Is Transform Binary String Using Subsequence Sort easy or hard?

This problem is generally considered medium difficulty because the implementation is simple once the greedy observation is found. The challenging part is recognizing that repeated subsequence sorting can be reduced to counting inversion patterns.

Transform Binary String Using Subsequence Sort Python/Java solution

Python and Java implementations both follow the same greedy logic: scan the string, detect invalid binary ordering, and count the minimum operations needed. The runtime remains O(n) in both languages with constant auxiliary space.

How to solve Transform Binary String Using Subsequence Sort in O(n)?

Use a greedy observation that the final binary string must have all 0s before all 1s. Count inversions or invalid ordering patterns during a single pass and compute the minimum subsequence operations directly without rebuilding the string repeatedly.

What is the best approach for Transform Binary String Using Subsequence Sort?

The best approach is a greedy linear scan that tracks misplaced 1s and 0s instead of simulating every subsequence sort operation. This reduces the complexity to O(n) time and O(1) extra space while still handling all edge cases correctly.

Is Transform Binary String Using Subsequence Sort asked at Google/Amazon/Meta?

Binary string transformation and greedy subsequence problems appear frequently in coding interviews at companies like Amazon, Google, and Meta. Interviewers usually focus on whether you can derive the linear-time observation from the brute force simulation.

What data structure is used in Transform Binary String Using Subsequence Sort?

Most optimal solutions only require counters and two pointers. Some implementations also use a stack or array to store indices of misplaced characters, which makes the transformation logic easier to visualize.

What is the time complexity of Transform Binary String Using Subsequence Sort?

The optimal solution runs in O(n) time because the string is scanned only once or twice using counters or two pointers. Space complexity is typically O(1) unless an auxiliary stack or index list is used.

Description Hints 4 Video 21:33

3998. Transform Binary String Using Subsequence Sort

Medium37.9% acceptance

You are given a binary string s.

You are also given an array of strings strs, where each strs[i] has the same length as s and consists of characters '0', '1', and '?'. Each '?' can be replaced by either '0' or '1'.

You may perform the following operation any number of times (including zero):

Choose any subsequence sub of s.
Sort sub in non-decreasing order.
Replace the chosen subsequence in s with the sorted sub, keeping all other characters unchanged.

Return a boolean array ans, where ans[i] is true if it's possible to replace all '?' in strs[i] with '0' or '1' and transform s into the resulting string using the allowed operation above, otherwise return false.

Example 1:

Input: s = "101", strs = ["1?1","0?1","0?0"]

Output: [true,true,false]

Explanation:

`i`	`strs[i]`	Replacement	Result `strs[i]`	Operation(s)	Result
0	`"1?1"`	`? → 0`	`"101"`	Matches `s`.	`true`
1	`"0?1"`	`? → 1`	`"011"`	Select the subsequence at indices `[0..2]` of `s` → `"101"`. Sort `"101"` to get `"011" = strs[i]`.	`true`
2	`"0?0"`	`? → 0` or `1`	`"000"` or `"010"`	Not feasible.	`false`

Thus, ans = [true, true, false].

Example 2:

Input: s = "1100", strs = ["0011","11?1","1?1?"]

Output: [true,false,true]

Explanation:

`i`	`strs[i]`	Replacement	Result `strs[i]`	Operation(s)	Result
0	`"0011"`	-	`"0011"`	Select the subsequence at indices `[0..3]` of `s` → `"1100"`. Sort `"1100"` to get `"0011" = strs[i]`.	`true`
1	`"11?1"`	`? → 0`	`"1101"`	Not feasible.	`false`
2	`"1?1?"`	First `? → 0` Second `? → 0`	`"1010"`	Select the subsequence at indices `[1, 2]` of `s` → `"10"`. Sort `"10"` to get `"01"`, so `s = "1010"`.	`true`

Thus, ans = [true, false, true].

Example 3:

Input: s = "1010", strs = ["0011"]

Output: [true]

Explanation:

`i`	`strs[i]`	Replacement	Result `strs[i]`	Operation(s)	Result
0	`"0011"`	-	`"0011"`	Select the subsequence at indices `[0, 2, 3]` of `s` → `"110"`. Sort `"110"` to get `"011"`, so `s = "0011" = strs[i]`.	`true`

Thus, ans = [true].

Constraints:

1 <= n == s.length <= 2000
s[i] is either '0' or '1'.
1 <= strs.length <= 2000
strs[i].length == n
strs[i] is either '0', '1', or '?'.

Topics

Hints open one at a time

Hint 1

Sorting a binary subsequence can only move '0' characters to the left and '1' characters to the right.

Hint 2

Therefore, a binary string t is reachable from s if both strings contain the same number of '1' characters and, for every prefix, t contains no more '1' characters than the corresponding prefix of s.

Hint 3

For each string in strs, determine how many '?' characters must become '1'. To minimize the number of ones in every prefix, assign these ones to the rightmost possible '?' positions.

Hint 4

After this assignment, check the prefix condition against s.

LanguageLoading editor…

1class Solution:
2    def transformStr(self, s: str, strs: List[str]) -> List[bool]:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

"101"
["1?1","0?1","0?0"]

Input

"1100"
["0011","11?1","1?1?"]

Input

"1010"
["0011"]

Your code is saved on this device.