the king here.

### Solution :

` ````
Solution in C++ :
int solve(vector<vector<int>>& nums) {
int n = nums.size(), m = nums[0].size();
vector<vector<int>> dist(n, vector<int>(m, INT_MAX));
queue<pair<int, int>> q;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (nums[i][j] == 2) {
dist[i][j] = 0;
q.push({i, j});
break;
}
}
}
if (q.empty()) return -1;
vector<int> xdir{1, 2, -1, 2, -1, -2, 1, -2};
vector<int> ydir{2, 1, 2, -1, -2, -1, -2, 1};
while (!q.empty()) {
auto [i, j] = q.front();
q.pop();
for (int k = 0; k < 8; k++) {
int x = i + xdir[k];
int y = j + ydir[k];
if (x >= 0 and x < n and y >= 0 and y < m and nums[x][y] != 2 and
dist[x][y] > dist[i][j] + 1) {
if (nums[x][y] == 1) return dist[i][j] + 1;
dist[x][y] = dist[i][j] + 1;
q.push({x, y});
}
}
}
return -1;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
int[] dr = {-1, -1, 1, 1, 2, 2, -2, -2};
int[] dc = {2, -2, 2, -2, 1, -1, 1, -1};
public int solve(int[][] board) {
LinkedList<Pair<Integer, Integer>> q = new LinkedList();
HashSet<Pair<Integer, Integer>> hs = new HashSet();
int m = board.length, n = board[0].length;
int king_r = 0, king_c = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (board[i][j] == 1) {
q.addLast(new Pair(i, j));
hs.add(new Pair(i, j));
} else if (board[i][j] == 2) {
king_r = i;
king_c = j;
}
}
}
int moves = 0;
while (q.size() > 0) {
int size = q.size();
for (int i = 0; i < size; i++) {
Pair<Integer, Integer> cur = q.removeFirst();
int cur_r = cur.getKey(), cur_c = cur.getValue();
if (cur_r == king_r && cur_c == king_c)
return moves;
for (int move = 0; move < 8; move++) {
int new_r = cur_r + dr[move];
int new_c = cur_c + dc[move];
if (new_r < 0 || new_c < 0 || new_c >= n || new_r >= m)
continue;
if (hs.contains(new Pair(new_r, new_c)))
continue;
hs.add(new Pair(new_r, new_c));
q.addLast(new Pair(new_r, new_c));
}
}
moves++;
}
return -1;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, board):
r = len(board)
c = len(board[0])
dist = []
for _ in range(r):
dist.append([r * c + 1] * c)
q = []
sx, sy = None, None
for i in range(r):
for j in range(c):
if board[i][j] == 1:
q.append((i, j))
dist[i][j] = 0
elif board[i][j] == 2:
sx, sy = i, j
for x, y in q:
for dx, dy in [(-2, -1), (-2, 1), (-1, 2), (-1, -2), (1, -2), (1, 2), (2, -1), (2, 1)]:
nx, ny = x + dx, y + dy
if 0 <= nx < r and 0 <= ny < c and dist[nx][ny] > 1 + dist[x][y]:
dist[nx][ny] = 1 + dist[x][y]
q.append((nx, ny))
ans = dist[sx][sy]
if ans > r * c:
ans = -1
return ans
```

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