Build confidence with an intuition-first walkthrough focused on stack fundamentals.
You are given a string s.
Your task is to remove all digits by doing this operation repeatedly:
Return the resulting string after removing all digits.
Note that the operation cannot be performed on a digit that does not have any non-digit character to its left.
Example 1:
Input: s = "abc"
Output: "abc"
Explanation:
There is no digit in the string.
Example 2:
Input: s = "cb34"
Output: ""
Explanation:
First, we apply the operation on s[2], and s becomes "c4".
Then we apply the operation on s[1], and s becomes "".
Constraints:
1 <= s.length <= 100s consists only of lowercase English letters and digits.Problem summary: You are given a string s. Your task is to remove all digits by doing this operation repeatedly: Delete the first digit and the closest non-digit character to its left. Return the resulting string after removing all digits. Note that the operation cannot be performed on a digit that does not have any non-digit character to its left.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Stack
"abc"
"cb34"
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3174: Clear Digits
class Solution {
public String clearDigits(String s) {
StringBuilder stk = new StringBuilder();
for (char c : s.toCharArray()) {
if (Character.isDigit(c)) {
stk.deleteCharAt(stk.length() - 1);
} else {
stk.append(c);
}
}
return stk.toString();
}
}
// Accepted solution for LeetCode #3174: Clear Digits
func clearDigits(s string) string {
stk := []byte{}
for i := range s {
if s[i] >= '0' && s[i] <= '9' {
stk = stk[:len(stk)-1]
} else {
stk = append(stk, s[i])
}
}
return string(stk)
}
# Accepted solution for LeetCode #3174: Clear Digits
class Solution:
def clearDigits(self, s: str) -> str:
stk = []
for c in s:
if c.isdigit():
stk.pop()
else:
stk.append(c)
return "".join(stk)
// Accepted solution for LeetCode #3174: Clear Digits
fn clear_digits(s: String) -> String {
let mut ret = String::new();
for c in s.chars() {
if c.is_digit(10) {
ret.pop();
} else {
ret.push(c);
}
}
ret
}
fn main() {
let ret = clear_digits("abcd9e".to_string());
println!("ret={ret}");
}
#[test]
fn test() {
assert_eq!(clear_digits("abc".to_string()), "abc");
assert_eq!(clear_digits("cb34".to_string()), "");
}
// Accepted solution for LeetCode #3174: Clear Digits
function clearDigits(s: string): string {
const stk: string[] = [];
for (const c of s) {
if (!isNaN(parseInt(c))) {
stk.pop();
} else {
stk.push(c);
}
}
return stk.join('');
}
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.