# Break String Into Words - Amazon Top Interview Questions

### Problem Statement :

```Given a list of lowercase alphabet strings words and a lowercase alphabet string s, return whether or not the string can be broken down using the list of words. It's not required to use all of the words and you can reuse words.

Example 1

Input

words = ["quick", "brown", "the", "fox"]

s = "thequickbrownfox"

Output

True

Explanation

We can break the string down into "the quick brown fox".

Example 2

Input

words = ["hello", "world"]

s = "hellofoobarworld"

Output

False```

### Solution :

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

bool solve(vector<string>& words, string s) {
bool dp[s.length() + 1] = {0};
dp = 1;
for (int i = 0; i < s.length(); i++)
if (dp[i])
for (auto x : words)
if (s.substr(i, x.length()) == x) dp[i + x.length()] = 1;

return dp[s.length()];
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public boolean solve(String[] words, String s) {
int[] dp = new int[s.length()];
Arrays.fill(dp, -1);
HashSet<String> set = new HashSet<>();
return rec(set, s, 0, dp);
}

public boolean rec(HashSet<String> words, String s, int startIdx, int[] dp) {
if (words.contains(s.substring(startIdx)) || s.length() == startIdx)
return true;
// System.out.println(startIdx);
if (dp[startIdx] != -1)
return dp[startIdx] == 1;
int i = startIdx + 1;
while (i < s.length()) {
if (words.contains(s.substring(startIdx, i)) && rec(words, s, i, dp))
return (dp[startIdx] = 1) == 1;
i++;
}
dp[startIdx] = 0;
return false;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, words, s):
a = ""
for i in range(len(s) - 1, -1, -1):
a = s[i] + a
if a in words:
a = ""
if a == "":
return True
else:
return False```
```

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