# 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```
```

## Jesse and Cookies

Jesse loves cookies. He wants the sweetness of all his cookies to be greater than value K. To do this, Jesse repeatedly mixes two cookies with the least sweetness. He creates a special combined cookie with: sweetness Least sweet cookie 2nd least sweet cookie). He repeats this procedure until all the cookies in his collection have a sweetness > = K. You are given Jesse's cookies. Print t

## Find the Running Median

The median of a set of integers is the midpoint value of the data set for which an equal number of integers are less than and greater than the value. To find the median, you must first sort your set of integers in non-decreasing order, then: If your set contains an odd number of elements, the median is the middle element of the sorted sample. In the sorted set { 1, 2, 3 } , 2 is the median.

## Minimum Average Waiting Time

Tieu owns a pizza restaurant and he manages it in his own way. While in a normal restaurant, a customer is served by following the first-come, first-served rule, Tieu simply minimizes the average waiting time of his customers. So he gets to decide who is served first, regardless of how sooner or later a person comes. Different kinds of pizzas take different amounts of time to cook. Also, once h

## Merging Communities

People connect with each other in a social network. A connection between Person I and Person J is represented as . When two persons belonging to different communities connect, the net effect is the merger of both communities which I and J belongs to. At the beginning, there are N people representing N communities. Suppose person 1 and 2 connected and later 2 and 3 connected, then ,1 , 2 and 3 w

## Components in a graph

There are 2 * N nodes in an undirected graph, and a number of edges connecting some nodes. In each edge, the first value will be between 1 and N, inclusive. The second node will be between N + 1 and , 2 * N inclusive. Given a list of edges, determine the size of the smallest and largest connected components that have or more nodes. A node can have any number of connections. The highest node valu

## Kundu and Tree

Kundu is true tree lover. Tree is a connected graph having N vertices and N-1 edges. Today when he got a tree, he colored each edge with one of either red(r) or black(b) color. He is interested in knowing how many triplets(a,b,c) of vertices are there , such that, there is atleast one edge having red color on all the three paths i.e. from vertex a to b, vertex b to c and vertex c to a . Note that