# 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)```
```

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