995. Minimum Number of K Consecutive Bit Flips

[mah-shamim]

Topics: Array, Bit Manipulation, Queue, Sliding Window, Prefix Sum

You are given a binary array nums and an integer k.

A k-bit flip is choosing a subarray of length k from nums and simultaneously changing every 0 in the subarray to 1, and every 1 in the subarray to 0.

Return the minimum number of k-bit flips required so that there is no 0 in the array. If it is not possible, return -1.

A subarray is a contiguous part of an array.

Example 1:

• Input: nums = [0,1,0], k = 1
• Output: 2
• Explanation: Flip nums[0], then flip nums[2].

Example 2:

• Input: nums = [1,1,0], k = 2
• Output: -1
• Explanation: No matter how we flip subarrays of size 2, we cannot make the array become [1,1,1].

Example 3:

• Input: nums = [0,0,0,1,0,1,1,0], k = 3
• Output: 3
• Explanation:
  Flip nums[0],nums[1],nums[2]: nums becomes [1,1,1,1,0,1,1,0]
  Flip nums[4],nums[5],nums[6]: nums becomes [1,1,1,1,1,0,0,0]
  Flip nums[5],nums[6],nums[7]: nums becomes [1,1,1,1,1,1,1,1]

Constraints:

• 1 <= nums.length <= 105
• 1 <= k <= nums.length

[mah-shamim]

We are given a binary array nums and an integer k. Our goal is to transform all 0s in the array into 1s using a minimum number of k-bit flips. A k-bit flip involves choosing a contiguous subarray of length k, and flipping every bit in that subarray: changing 0 to 1 and 1 to 0. The task is to find the minimum number of k-bit flips required to make the entire array consist of 1s. If it's not possible, we should return -1.

Key Points:

• k-bit flip: A flip operation on a contiguous subarray of length k that reverses the bits (i.e., 0 becomes 1, and 1 becomes 0).
• The aim is to find the minimum number of flips that convert all 0s to 1s.
• If it's not possible to flip enough bits due to constraints (such as exceeding the array length), the function should return -1.

Approach:

1. Sliding Window Technique:

• We need to traverse the array and flip the bits when required.
• Instead of directly flipping bits, we use a flip tracking mechanism to efficiently manage flips over the array.
• We utilize a window of size k to track valid flips, maintaining a count of flips that affect the current position.

2. Flip Tracking:

• We use an array flipped to keep track of whether a position has been flipped.
• A variable validFlipsFromPastWindow is used to track the cumulative effect of flips in the current window.
• For each position in the array:
  • If the current bit is incorrect (i.e., it doesn't match the desired 1), we flip the window starting at that position.
  • We maintain the effect of flips by updating the flip count and adjusting the valid flips from past windows.
• If a flip goes beyond the length of the array, return -1.

Plan:

1. Initialize an array flipped to track which positions have been flipped.
2. Use a variable validFlipsFromPastWindow to track the number of flips that affect the current position.
3. Traverse the array and for each element:

• Determine whether a flip is required.
• If necessary, flip a subarray of size k and update the tracking variables.

4. At the end of the traversal, return the total number of flips made.
5. If a flip is not possible (due to bounds constraints), return -1.

Let's implement this solution in PHP: 995. Minimum Number of K Consecutive Bit Flips

<?php
/**
 * @param Integer[] $nums
 * @param Integer $k
 * @return Integer
 */
function minKBitFlips($nums, $k) {
    // Keeps track of flipped states
    $flipped = array_fill(0, count($nums), false);
    // Tracks valid flips within the past window
    $validFlipsFromPastWindow = 0;
    // Counts total flips needed
    $flipCount = 0;

    for ($i = 0; $i < count($nums); $i++) {
        if ($i >= $k) {
            // Decrease count of valid flips from the past window if needed
            if ($flipped[$i - $k]) {
                $validFlipsFromPastWindow--;
            }
        }

        // Check if current bit needs to be flipped
        if ($validFlipsFromPastWindow % 2 == $nums[$i]) {
            // If flipping the window extends beyond the array length,
            // return -1
            if ($i + $k > count($nums)) {
                return -1;
            }
            // Increment the count of valid flips and mark current as
            // flipped
            $validFlipsFromPastWindow++;
            $flipped[$i] = true;
            $flipCount++;
        }
    }

    return $flipCount;
}

// Example 1
$nums = [0,1,0];
$k = 1;
echo minKBitFlips($nums, $k);  // Output: 2

// Example 2
$nums = [1,1,0];
$k = 2;
echo minKBitFlips($nums, $k);  // Output: -1

// Example 3
$nums = [0,0,0,1,0,1,1,0];
$k = 3;
echo minKBitFlips($nums, $k);  // Output: 3
?>

Explanation:

• We traverse the array while keeping track of the cumulative effect of flips in a sliding window of size k.
• Each time we encounter a 0 (or a flipped 0), we decide whether a flip is required by checking if the flip count up to that point is even or odd.
• If a flip goes beyond the array length, we return -1 as it's impossible to flip.

Example Walkthrough:

Let's walk through Example 1:

Input: nums = [0, 1, 0], k = 1

• Start with flipped = [false, false, false] and validFlipsFromPastWindow = 0.
• For i = 0:
  • The bit is 0 and no flip has occurred yet, so we flip the subarray starting at index 0.
  • The array becomes [1, 1, 0], flipped = [true, false, false], and flipCount = 1.
  • Now, the bit at index 0 is correct.
• For i = 2:
  • The bit is 0 (after the first flip) and requires another flip.
  • The array becomes [1, 1, 1], flipped = [true, false, true], and flipCount = 2.
  • Now, the bit at index 2 is correct.
• The final result is flipCount = 2.

Time Complexity:

• Time Complexity: O(n) where n is the length of the nums array. We perform a single pass over the array and use constant-time operations for each element.
• Space Complexity: O(n) for the flipped array used to track the flip states.

Output for Example:

• For the input nums = [0,1,0] and k = 1, the output is 2.

This approach efficiently solves the problem by using a sliding window technique combined with flip tracking. By ensuring that we only flip the necessary bits and adjusting the flips dynamically, we can minimize the number of flips required. If the problem constraints are such that it's impossible to achieve the desired result, we handle that with an early exit. This method ensures optimal time and space complexity for large inputs.

[basharul-siddike]

Thanks to solution for the "Minimum Number of K Consecutive Bit Flips" problem

[mah-shamim]

Thanks