Leetcode #616: Add Bold Tag in String
In this guide, we solve Leetcode #616 Add Bold Tag in String in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a string s and an array of strings words. You should add a closed pair of bold tag and to wrap the substrings in s that exist in words.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Trie, Array, Hash Table, String, String Matching
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: s = "abcxyz123", words = ["abc","123"]
Output: "<b>abc</b>xyz<b>123</b>"
Explanation: The two strings of words are substrings of s as following: "abcxyz123".
We add <b> before each substring and </b> after each substring.
Python Solution
class Trie:
def __init__(self):
self.children = [None] * 128
self.is_end = False
def insert(self, word):
node = self
for c in word:
idx = ord(c)
if node.children[idx] is None:
node.children[idx] = Trie()
node = node.children[idx]
node.is_end = True
class Solution:
def addBoldTag(self, s: str, words: List[str]) -> str:
trie = Trie()
for w in words:
trie.insert(w)
n = len(s)
pairs = []
for i in range(n):
node = trie
for j in range(i, n):
idx = ord(s[j])
if node.children[idx] is None:
break
node = node.children[idx]
if node.is_end:
pairs.append([i, j])
if not pairs:
return s
st, ed = pairs[0]
t = []
for a, b in pairs[1:]:
if ed + 1 < a:
t.append([st, ed])
st, ed = a, b
else:
ed = max(ed, b)
t.append([st, ed])
ans = []
i = j = 0
while i < n:
if j == len(t):
ans.append(s[i:])
break
st, ed = t[j]
if i < st:
ans.append(s[i:st])
ans.append('<b>')
ans.append(s[st : ed + 1])
ans.append('</b>')
j += 1
i = ed + 1
return ''.join(ans)
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.