0402.Remove K Digits

Link: 402. Remove K Digits - Medium
Track: NeetCode150

Question

Restate the problem

• Given string num that represents a non-negative integer (like "1432219") and an integer k
• Return the answer as a string, and it must not have leading zeros
• remove exactly k digits, so the resulting number is as small as possible.

Whenever the next digit is smaller than the previous digit, we should remove the previous larger digit (if we still have k removals) to make the number as small as possible.

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Method 1
Method 2

Method - Stack

Approach - monotonic increasing stack

• Use stack to keep digits:
  • Scan digits left → right.
  • While k > 0 and stack[-1] > ch, pop the stack (remove that larger digit), decrement k.
  • Push ch.
• If k is still > 0 after the scan, remove from the end (stack[:-k]) because the number is already increasing, so the tail digits are the best to delete.
• Result: res = ''.join(stack).lstrip('0'); if empty return "0"

Complexity

• Time Complexity: O(n)
• Space Complexity: O(n)

Edge Case

• k == len(num), return “0”

Code

class Solution:
    def removeKdigits(self, num: str, k: int) -> str:
        stack = []
        for char in num:
            while stack and stack[-1] > char and k != 0:
                k -= 1
                stack.pop()
            
            stack.append(char)
        
        if k > 0:
            stack = stack[:-k]
 
        tmp = ("").join(stack)
        res = tmp.lstrip('0')
        
        return str(res) if res else "0"   

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History

Jan-25-2026 Solved w bug

• stack = stack[:k] to slice

Jan-24-2026 Solved w bug

• lstrip only works on str

Jan-22-2026 Solved w bug, took too long time

• digital string can compare
• use while to deal with stack
• str.join() works with any iterable of strings
• lstrip() only work with strings, to eliminate '0'