Move from brute-force thinking to an efficient approach using dynamic programming strategy.
You have intercepted a secret message encoded as a string of numbers. The message is decoded via the following mapping:
"1" -> 'A'
"2" -> 'B'
...
"25" -> 'Y'
"26" -> 'Z'
However, while decoding the message, you realize that there are many different ways you can decode the message because some codes are contained in other codes ("2" and "5" vs "25").
For example, "11106" can be decoded into:
"AAJF" with the grouping (1, 1, 10, 6)"KJF" with the grouping (11, 10, 6)(1, 11, 06) is invalid because "06" is not a valid code (only "6" is valid).Note: there may be strings that are impossible to decode.
Given a string s containing only digits, return the number of ways to decode it. If the entire string cannot be decoded in any valid way, return 0.
The test cases are generated so that the answer fits in a 32-bit integer.
Example 1:
Input: s = "12"
Output: 2
Explanation:
"12" could be decoded as "AB" (1 2) or "L" (12).
Example 2:
Input: s = "226"
Output: 3
Explanation:
"226" could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).
Example 3:
Input: s = "06"
Output: 0
Explanation:
"06" cannot be mapped to "F" because of the leading zero ("6" is different from "06"). In this case, the string is not a valid encoding, so return 0.
Constraints:
1 <= s.length <= 100s contains only digits and may contain leading zero(s).Problem summary: You have intercepted a secret message encoded as a string of numbers. The message is decoded via the following mapping: "1" -> 'A' "2" -> 'B' ... "25" -> 'Y' "26" -> 'Z' However, while decoding the message, you realize that there are many different ways you can decode the message because some codes are contained in other codes ("2" and "5" vs "25"). For example, "11106" can be decoded into: "AAJF" with the grouping (1, 1, 10, 6) "KJF" with the grouping (11, 10, 6) The grouping (1, 11, 06) is invalid because "06" is not a valid code (only "6" is valid). Note: there may be strings that are impossible to decode. Given a string s containing only digits, return the number of ways to decode it. If the entire string cannot be decoded in any valid way, return 0. The test cases are generated so that the answer fits in a 32-bit integer.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming
"12"
"226"
"06"
decode-ways-ii)number-of-ways-to-separate-numbers)count-number-of-texts)class Solution {
public int numDecodings(String s) {
if (s.isEmpty() || s.charAt(0) == '0') return 0;
int prev2 = 1, prev1 = 1;
for (int i = 1; i < s.length(); i++) {
int cur = 0;
if (s.charAt(i) != '0') cur += prev1;
int two = (s.charAt(i - 1) - '0') * 10 + (s.charAt(i) - '0');
if (two >= 10 && two <= 26) cur += prev2;
prev2 = prev1;
prev1 = cur;
}
return prev1;
}
}
func numDecodings(s string) int {
if len(s) == 0 || s[0] == '0' {
return 0
}
prev2, prev1 := 1, 1
for i := 1; i < len(s); i++ {
cur := 0
if s[i] != '0' {
cur += prev1
}
two := int(s[i-1]-'0')*10 + int(s[i]-'0')
if two >= 10 && two <= 26 {
cur += prev2
}
prev2, prev1 = prev1, cur
}
return prev1
}
class Solution:
def numDecodings(self, s: str) -> int:
if not s or s[0] == '0':
return 0
prev2, prev1 = 1, 1
for i in range(1, len(s)):
cur = 0
if s[i] != '0':
cur += prev1
two = int(s[i - 1:i + 1])
if 10 <= two <= 26:
cur += prev2
prev2, prev1 = prev1, cur
return prev1
impl Solution {
pub fn num_decodings(s: String) -> i32 {
let bytes = s.as_bytes();
if bytes.is_empty() || bytes[0] == b'0' {
return 0;
}
let mut prev2 = 1;
let mut prev1 = 1;
for i in 1..bytes.len() {
let mut cur = 0;
if bytes[i] != b'0' {
cur += prev1;
}
let two = (bytes[i - 1] - b'0') as i32 * 10 + (bytes[i] - b'0') as i32;
if (10..=26).contains(&two) {
cur += prev2;
}
prev2 = prev1;
prev1 = cur;
}
prev1
}
}
function numDecodings(s: string): number {
if (!s || s[0] === '0') return 0;
let prev2 = 1;
let prev1 = 1;
for (let i = 1; i < s.length; i++) {
let cur = 0;
if (s[i] !== '0') cur += prev1;
const two = Number(s.slice(i - 1, i + 1));
if (two >= 10 && two <= 26) cur += prev2;
prev2 = prev1;
prev1 = cur;
}
return prev1;
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.