Problem

Given a positive integer num represented as a string, return the integernum without trailing zeros as a string.

Examples

Example 1

1
2
3

Input: num = "51230100"
Output: "512301"
Explanation: Integer "51230100" has 2 trailing zeros, we remove them and return integer "512301".

Example 2

1
2
3

Input: num = "123"
Output: "123"
Explanation: Integer "123" has no trailing zeros, we return integer "123".

Constraints

  • 1 <= num.length <= 1000
  • num consists of only digits.
  • num doesn’t have any leading zeros.

Solution

Method 1 - String Rstrip or Manual Scan

Intuition

We want to remove all trailing zeros from the string. This can be done by scanning from the end or using built-in string methods.

Approach

  1. Start from the end of the string and move left until a non-zero is found.
  2. Return the substring up to that point.
  3. Or, use rstrip(‘0’) in Python.

Code

1
2
3
4
5
6
7
8

#include <string>
using namespace std;

string removeTrailingZeros(string num) {
    int i = num.size() - 1;
    while (i >= 0 && num[i] == '0') i--;
    return num.substr(0, i+1);
}
1
2
3
4
5
6
7

func removeTrailingZeros(num string) string {
    i := len(num) - 1
    for i >= 0 && num[i] == '0' {
        i--
    }
    return num[:i+1]
}
1
2
3
4
5
6
7

public class Solution {
    public String removeTrailingZeros(String num) {
        int i = num.length() - 1;
        while (i >= 0 && num.charAt(i) == '0') i--;
        return num.substring(0, i+1);
    }
}
1
2
3
4
5

fun removeTrailingZeros(num: String): String {
    var i = num.length - 1
    while (i >= 0 && num[i] == '0') i--
    return num.substring(0, i+1)
}
1
2

def removeTrailingZeros(num: str) -> str:
    return num.rstrip('0')
1
2
3
4
5
6
7
8

fn remove_trailing_zeros(num: &str) -> String {
    let mut end = num.len();
    let bytes = num.as_bytes();
    while end > 0 && bytes[end-1] == b'0' {
        end -= 1;
    }
    num[..end].to_string()
}
1
2
3
4
5

function removeTrailingZeros(num: string): string {
    let i = num.length - 1;
    while (i >= 0 && num[i] === '0') i--;
    return num.slice(0, i + 1);
}

Complexity

  • ⏰ Time complexity: O(N), where N is the length of num.
  • 🧺 Space complexity: O(N), for the output string.