Number of Times Binary String Is Prefix-Aligned

Problem

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**.

Example 1:

Input: flips = [3,2,4,1,5]
Output: 2
Explanation: The binary string is initially "00000".
After applying step 1: The string becomes "00100", which is not prefix-aligned.
After applying step 2: The string becomes "01100", which is not prefix-aligned.
After applying step 3: The string becomes "01110", which is not prefix-aligned.
After applying step 4: The string becomes "11110", which is prefix-aligned.
After applying step 5: The string becomes "11111", which is prefix-aligned.
We can see that the string was prefix-aligned 2 times, so we return 2.

Example 2:

Input: flips = [4,1,2,3]
Output: 1
Explanation: The binary string is initially "0000".
After applying step 1: The string becomes "0001", which is not prefix-aligned.
After applying step 2: The string becomes "1001", which is not prefix-aligned.
After applying step 3: The string becomes "1101", which is not prefix-aligned.
After applying step 4: The string becomes "1111", which is prefix-aligned.
We can see that the string was prefix-aligned 1 time, so we return 1.

Constraints:

n == flips.length
1 <= n <= 5 * 10^4
flips is a permutation of the integers in the range [1, n].

Solution (Java)

class Solution {
    public int numTimesAllBlue(int[] flips) {
        int ans = 0;
        int max = 0;
        for (int i = 0; i < flips.length; i++) {
            max = Math.max(max, flips[i]);
            if (max == i + 1) {
                ++ans;
            }
        }
        return ans;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).