Introduction

In this blog post, we will explore how to solve the LeetCode problem "402" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given string num representing a non-negative integer num, and an integer k, return the smallest possible integer after removing k digits from num. Example 1: Input: num = "1432219", k = 3 Output: "1219" Explanation: Remove the three digits 4, 3, and 2 to form the new number 1219 which is the smallest. Example 2: Input: num = "10200", k = 1 Output: "200" Explanation: Remove the leading 1 and the number is 200. Note that the output must not contain leading zeroes. Example 3: Input: num = "10", k = 2 Output: "0" Explanation: Remove all the digits from the number and it is left with nothing which is 0. Constraints: 1 <= k <= num.length <= 105 num consists of only digits. num does not have any leading zeros except for the zero itself.

Explanation

Here's the solution to the problem, following the requested format:

• High-level approach:

  • Use a stack to maintain a monotonically increasing sequence of digits.
  • Iterate through the digits of the input number. If the current digit is smaller than the top of the stack and we still have k digits to remove, pop the top of the stack and decrement k.
  • After iterating through all digits, remove the remaining k digits from the end of the stack.

• Complexity:

  • Runtime: O(n), where n is the length of the input string num.
  • Storage: O(n), for the stack.

• Python Code:

Code

    def remove_kdigits(num: str, k: int) -> str:
    """
    Given string num representing a non-negative integer num, and an integer k,
    return the smallest possible integer after removing k digits from num.
    """
    stack = []
    for digit in num:
        while stack and k > 0 and stack[-1] > digit:
            stack.pop()
            k -= 1
        stack.append(digit)

    # Remove remaining digits from the end
    while k > 0:
        stack.pop()
        k -= 1

    # Remove leading zeros
    result = "".join(stack).lstrip("0")

    return result if result else "0"