You have a 1-indexed binary string of length n where all the bits are 0 initially. We will flip all the bits of this binary string (i.e., change them from 0 to 1) one by one. You are given a 1-indexed integer array flips where flips[i] indicates that the bit at index i will be flipped in the ith step.
A binary string is prefix-aligned if, after the ith step, all the bits in the inclusive range [1, i] are ones and all the other bits are zeros.
Return the number of times the binary string is prefix-aligned during the flipping process.
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Example 1:
\n\n\nInput: flips = [3,2,4,1,5]\nOutput: 2\nExplanation: The binary string is initially "00000".\nAfter applying step 1: The string becomes "00100", which is not prefix-aligned.\nAfter applying step 2: The string becomes "01100", which is not prefix-aligned.\nAfter applying step 3: The string becomes "01110", which is not prefix-aligned.\nAfter applying step 4: The string becomes "11110", which is prefix-aligned.\nAfter applying step 5: The string becomes "11111", which is prefix-aligned.\nWe can see that the string was prefix-aligned 2 times, so we return 2.\n\n\n
Example 2:
\n\n\nInput: flips = [4,1,2,3]\nOutput: 1\nExplanation: The binary string is initially "0000".\nAfter applying step 1: The string becomes "0001", which is not prefix-aligned.\nAfter applying step 2: The string becomes "1001", which is not prefix-aligned.\nAfter applying step 3: The string becomes "1101", which is not prefix-aligned.\nAfter applying step 4: The string becomes "1111", which is prefix-aligned.\nWe can see that the string was prefix-aligned 1 time, so we return 1.\n\n\n
\n
Constraints:
\n\nn == flips.length1 <= n <= 5 * 104flips is a permutation of the integers in the range [1, n].给你一个长度为 n 、下标从 1 开始的二进制字符串，所有位最开始都是 0 。我们会按步翻转该二进制字符串的所有位（即，将 0 变为 1）。
给你一个下标从 1 开始的整数数组 flips ，其中 flips[i] 表示对应下标 i 的位将会在第 i 步翻转。
二进制字符串 前缀一致 需满足：在第 i 步之后，在 闭 区间 [1, i] 内的所有位都是 1 ，而其他位都是 0 。
返回二进制字符串在翻转过程中 前缀一致 的次数。
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示例 1：
\n\n\n输入：flips = [3,2,4,1,5]\n输出：2\n解释：二进制字符串最开始是 \"00000\" 。\n执行第 1 步：字符串变为 \"00100\" ，不属于前缀一致的情况。\n执行第 2 步：字符串变为 \"01100\" ，不属于前缀一致的情况。\n执行第 3 步：字符串变为 \"01110\" ，不属于前缀一致的情况。\n执行第 4 步：字符串变为 \"11110\" ，属于前缀一致的情况。\n执行第 5 步：字符串变为 \"11111\" ，属于前缀一致的情况。\n在翻转过程中，前缀一致的次数为 2 ，所以返回 2 。\n\n\n
示例 2：
\n\n\n输入：flips = [4,1,2,3]\n输出：1\n解释：二进制字符串最开始是 \"0000\" 。\n执行第 1 步：字符串变为 \"0001\" ，不属于前缀一致的情况。\n执行第 2 步：字符串变为 \"1001\" ，不属于前缀一致的情况。\n执行第 3 步：字符串变为 \"1101\" ，不属于前缀一致的情况。\n执行第 4 步：字符串变为 \"1111\" ，属于前缀一致的情况。\n在翻转过程中，前缀一致的次数为 1 ，所以返回 1 。\n\n
\n\n
提示：
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collections<\\/a>\\u3002\\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u6ce8\\u610f Python 2.7 \\u5c06\\u57282020\\u5e74\\u540e\\u4e0d\\u518d\\u7ef4\\u62a4<\\/a>\\u3002 \\u5982\\u60f3\\u4f7f\\u7528\\u6700\\u65b0\\u7248\\u7684Python\\uff0c\\u8bf7\\u9009\\u62e9Python 3\\u3002<\\/p>\"],\"c\":[\"C\",\" \\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5982\\u60f3\\u4f7f\\u7528\\u54c8\\u5e0c\\u8868\\u8fd0\\u7b97, \\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 uthash<\\/a>\\u3002 \\\"uthash.h\\\"\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5bfc\\u5165\\u3002\\u8bf7\\u770b\\u5982\\u4e0b\\u793a\\u4f8b:<\\/p>\\r\\n\\r\\n 1. 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