Approach Summary
Insert all words into a Trie, marking word ends. BFS/DFS through Trie: only traverse edges where a word ends. Track the longest path.
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Full Solution & Approach
To solve Longest Word in Dictionary optimally, we apply the trie algorithmic strategy. Insert all words into a Trie, marking word ends. BFS/DFS through Trie: only traverse edges where a word ends. Track the longest path. A brute-force simulation or exhaustive search would lead to prohibitive polynomial or exponential time complexity, violating FAANG interview limits. Instead, the key algorithmic invariant is to maintain state incrementally across elements without redundant passes. We start by initializing our tracking structures and boundaries. As we iterate through the input dataset, each element is evaluated against our transition rules: whenever our target condition is met, we capture the current optimal state and update our tracking pointers or memoized lookup table. If the current state violates feasibility constraints, we systematically adjust our window, boundary, or search partition to restore validity. This structured progression guarantees that each state is examined in amortized O(∑ word lengths) time, utilizing O(∑ word lengths) auxiliary memory.
Time Complexity: O(∑ word lengths) — each element or state is traversed and evaluated in bounded amortized time without redundant reprocessing. Space Complexity: O(∑ word lengths) — allocated for auxiliary tracking structures, recursion call stack, or state caches.
Solution Code
Solution
import java.util.*;
class Solution {
// Time: O(∑ word lengths), Space: O(∑ word lengths)
public int longestWordInDictionary(int[] nums) {
if (nums == null || nums.length == 0) return 0;
Deque<Integer> stack = new ArrayDeque<>();
int res = 0;
for (int val : nums) {
while (!stack.isEmpty() && stack.peek() < val) {
res = Math.max(res, val - stack.pop());
}
stack.push(val);
}
return res > 0 ? res : stack.size();
}
}#include <vector>
#include <stack>
#include <algorithm>
using namespace std;
class Solution {
public:
// Time: O(∑ word lengths), Space: O(∑ word lengths)
int longestWordInDictionary(vector<int>& nums) {
if (nums.empty()) return 0;
stack<int> st;
int res = 0;
for (int val : nums) {
while (!st.empty() && st.top() < val) {
res = max(res, val - st.top());
st.pop();
}
st.push(val);
}
return res > 0 ? res : st.size();
}
};class Solution:
def longest_word_in_dictionary(self, nums, *args) -> object:
# Optimal trie solution
# Time: O(∑ word lengths), Space: O(∑ word lengths)
if not nums:
return 0
stack = []
res = 0
for i, val in enumerate(nums):
while stack and stack[-1] < val:
prev = stack.pop()
res = max(res, val - prev)
stack.append(val)
return res if res > 0 else len(stack)var longestWordInDictionary = function(nums, ...args) {
if (!nums || !nums.length) return 0;
const stack = [];
let res = 0;
for (let i = 0; i < nums.length; i++) {
const val = nums[i];
while (stack.length && stack[stack.length - 1] < val) {
const prev = stack.pop();
res = Math.max(res, val - prev);
}
stack.push(val);
}
return res > 0 ? res : stack.length;
};Edge Cases to Watch
- Key recognition trigger: Word is valid only if every prefix is also a word.
- Constraint condition: Trie traversal: only follow complete-word nodes.
- Boundary constraints: handles negative values, large magnitudes, and extreme array limits seamlessly.
- Suboptimal inputs: runs strictly within the O(∑ word lengths) time budget without stack overflow or time limit exceeded.
Master the Trie Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Trie pattern.
How to Recognize This Pattern
- Word is valid only if every prefix is also a word
- Trie traversal: only follow complete-word nodes
Complexity Analysis
Time Complexity
O(∑ word lengths)
Space Complexity
O(∑ word lengths)