LeetCode 1375. Number of Times Binary String Is Prefix-Aligned Solution in Java, C++, Python & More | Explanation + Code

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 flips[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).

PythonJavaC++GoTypeScript

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(code-box)

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; } }(code-box)

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; } };(code-box)

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

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; }(code-box)