Remove Trailing Zeros From a String

Difficulty: Easy | Acceptance: 79.60% | Paid: No Topics: String

Given a positive integer num represented as a string, return the string representation of the integer without any trailing zeros.

Table of Contents

• Examples
• Constraints
• Two Pointers Approach
• Built-in String Methods
• Iterative from End

Examples

Example 1:

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

Example 2:

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

Constraints

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

Two Pointers Approach

Intuition Use two pointers to find the position of the last non-zero character, then extract the substring from start to that position.

Steps

• Initialize a pointer at the end of the string
• Move the pointer backwards while the current character is ‘0’
• Return the substring from index 0 to the pointer position (inclusive)

class Solution:
    def removeTrailingZeros(self, num: str) -> str:
        n = len(num)
        right = n - 1
        while right >= 0 and num[right] == '0':
            right -= 1
        return num[:right + 1]

Complexity

• Time: O(n) where n is the length of the string
• Space: O(n) for the resulting substring
• Notes: Simple and efficient, uses minimal extra space

Built-in String Methods

Intuition Leverage built-in string manipulation functions to remove trailing zeros directly.

Steps

• Use language-specific methods to strip trailing zeros
• In Python, use rstrip() method
• In Java, use lastIndexOf() or regex
• In JavaScript/TypeScript, use replaceEnd() or regex

class Solution:
    def removeTrailingZeros(self, num: str) -> str:
        return num.rstrip('0')

Complexity

• Time: O(n) where n is the length of the string
• Space: O(n) for the resulting string
• Notes: Most concise solution, leverages built-in optimizations

Iterative from End

Intuition Iterate from the end of the string and build the result by counting trailing zeros, then slice accordingly.

Steps

• Count the number of trailing zeros by iterating from the end
• Use the count to determine the end index for slicing
• Return the substring excluding trailing zeros

class Solution:
    def removeTrailingZeros(self, num: str) -> str:
        count = 0
        for i in range(len(num) - 1, -1, -1):
            if num[i] == '0':
                count += 1
            else:
                break
        return num[:len(num) - count]

Complexity

• Time: O(n) where n is the length of the string
• Space: O(n) for the resulting string
• Notes: Clear logic with explicit counting, easy to understand