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1375. Number of Times Binary String Is Prefix-Aligned

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Description

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 * 104
• flips is a permutation of the integers in the range [1, n].

Solutions

Solution 1: Direct Traversal

We can traverse the array $flips$, keeping track of the maximum value $mx$ of the elements we have traversed so far. If $mx$ equals the current index $i$ we are traversing, it means that the first $i$ elements have all been flipped, i.e., the prefix is consistent, and we increment the answer.

After the traversal is finished, we return the answer.

The time complexity is $O(n)$, where $n$ is the length of the array $flips$. The space complexity is $O(1)$.

class Solution:
    def numTimesAllBlue(self, flips: List[int]) -> int:
        ans = mx = 0
        for i, x in enumerate(flips, 1):
            mx = max(mx, x)
            ans += mx == i
        return ans

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

class Solution {
public:
    int numTimesAllBlue(vector<int>& flips) {
        int ans = 0, mx = 0;
        for (int i = 1; i <= flips.size(); ++i) {
            mx = max(mx, flips[i - 1]);
            ans += mx == i;
        }
        return ans;
    }
};

func numTimesAllBlue(flips []int) (ans int) {
	mx := 0
	for i, x := range flips {
		mx = max(mx, x)
		if mx == i+1 {
			ans++
		}
	}
	return
}

function numTimesAllBlue(flips: number[]): number {
    let ans = 0;
    let mx = 0;
    for (let i = 1; i <= flips.length; ++i) {
        mx = Math.max(mx, flips[i - 1]);
        if (mx === i) {
            ++ans;
        }
    }
    return ans;
}