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.
Build confidence with an intuition-first walkthrough focused on hash map fundamentals.
You are given the strings key and message, which represent a cipher key and a secret message, respectively. The steps to decode message are as follows:
key as the order of the substitution table.message is then substituted using the table.' ' are transformed to themselves.key = "happy boy" (actual key would have at least one instance of each letter in the alphabet), we have the partial substitution table of ('h' -> 'a', 'a' -> 'b', 'p' -> 'c', 'y' -> 'd', 'b' -> 'e', 'o' -> 'f').Return the decoded message.
Example 1:
Input: key = "the quick brown fox jumps over the lazy dog", message = "vkbs bs t suepuv" Output: "this is a secret" Explanation: The diagram above shows the substitution table. It is obtained by taking the first appearance of each letter in "the quick brown fox jumps over the lazy dog".
Example 2:
Input: key = "eljuxhpwnyrdgtqkviszcfmabo", message = "zwx hnfx lqantp mnoeius ycgk vcnjrdb" Output: "the five boxing wizards jump quickly" Explanation: The diagram above shows the substitution table. It is obtained by taking the first appearance of each letter in "eljuxhpwnyrdgtqkviszcfmabo".
Constraints:
26 <= key.length <= 2000key consists of lowercase English letters and ' '.key contains every letter in the English alphabet ('a' to 'z') at least once.1 <= message.length <= 2000message consists of lowercase English letters and ' '.Problem summary: You are given the strings key and message, which represent a cipher key and a secret message, respectively. The steps to decode message are as follows: Use the first appearance of all 26 lowercase English letters in key as the order of the substitution table. Align the substitution table with the regular English alphabet. Each letter in message is then substituted using the table. Spaces ' ' are transformed to themselves. For example, given key = "happy boy" (actual key would have at least one instance of each letter in the alphabet), we have the partial substitution table of ('h' -> 'a', 'a' -> 'b', 'p' -> 'c', 'y' -> 'd', 'b' -> 'e', 'o' -> 'f'). Return the decoded message.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Hash Map
"the quick brown fox jumps over the lazy dog" "vkbs bs t suepuv"
"eljuxhpwnyrdgtqkviszcfmabo" "zwx hnfx lqantp mnoeius ycgk vcnjrdb"
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2325: Decode the Message
class Solution {
public String decodeMessage(String key, String message) {
char[] d = new char[128];
d[' '] = ' ';
for (int i = 0, j = 0; i < key.length(); ++i) {
char c = key.charAt(i);
if (d[c] == 0) {
d[c] = (char) ('a' + j++);
}
}
char[] ans = message.toCharArray();
for (int i = 0; i < ans.length; ++i) {
ans[i] = d[ans[i]];
}
return String.valueOf(ans);
}
}
// Accepted solution for LeetCode #2325: Decode the Message
func decodeMessage(key string, message string) string {
d := [128]byte{}
d[' '] = ' '
for i, j := 0, 0; i < len(key); i++ {
if d[key[i]] == 0 {
d[key[i]] = byte('a' + j)
j++
}
}
ans := []byte(message)
for i, c := range ans {
ans[i] = d[c]
}
return string(ans)
}
# Accepted solution for LeetCode #2325: Decode the Message
class Solution:
def decodeMessage(self, key: str, message: str) -> str:
d = {" ": " "}
i = 0
for c in key:
if c not in d:
d[c] = ascii_lowercase[i]
i += 1
return "".join(d[c] for c in message)
// Accepted solution for LeetCode #2325: Decode the Message
use std::collections::HashMap;
impl Solution {
pub fn decode_message(key: String, message: String) -> String {
let mut d = HashMap::new();
for c in key.as_bytes() {
if *c == b' ' || d.contains_key(c) {
continue;
}
d.insert(c, char::from((97 + d.len()) as u8));
}
message
.as_bytes()
.iter()
.map(|c| d.get(c).unwrap_or(&' '))
.collect()
}
}
// Accepted solution for LeetCode #2325: Decode the Message
function decodeMessage(key: string, message: string): string {
const d = new Map<string, string>();
for (const c of key) {
if (c === ' ' || d.has(c)) {
continue;
}
d.set(c, String.fromCharCode('a'.charCodeAt(0) + d.size));
}
d.set(' ', ' ');
return [...message].map(v => d.get(v)).join('');
}
Use this to step through a reusable interview workflow for this problem.
For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.
One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.
Review these before coding to avoid predictable interview regressions.
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.