Remove Trailing Zeros From a String

Remove Trailing Zeros From a String - Problem

String Easy

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

Trailing zeros are the zeros at the end of a number. For example, the number "12300" has two trailing zeros.

Input & Output

Example 1 — Basic Case

$ Input: num = "51230"

› Output: "5123"

💡 Note: The string has one trailing zero. Remove it to get "5123".

Example 2 — Multiple Trailing Zeros

$ Input: num = "40300"

› Output: "403"

💡 Note: The string has two trailing zeros. Remove both to get "403".

Example 3 — No Trailing Zeros

$ Input: num = "123"

› Output: "123"

💡 Note: No trailing zeros to remove, return the original string.

Constraints

1 ≤ num.length ≤ 1000
num consists of only digits
num does not contain leading zeros

Visualization

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Asked in

a Amazon 25 M Microsoft 18

The key insight is to work backwards from the end of the string to find the last non-zero character. Best approach uses a single pass to find the cutoff position and substring. Time: O(n), Space: O(1)

Common Approaches

✓ Character by Character Removal

⏱️ Time: O(k) Space: O(1)

Start from the end of the string and keep removing characters as long as they are '0'. Stop when we encounter a non-zero character or reach the beginning of the string.

Find Last Non-Zero Position

⏱️ Time: O(n) Space: O(1)

Iterate from the end of the string backwards to find the last non-zero character. Once found, return the substring from the beginning up to and including that position. This avoids multiple string operations.

Character by Character Removal — Algorithm Steps

Start with the original string
Check if last character is '0'
If yes, remove it and repeat
Continue until last character is not '0'

Visualization

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Step-by-Step Walkthrough

Start

Original string with trailing zeros

Remove

Remove zeros one by one from the end

Result

Stop when non-zero character reached

Code -

solution.c — C

#include <stdio.h>
#include <string.h>

void solution(char* num) {
    int len = strlen(num);
    while (len > 1 && num[len - 1] == '0') {
        num[len - 1] = '\0';
        len--;
    }
}

int main() {
    char num[1000];
    fgets(num, sizeof(num), stdin);
    num[strcspn(num, "\n")] = 0;
    solution(num);
    printf("%s\n", num);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(k)

Where k is the number of trailing zeros to remove

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables, no extra data structures

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Character by Character Removal — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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