How do you solve Minimum Number of K Consecutive Bit Flips on LeetCode?

Minimum Number of K Consecutive Bit Flips (LeetCode #995) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(k) space complexity. Key insight: Flipping bits affects subsequent bits, so we need to track the effect of flips.

What is the time complexity of Minimum Number of K Consecutive Bit Flips?

The optimal solution for Minimum Number of K Consecutive Bit Flips runs in O(n) time and uses O(k) space. The time complexity is O(n) because we only pass through the array once, and the space complexity is O(k) due to the queue used to track the flips.

What is the brute force solution for Minimum Number of K Consecutive Bit Flips?

The brute force approach for Minimum Number of K Consecutive Bit Flips is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible subarray of length k and flipping the bits. This method is straightforward but inefficient, as it requires checking multiple overlapping subarrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Number of K Consecutive Bit Flips?

Minimum Number of K Consecutive Bit Flips uses the following concepts: Array, Bit Manipulation, Queue, Sliding Window, Prefix Sum. The recommended patterns to study are: Sliding Window, Greedy.

What are the interview tips for Minimum Number of K Consecutive Bit Flips?

Always clarify the problem constraints and edge cases before diving into coding. Explain your thought process clearly while coding; this helps interviewers understand your approach. Practice writing clean and efficient code, as readability is often as important as functionality.

What are common mistakes when solving Minimum Number of K Consecutive Bit Flips?

Failing to check if the flip goes out of bounds of the array. Not considering the cumulative effect of flips when checking the current bit.

#995

Minimum Number of K Consecutive Bit Flips

Hard

ArrayBit ManipulationQueueSliding WindowPrefix SumSliding WindowGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(k)

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Intuition

Time O(n)Space O(k)

The optimal approach uses a greedy strategy with a sliding window technique to track the flips efficiently. Instead of flipping bits directly, we keep track of the number of flips that affect each position, allowing us to determine when to flip without modifying the array repeatedly.

⚙️

Algorithm

4 steps

1Step 1: Initialize a flip count and a variable to track the current effect of flips.
2Step 2: Iterate through the array, and for each bit, check if it's affected by previous flips.
3Step 3: If the current bit is 0 and it's not flipped, perform a flip and update the count.
4Step 4: Use a queue to manage the flips' effects over the last k elements.

solution.py16 lines

1def minKBitFlips(nums, k):
2    flips = 0
3    n = len(nums)
4    flip_effect = 0
5    flip_queue = []
6    for i in range(n):
7        if flip_queue and flip_queue[0] <= i:
8            flip_effect -= 1
9            flip_queue.pop(0)
10        if (nums[i] + flip_effect) % 2 == 0:
11            if i + k > n:
12                return -1
13            flips += 1
14            flip_effect += 1
15            flip_queue.append(i + k)
16    return flips

ℹ

Complexity note: The time complexity is O(n) because we only pass through the array once, and the space complexity is O(k) due to the queue used to track the flips.

1Flipping bits affects subsequent bits, so we need to track the effect of flips.
2Using a queue allows us to efficiently manage the range of flips without modifying the original array.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.