LeetCode #820 — MEDIUM

Short Encoding of Words

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

A valid encoding of an array of words is any reference string s and array of indices indices such that:

words.length == indices.length
The reference string s ends with the '#' character.
For each index indices[i], the substring of s starting from indices[i] and up to (but not including) the next '#' character is equal to words[i].

Given an array of words, return the length of the shortest reference string s possible of any valid encoding of words.

Example 1:

Input: words = ["time", "me", "bell"]
Output: 10
Explanation: A valid encoding would be s = "time#bell#" and indices = [0, 2, 5].
words[0] = "time", the substring of s starting from indices[0] = 0 to the next '#' is underlined in "time#bell#"
words[1] = "me", the substring of s starting from indices[1] = 2 to the next '#' is underlined in "time#bell#"
words[2] = "bell", the substring of s starting from indices[2] = 5 to the next '#' is underlined in "time#bell#"

Example 2:

Input: words = ["t"]
Output: 2
Explanation: A valid encoding would be s = "t#" and indices = [0].

Constraints:

1 <= words.length <= 2000
1 <= words[i].length <= 7
words[i] consists of only lowercase letters.

Step 01

Brute Force Baseline

Problem summary: A valid encoding of an array of words is any reference string s and array of indices indices such that: words.length == indices.length The reference string s ends with the '#' character. For each index indices[i], the substring of s starting from indices[i] and up to (but not including) the next '#' character is equal to words[i]. Given an array of words, return the length of the shortest reference string s possible of any valid encoding of words.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Trie

Example 1

["time","me","bell"]

Example 2

["t"]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #820: Short Encoding of Words
class Trie {
    Trie[] children = new Trie[26];
}

class Solution {
    public int minimumLengthEncoding(String[] words) {
        Trie root = new Trie();
        for (String w : words) {
            Trie cur = root;
            for (int i = w.length() - 1; i >= 0; i--) {
                int idx = w.charAt(i) - 'a';
                if (cur.children[idx] == null) {
                    cur.children[idx] = new Trie();
                }
                cur = cur.children[idx];
            }
        }
        return dfs(root, 1);
    }

    private int dfs(Trie cur, int l) {
        boolean isLeaf = true;
        int ans = 0;
        for (int i = 0; i < 26; i++) {
            if (cur.children[i] != null) {
                isLeaf = false;
                ans += dfs(cur.children[i], l + 1);
            }
        }
        if (isLeaf) {
            ans += l;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #820: Short Encoding of Words
type trie struct {
	children [26]*trie
}

func minimumLengthEncoding(words []string) int {
	root := new(trie)
	for _, w := range words {
		cur := root
		for i := len(w) - 1; i >= 0; i-- {
			if cur.children[w[i]-'a'] == nil {
				cur.children[w[i]-'a'] = new(trie)
			}
			cur = cur.children[w[i]-'a']
		}
	}
	return dfs(root, 1)
}

func dfs(cur *trie, l int) int {
	isLeaf, ans := true, 0
	for i := 0; i < 26; i++ {
		if cur.children[i] != nil {
			isLeaf = false
			ans += dfs(cur.children[i], l+1)
		}
	}
	if isLeaf {
		ans += l
	}
	return ans
}

# Accepted solution for LeetCode #820: Short Encoding of Words
class Trie:
    def __init__(self) -> None:
        self.children = [None] * 26


class Solution:
    def minimumLengthEncoding(self, words: List[str]) -> int:
        root = Trie()
        for w in words:
            cur = root
            for c in w[::-1]:
                idx = ord(c) - ord("a")
                if cur.children[idx] == None:
                    cur.children[idx] = Trie()
                cur = cur.children[idx]
        return self.dfs(root, 1)

    def dfs(self, cur: Trie, l: int) -> int:
        isLeaf, ans = True, 0
        for i in range(26):
            if cur.children[i] != None:
                isLeaf = False
                ans += self.dfs(cur.children[i], l + 1)
        if isLeaf:
            ans += l
        return ans

// Accepted solution for LeetCode #820: Short Encoding of Words
struct Solution;

use std::collections::HashMap;

#[derive(PartialEq, Eq, Default)]
struct Trie {
    children: HashMap<char, Trie>,
    end: bool,
}

impl Trie {
    fn new() -> Self {
        Self::default()
    }
    fn insert<I>(&mut self, iter: I)
    where
        I: Iterator<Item = char>,
    {
        let mut link = self;
        for c in iter {
            link = link.children.entry(c).or_default();
        }
        link.end = true;
    }

    fn dfs(&self, length: usize, sum: &mut usize) {
        let mut is_leaf = true;
        for link in self.children.values() {
            is_leaf = false;
            link.dfs(length + 1, sum);
        }
        if is_leaf {
            *sum += length + 1;
        }
    }
}

impl Solution {
    fn minimum_length_encoding(words: Vec<String>) -> i32 {
        let mut trie = Trie::new();
        for word in words {
            trie.insert(word.chars().rev());
        }
        let mut res = 0;
        trie.dfs(0, &mut res);
        res as i32
    }
}

#[test]
fn test() {
    let words = vec_string!["time", "me", "bell"];
    let res = 10;
    assert_eq!(Solution::minimum_length_encoding(words), res);
}

// Accepted solution for LeetCode #820: Short Encoding of Words
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #820: Short Encoding of Words
// class Trie {
//     Trie[] children = new Trie[26];
// }
// 
// class Solution {
//     public int minimumLengthEncoding(String[] words) {
//         Trie root = new Trie();
//         for (String w : words) {
//             Trie cur = root;
//             for (int i = w.length() - 1; i >= 0; i--) {
//                 int idx = w.charAt(i) - 'a';
//                 if (cur.children[idx] == null) {
//                     cur.children[idx] = new Trie();
//                 }
//                 cur = cur.children[idx];
//             }
//         }
//         return dfs(root, 1);
//     }
// 
//     private int dfs(Trie cur, int l) {
//         boolean isLeaf = true;
//         int ans = 0;
//         for (int i = 0; i < 26; i++) {
//             if (cur.children[i] != null) {
//                 isLeaf = false;
//                 ans += dfs(cur.children[i], l + 1);
//             }
//         }
//         if (isLeaf) {
//             ans += l;
//         }
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(L)

Space

O(N × L)

Approach Breakdown

HASH SET

O(N × L) time

O(N × L) space

Store all N words in a hash set. Each insert/lookup hashes the entire word of length L, giving O(L) per operation. Prefix queries require checking every stored word against the prefix — O(N × L) per prefix search. Space is O(N × L) for storing all characters.

TRIE

O(L) time

O(N × L) space

Each operation (insert, search, prefix) takes O(L) time where L is the word length — one node visited per character. Total space is bounded by the sum of all stored word lengths. Tries win over hash sets when you need prefix matching: O(L) prefix search vs. checking every stored word.

Shortcut: One node per character → O(L) per operation. Prefix queries are what make tries worthwhile.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.