Blocks to Spell Word - Google Top Interview Questions

Problem Statement :

```You are given a list of lowercase alphabet strings words and a string target. Assuming you can pick at most one character from each string in words, return whether you can spell out target in any order.

Constraints

n ≤ 12 where n is the length of target

m ≤ 12 where m is the length of words

Example 1
Input

words = ["how", "do", "i", "shot", "web"]

target = "wow"

Output

True

Explanation

We can pick "w" from "how", "o" from "do" and "w" from "web".

Example 2

Input

words = ["bi", "n", "a", "r", "y"]

target = "binary"

Output

False

Explanation

There's no way to pick both "b" and "i".```

Solution :

```                        ```Solution in C++ :

struct Graph {
// number of verticies
int V;
// number of removed verticies
int removed;
// rem[i] = true if node i was removed from graph
bool* rem;

Graph(int V) {
this->V = V;
removed = 0;
rem = new bool[V];
}

void addEdge(int vFrom, int vTo) {
}

// removes a directed edge
void removeEdge(int vFrom, int vTo) {
// iterate to find vTo in vFrom's adjacency list
// found vTo in vFrom's adjacency list
if (*it2 == vTo) {
removed += (rem[vFrom] = adjs[vFrom].empty());  // true if vFrom now has degree 0
break;
}
}
}
void print() {
for (int i = 0; i < V; ++i) {
cerr << '
' << i << "-> ";
if (rem[i]) continue;
}
cerr << '
';
}
};

bool solve(vector<string>& words, string target) {
const int W = words.size(), T = target.size();
if (words.size() < target.length()) return false;

// bipartite graph
Graph g(W + T);
// the degree of the node with the smallest degree
int min_degree = INT_MAX, n1;
// iterate over all words and form edges
for (int i = 0; i < W; ++i) {
for (int t = 0; t < T; ++t) {
if (words[i].find(target[t]) != -1) {
}
}
n1 = i;
}
}

// removing words that cannot contribute to the target
// returning false if a target character could not be found
for (int i = 0; i < g.V; ++i) {
// if a letter in the target could not find a match in words
if (i > W && g.rem[i]) return false;
}

// cerr << "start conditions";
// g.print();
int pairs = 0;
int itr = 0;
// while not all edges have been removed
while (g.removed != g.V) {
// n1's first neighbor
// mark nodes n1 and n2 as removed and remove edges from other nodes pointing to them
g.rem[n1] = true;
g.rem[n2] = true;
g.removed += 2;
// cerr << '
' << "match " << ++itr << ": (" << min(n1, n2) << ", " << max(n1,n2) << ')';
// g.print();
// disconnect all edges n->n1
for (int n = 0; n < g.V; ++n)
if (!g.rem[n]) g.removeEdge(n, n1);
// disconnect all edges n->n2
// and calculate new min_degree and n1
min_degree = INT_MAX;
for (int n = 0; n < g.V; ++n) {
if (g.rem[n]) continue;
g.removeEdge(n, n2);
n1 = n;
}
}
// indicate a new pair was made
++pairs;
}
return pairs == T;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public boolean solve(String[] words, String target) {
int N = words.length;
int M = target.length();
boolean[][] dp = new boolean[1 << M][N + 1];
dp[0][0] = true;
for (int bit = 0; bit < (1 << M); bit++) {
for (int i = 1; i <= N; i++) {
dp[bit][i] = dp[bit][i - 1];
for (int j = 0; j < M; j++) {
if ((bit & (1 << j)) > 0) {
String use = Character.toString(target.charAt(j));
if (words[i - 1].contains(use))
dp[bit][i] |= dp[bit - (1 << j)][i - 1];
}
}
}
}
return dp[(1 << M) - 1][N];
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, words, target):
@cache
if i == len(target):
return True

for j, x in enumerate(words):
if (mask & (1 << j)) == 0 and target[i] in x:
if go(i + 1, mask | (1 << j)):
return True

return False

return go(0, 0)```
```

Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F

Binary Search Tree : Insertion

You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <

Tree: Huffman Decoding

Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only t

Binary Search Tree : Lowest Common Ancestor

You are given pointer to the root of the binary search tree and two values v1 and v2. You need to return the lowest common ancestor (LCA) of v1 and v2 in the binary search tree. In the diagram above, the lowest common ancestor of the nodes 4 and 6 is the node 3. Node 3 is the lowest node which has nodes and as descendants. Function Description Complete the function lca in the editor b

Swap Nodes [Algo]

A binary tree is a tree which is characterized by one of the following properties: It can be empty (null). It contains a root node only. It contains a root node with a left subtree, a right subtree, or both. These subtrees are also binary trees. In-order traversal is performed as Traverse the left subtree. Visit root. Traverse the right subtree. For this in-order traversal, start from

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