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.
\n
Example 1:
\n\nInput: s = "110", k = 1
\n\nOutput: 1
\n\nExplanation:
\n\n'0' in s.k = 1, we can flip it directly in one operation.Example 2:
\n\nInput: s = "0101", k = 3
\n\nOutput: 2
\n\nExplanation:
\n\nOne optimal set of operations choosing k = 3 indices in each operation is:
[0, 1, 3]. s changes from "0101" to "1000".[1, 2, 3]. s changes from "1000" to "1111".Thus, the minimum number of operations is 2.
\nExample 3:
\n\nInput: s = "101", k = 2
\n\nOutput: -1
\n\nExplanation:
\n\nSince 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.
\n
Constraints:
\n\n1 <= s.length <= 105s[i] is either '0' or '1'.1 <= k <= s.length给你一个二进制字符串 s 和一个整数 k。
在一次操作中，你必须选择 恰好 k 个 不同的 下标，并将每个 '0' 翻转 为 '1'，每个 '1' 翻转为 '0'。
返回使字符串中所有字符都等于 '1' 所需的 最少 操作次数。如果不可能，则返回 -1。
\n\n
示例 1:
\n\n输入： s = \"110\", k = 1
\n\n输出： 1
\n\n解释：
\n\ns 中有一个 '0'。k = 1，我们可以直接在一次操作中翻转它。示例 2:
\n\n输入： s = \"0101\", k = 3
\n\n输出： 2
\n\n解释：
\n\n每次操作选择 k = 3 个下标的一种最优操作方案是：
[0, 1, 3]。s 从 \"0101\" 变为 \"1000\"。[1, 2, 3]。s 从 \"1000\" 变为 \"1111\"。因此，最少操作次数为 2。
\n示例 3:
\n\n输入： s = \"101\", k = 2
\n\n输出： -1
\n\n解释：
\n\n由于 k = 2 且 s 中只有一个 '0'，因此不可能通过翻转恰好 k 个位来使所有字符变为 '1'。因此，答案是 -1。
\n\n
提示:
\n\n1 <= s.length <= 105s[i] 的值为 '0' 或 '1'。1 <= k <= s.lengthz = 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.",
"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.",
"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."
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