Move from brute-force thinking to an efficient approach using stack strategy.
Given an encoded string, return its decoded string.
The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer.
You may assume that the input string is always valid; there are no extra white spaces, square brackets are well-formed, etc. Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, k. For example, there will not be input like 3a or 2[4].
The test cases are generated so that the length of the output will never exceed 105.
Example 1:
Input: s = "3[a]2[bc]" Output: "aaabcbc"
Example 2:
Input: s = "3[a2[c]]" Output: "accaccacc"
Example 3:
Input: s = "2[abc]3[cd]ef" Output: "abcabccdcdcdef"
Constraints:
1 <= s.length <= 30s consists of lowercase English letters, digits, and square brackets '[]'.s is guaranteed to be a valid input.s are in the range [1, 300].Problem summary: Given an encoded string, return its decoded string. The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer. You may assume that the input string is always valid; there are no extra white spaces, square brackets are well-formed, etc. Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, k. For example, there will not be input like 3a or 2[4]. The test cases are generated so that the length of the output will never exceed 105.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Stack
"3[a]2[bc]"
"3[a2[c]]"
"2[abc]3[cd]ef"
encode-string-with-shortest-length)number-of-atoms)brace-expansion)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #394: Decode String
class Solution {
public String decodeString(String s) {
Deque<Integer> s1 = new ArrayDeque<>();
Deque<String> s2 = new ArrayDeque<>();
int num = 0;
String res = "";
for (char c : s.toCharArray()) {
if ('0' <= c && c <= '9') {
num = num * 10 + c - '0';
} else if (c == '[') {
s1.push(num);
s2.push(res);
num = 0;
res = "";
} else if (c == ']') {
StringBuilder t = new StringBuilder();
for (int i = 0, n = s1.pop(); i < n; ++i) {
t.append(res);
}
res = s2.pop() + t.toString();
} else {
res += String.valueOf(c);
}
}
return res;
}
}
// Accepted solution for LeetCode #394: Decode String
// Auto-generated Go example from java.
func exampleSolution() {
}
// Reference (java):
// // Accepted solution for LeetCode #394: Decode String
// class Solution {
// public String decodeString(String s) {
// Deque<Integer> s1 = new ArrayDeque<>();
// Deque<String> s2 = new ArrayDeque<>();
// int num = 0;
// String res = "";
// for (char c : s.toCharArray()) {
// if ('0' <= c && c <= '9') {
// num = num * 10 + c - '0';
// } else if (c == '[') {
// s1.push(num);
// s2.push(res);
// num = 0;
// res = "";
// } else if (c == ']') {
// StringBuilder t = new StringBuilder();
// for (int i = 0, n = s1.pop(); i < n; ++i) {
// t.append(res);
// }
// res = s2.pop() + t.toString();
// } else {
// res += String.valueOf(c);
// }
// }
// return res;
// }
// }
# Accepted solution for LeetCode #394: Decode String
class Solution:
def decodeString(self, s: str) -> str:
s1, s2 = [], []
num, res = 0, ''
for c in s:
if c.isdigit():
num = num * 10 + int(c)
elif c == '[':
s1.append(num)
s2.append(res)
num, res = 0, ''
elif c == ']':
res = s2.pop() + res * s1.pop()
else:
res += c
return res
// Accepted solution for LeetCode #394: Decode String
impl Solution {
pub fn decode_string(s: String) -> String {
let mut stack = vec![];
for ch in s.chars() {
if ch != ']' {
stack.push(ch.to_string());
} else {
let mut substr = String::new();
while stack.last() != Some(&"[".to_string()) {
substr = stack.pop().unwrap().to_string() + &substr;
}
stack.pop();
let mut multiplier = String::new();
while !stack.is_empty() && stack.last().unwrap().parse::<i32>().is_ok() {
multiplier = stack.pop().unwrap().to_string() + &multiplier;
}
stack.push(substr.repeat(multiplier.parse::<usize>().unwrap()));
}
}
stack.join("")
}
}
// Accepted solution for LeetCode #394: Decode String
function decodeString(s: string): string {
let ans = '';
let stack = [];
let count = 0; // repeatCount
for (let cur of s) {
if (/[0-9]/.test(cur)) {
count = count * 10 + Number(cur);
} else if (/[a-z]/.test(cur)) {
ans += cur;
} else if ('[' == cur) {
stack.push([ans, count]);
// reset
ans = '';
count = 0;
} else {
// match ']'
let [pre, count] = stack.pop();
ans = pre + ans.repeat(count);
}
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.
Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.
Review these before coding to avoid predictable interview regressions.
Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.
Usually fails on: Indices point to blocked elements and outputs shift.
Fix: Pop while invariant is violated before pushing current element.