Last updated on February 9th, 2025 at 07:10 am
Here, we see the Replace Words LeetCode Solution. This Leetcode problem is solved using different approaches in many programming languages, such as C++, Java, JavaScript, Python, etc.
List of all LeetCode Solution
Topics
Hash Table, Trie
Companies
Uber
Level of Question
Medium
Replace Words LeetCode Solution
Table of Contents
1. Problem Statement
In English, we have a concept called root, which can be followed by some other word to form another longer word – let’s call this word successor. For example, when the root “an” is followed by the successor word “other”, we can form a new word “another”.
Given a dictionary consisting of many roots and a sentence consisting of words separated by spaces, replace all the successors in the sentence with the root forming it. If a successor can be replaced by more than one root, replace it with the root that has the shortest length.
Return the sentence after the replacement.
Example 1:
Input: dictionary = [“cat”,”bat”,”rat”], sentence = “the cattle was rattled by the battery”
Output: “the cat was rat by the bat”
Example 2:
Input: dictionary = [“a”,”b”,”c”], sentence = “aadsfasf absbs bbab cadsfafs”
Output: “a a b c”
2. Coding Pattern Used in Solution
The coding pattern used in all the provided implementations is “Prefix Matching”, where we need to replace words or find the shortest prefix match for a given word. The goal is to replace words in a sentence with their shortest prefix from a dictionary.
3. Code Implementation in Different Languages
3.1 Replace Words C++
class Solution {
public:
string replaceWords(vector<string>& dictionary, string sentence) {
string ans="";
unordered_set<string>s(dictionary.begin(),dictionary.end());
string x="";
for(int i=0;i<sentence.size();)
{
if(sentence[i]==' ')
{
ans+=x;
ans+=' ';
x="";
}
else
{
x+=sentence[i];
if(s.find(x)!=s.end())
{
while(i<sentence.size() && sentence[i]!=' ')
{
i++;
}
ans+=x;
x="";
continue;
}
}
i++;
}
if(x.size()!=0)ans+=x;
return ans;
}
};
3.2 Replace Words Java
class Solution {
public String replaceWords(List<String> dictionary, String sentence) {
Set<String> set = new HashSet();
for (String root: dictionary)
set.add(root);
StringBuilder ans = new StringBuilder();
String word[] = sentence.split(" ");
for (int j=0;j<word.length;j++) {
String prefix = "";
for (int i=1; i<=word[j].length();++i) {
prefix = word[j].substring(0, i);
if (set.contains(prefix))
break;
}
if(ans.length()>0)
ans.append(" ");
ans.append(prefix);
}
return ans.toString();
}
}
3.3 Replace Words JavaScript
var replaceWords = function(dictionary, sentence) {
let wordArr = sentence.split(" ");
wordArr = wordArr.map(w => {
for(let i = 0; i <= dictionary.length - 1; i++) {
if (w.indexOf(dictionary[i]) === 0) {
w = dictionary[i];
}
}
return w;
})
return wordArr.reduce((str, word) => str += `${word} `, "").trim();
};
3.4 Replace Words Python
class Solution(object):
def replaceWords(self, dictionary, sentence):
setenceAsList = sentence.split(" ")
for i in range(len(setenceAsList)):
for j in dictionary:
if setenceAsList[i].startswith(j):
setenceAsList[i] = j
return " ".join(setenceAsList)
4. Time and Space Complexity
| Time Complexity | Space Complexity | |
| C++ | O(n * m + k) | O(k + n) |
| Java | O(n * m + k) | O(k + n) |
| JavaScript | O(n * m * k) | O(n) |
| Python | O(n * m * k) | O(n) |
Here,m = avg word lengthn = number of words in the sentencek = dictionary size
- The C++ and Java implementations are more efficient due to the use of hash-based data structures (
unordered_setandHashSet) for prefix lookups. - The JavaScript and Python implementations are less efficient because they iterate through the dictionary for each word in the sentence, leading to an additional factor of
kin the time complexity. - All implementations have similar space complexity, dominated by the size of the result string and the dictionary storage (if applicable).