Unidirectional Word Search - Amazon Top Interview Questions
Problem Statement :
Given a two-dimensional matrix of characters board and a string target, return whether the target can be found in the matrix by going left-to-right, or up-to-down unidirectionally. Constraints n, m ≤ 250 where n is the number of rows and columns in board k ≤ 250 where k is the length of word Example 1 Input board = [ ["H", "E", "L", "L", "O"], ["A", "B", "C", "D", "E"] ] word = "HELLO" Output True Example 2 Input board = [ ["x", "z", "d", "x"], ["p", "g", "u", "x"], ["k", "j", "z", "d"] ] word = "xgz" Output False Explanation You can't make "xgz" going left-to-right or up-to-down.
Solution :
Solution in C++ :
bool solve(vector<vector<string>>& board, string word) {
int n = board.size();
int m = board[0].size();
vector<vector<char>> adj(n);
char ch = word[0];
queue<pair<int, int>> q;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
adj[i].push_back(board[i][j][0]);
if (board[i][j][0] == ch) q.push({i, j});
}
}
while (!q.empty()) {
int i = q.front().first;
int j = q.front().second;
q.pop();
int len = 0;
for (int k = j; k < m; k++) {
if (adj[i][k] == word[len])
len++;
else
break;
}
if (len == word.length()) return true;
len = 0;
for (int k = i; k < n; k++) {
if (adj[k][j] == word[len])
len++;
else
break;
}
if (len == word.length()) return true;
}
return false;
}
Solution in Java :
import java.util.*;
class Solution {
public boolean solve(String[][] board, String word) {
for (int i = 0; i < board.length; i++) {
StringBuilder sbr = new StringBuilder();
for (int j = 0; j < board[0].length; j++) sbr.append(board[i][j]);
if (sbr.toString().contains(word))
return true;
}
for (int i = 0; i < board[0].length; i++) {
StringBuilder sbr = new StringBuilder();
for (int j = 0; j < board.length; j++) sbr.append(board[j][i]);
if (sbr.toString().contains(word))
return true;
}
return false;
}
}
Solution in Python :
class Solution:
def solve(self, board, word):
r = len(board)
c = len(board[0])
for i in range(c):
k = 0
for j in range(r):
if board[j][i] == word[k]:
k += 1
if k == len(word):
return True
for b in board:
p = 0
for i, c in enumerate(b):
if c == word[p]:
p += 1
if p == len(word):
return True
return False
View More Similar Problems
Kitty's Calculations on a Tree
Kitty has a tree, T , consisting of n nodes where each node is uniquely labeled from 1 to n . Her friend Alex gave her q sets, where each set contains k distinct nodes. Kitty needs to calculate the following expression on each set: where: { u ,v } denotes an unordered pair of nodes belonging to the set. dist(u , v) denotes the number of edges on the unique (shortest) path between nodes a
View Solution →Is This a Binary Search Tree?
For the purposes of this challenge, we define a binary tree to be a binary search tree with the following ordering requirements: The data value of every node in a node's left subtree is less than the data value of that node. The data value of every node in a node's right subtree is greater than the data value of that node. Given the root node of a binary tree, can you determine if it's also a
View Solution →Square-Ten Tree
The square-ten tree decomposition of an array is defined as follows: The lowest () level of the square-ten tree consists of single array elements in their natural order. The level (starting from ) of the square-ten tree consists of subsequent array subsegments of length in their natural order. Thus, the level contains subsegments of length , the level contains subsegments of length , the
View Solution →Balanced Forest
Greg has a tree of nodes containing integer data. He wants to insert a node with some non-zero integer value somewhere into the tree. His goal is to be able to cut two edges and have the values of each of the three new trees sum to the same amount. This is called a balanced forest. Being frugal, the data value he inserts should be minimal. Determine the minimal amount that a new node can have to a
View Solution →Jenny's Subtrees
Jenny loves experimenting with trees. Her favorite tree has n nodes connected by n - 1 edges, and each edge is ` unit in length. She wants to cut a subtree (i.e., a connected part of the original tree) of radius r from this tree by performing the following two steps: 1. Choose a node, x , from the tree. 2. Cut a subtree consisting of all nodes which are not further than r units from node x .
View Solution →Tree Coordinates
We consider metric space to be a pair, , where is a set and such that the following conditions hold: where is the distance between points and . Let's define the product of two metric spaces, , to be such that: , where , . So, it follows logically that is also a metric space. We then define squared metric space, , to be the product of a metric space multiplied with itself: . For
View Solution →