What is the brute force approach and complexity for Remove K Digits?

k times: scan for the first digit bigger than the next one and erase it with time complexity O(N · K) and space complexity O(N).

What is the optimal approach and complexity for Remove K Digits?

Pop a bigger top while a smaller digit arrives (and k > 0); trim the tail; strip zeros with optimal time complexity O(N) and space complexity O(N).

How do you identify the pattern for Remove K Digits?

Look for key signals in the prompt: "remove k digits" indicates We drop items, and a new item can make earlier ones worth dropping → pop from a stack.; "smallest number" indicates Smaller digits should come first → keep the kept digits increasing: a bigger digit before a smaller one is popped.; "original order" indicates We keep a subsequence, so a left-to-right scan with a stack fits..

Remove K Digits

Medium·LeetCode #402 Pro OnlyPattern Transfer

LeetCode

Problem Statement

You are given a non-negative number as a string num, and an integer k.

Remove exactly k digits from num. The other digits stay in their original order. Return the smallest number you can get, as a string.

The result should not have leading zeros. If nothing is left, or only zeros are left, return "0".

Input

A digit string num and an integer k.

Output

The smallest possible number after removing k digits, as a string without leading zeros.

Example 1

Input: num = "1432219", k = 3

Output: "1219"

Remove 4, 3 and the first 2. What is left is 1219.

Example 2

Input: num = "10200", k = 1

Output: "200"

Remove the 1 to get "0200". Leading zeros are dropped, so the answer is "200".

Example 3

Input: num = "10", k = 2

Output: "0"

Both digits are removed. Nothing is left, so the answer is "0".

Constraints

• 1 ≤ k ≤ num.length ≤ 10^5
• `num` has only digits
• `num` has no leading zeros, except when the number is just "0"

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“remove k digits”

→ We drop items, and a new item can make earlier ones worth dropping → pop from a stack.

“smallest number”

→ Smaller digits should come first → keep the kept digits increasing: a bigger digit before a smaller one is popped.

“original order”

→ We keep a subsequence, so a left-to-right scan with a stack fits.

Approach Complexity Ladder

Brute Force — remove one digit at a timeO(N · K)O(N)

k times: scan for the first digit bigger than the next one and erase it

Optimal — increasing stack of kept digitsO(N)O(N)

Pop a bigger top while a smaller digit arrives (and k > 0); trim the tail; strip zeros

Brute Force Intuition

Make the k removals one by one. Each time, scan from the left for the first digit that is bigger than the digit right after it, and erase it. If the digits never go down, erase the last digit. At the end, strip leading zeros.

Recognition rule for this sub-pattern:

Need the first smaller value. Keep an increasing stack; greedy "remove bigger digits before me" problems also live here.

Remember this · Monotonic Stack

Trigger

Next/previous greater or smaller element, spans, histogram areas, "for each element, how far does it dominate?"

Think

When a new element arrives, which older elements can never be the answer again?

Key question

Is my stack increasing or decreasing, and what do I record when I pop?

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