Is Minimum Operations to Equalize Binary String easy or hard?

Minimum Operations to Equalize Binary String is categorized as a Hard problem. It combines string processing with graph traversal and advanced data structures such as ordered sets or union find. The difficulty comes from modeling the operations efficiently and avoiding exponential state exploration.

Minimum Operations to Equalize Binary String Python/Java solution

Python implementations commonly use collections.deque for BFS and sorted containers or tree-like structures to simulate an ordered set. Java solutions typically rely on TreeSet for ordered index management along with a queue for BFS traversal. Both achieve O(n log n) time complexity.

How to solve Minimum Operations to Equalize Binary String in O(n)?

Pure O(n) solutions are difficult because efficient index lookup and removal are required during state expansion. Most optimal implementations rely on ordered sets or balanced trees, leading to O(n log n) complexity. BFS ensures minimal operations while the ordered structure maintains the next unresolved indices efficiently.

What is the best approach for Minimum Operations to Equalize Binary String?

The most efficient approach uses Breadth-First Search combined with an ordered set to track unresolved indices. BFS guarantees the minimum number of operations, while the ordered set allows fast insertion, deletion, and neighbor lookup. This combination processes each index once and results in O(n log n) time complexity with O(n) space.

Is Minimum Operations to Equalize Binary String asked at Google/Amazon/Meta?

Hard string and graph-style transformation problems similar to this appear frequently in Google, Amazon, and Meta interview rounds. They test your ability to model a problem as a graph, apply BFS for shortest operations, and optimize index tracking with advanced data structures.

What data structure is used in Minimum Operations to Equalize Binary String?

The optimized solution typically uses a BFS queue plus an ordered set or balanced tree structure. The queue handles level-by-level exploration of operations, while the ordered set stores unresolved indices and supports fast range queries and deletions in O(log n).

What is the time complexity of Minimum Operations to Equalize Binary String?

The optimal solution runs in O(n log n) time using BFS with an ordered set. Each index is inserted and removed from the ordered structure at most once, and each such operation costs O(log n). The overall space complexity is O(n) for the queue and index tracking structures.

Description Hints 3 Solution Video 1:13:04

3666. Minimum Operations to Equalize Binary String

Hard45.2% acceptance4 companies

You are given a binary string s, and an integer k.

In one operation, you must choose exactly k different indices and flip each '0' to '1' and each '1' to '0'.

Return the minimum number of operations required to make all characters in the string equal to '1'. If it is not possible, return -1.

Example 1:

Input: s = "110", k = 1

Output: 1

Explanation:

There is one '0' in s.
Since k = 1, we can flip it directly in one operation.

Example 2:

Input: s = "0101", k = 3

Output: 2

Explanation:

One optimal set of operations choosing k = 3 indices in each operation is:

Operation 1: Flip indices [0, 1, 3]. s changes from "0101" to "1000".
Operation 2: Flip indices [1, 2, 3]. s changes from "1000" to "1111".

Thus, the minimum number of operations is 2.

Example 3:

Input: s = "101", k = 2

Output: -1

Explanation:

Since k = 2 and s has only one '0', it is impossible to flip exactly k indices to make all '1'. Hence, the answer is -1.

Constraints:

1 <= s.length <= 10⁵
s[i] is either '0' or '1'.
1 <= k <= s.length

Asked at 4 companies

Infosys

Google Amazon Microsoft

Topics

Hints open one at a time

Hint 1

Model state as z = number of zeros; flipping k picks i zeros (i between max(0, k - (n - z)) and min(k, z)) and transforms z to z' = z + k - 2 * i, so z' lies in a contiguous range and has parity (z + k) % 2.

Hint 2

Build a graph on states 0..n and run BFS from initial z to reach 0; each edge from z goes to all z' in that computed interval.

Hint 3

For speed, keep two ordered sets of unvisited states by parity and erase ranges with lower_bound while BFSing to achieve near O(n log n) time.

1:13:04

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1class Solution:
2    def minOperations(self, s: str, k: int) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

"110"
1

Input

"0101"
3

Input

"101"
2

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