SAMSUNG 49-Inch Odyssey G9

Because earth is not flat

Intuition

To solve the problem of flipping k-bit subarrays in a binary array nums so that there are no 0s left, we need to determine the minimum number of flips required. If it's not possible to achieve this, we return -1.

The key intuition is to track the flip operations using a sliding window technique. By keeping track of the current state of flips, we can efficiently decide whether to flip the subarray starting at the current index or not.

Approach

1. Initialize Variables:

   • flag to keep track of the cumulative flip state.
   • res to count the number of flips performed.
   • n to store the length of the input array nums.

2. Iterate Through the Array:

   • For each element nums[i]:
     • Calculate the current state cur by XORing nums[i] with flag.
     • If cur is 0, it means the current bit is 0 and needs to be flipped.
       • If flipping a subarray starting at i would exceed the bounds of the array (i + k > n), return -1.
       • Otherwise, increment the flip counter res and toggle the flag to indicate a new flip operation.
     • After potentially flipping, set nums[i] to the flipped value (1 - cur).

3. Adjust Flip State:

   • After processing the current element, update the flag based on the element that falls out of the flip window (i.e., i + 1 - k).

4. Return the Result:

   • Return the total number of flips res.

Complexity

• Time Complexity: O(n), where n is the length of the array nums. We iterate through the array once, and each operation inside the loop is O(1).
• Space Complexity: O(1), since we use a constant amount of extra space regardless of the input size.

Code Explanation

Function Signature

var minKBitFlips = function (nums, k) {

• nums: An array of binary digits (0s and 1s).
• k: The length of the subarray to flip.

Variables

var flag = 0;
var res = 0;
var n = nums.length;

• flag: Keeps track of whether the current bit has been flipped an odd or even number of times.
• res: Counts the number of flips performed.
• n: The length of the input array nums.

Main Loop

for (var i = 0; i < n; i++) {

The loop iterates through each element of the array nums.

Determine Current Bit State

var cur = nums[i] ^ flag;

• cur represents the current state of the bit after accounting for the flips indicated by flag.
• XOR (^) operation is used to determine the current state: if flag is 1, it means the bit has been flipped an odd number of times, otherwise even.

Check If Flip is Needed

if (cur === 0) {
    if (i + k > n) return -1;
    flag ^= 1;
    res++;
}

• If cur is 0 (meaning the current bit is 0 and needs to be flipped):
  • Check if there are enough bits remaining to perform a k-bit flip (i + k > n). If not, return -1 (not possible to flip).
  • Toggle the flag to indicate a flip.
  • Increment the flip count res.

Mark the Flip

nums[i] = 1 - cur;

• Change the current bit to 1 to mark that this position has been considered for a flip.

Maintain Flip State

if (i + 1 - k >= 0) flag ^= nums[i + 1 - k];

• Once a window of k bits has passed, adjust the flag back by XORing it with the bit that is k positions behind the current position to maintain the correct flip state for the rest of the array.

Return Result

return res;

• Finally, return the total number of flips performed.

This function efficiently determines the minimum number of k-bit flips required to ensure there are no 0s in the array nums, or returns -1 if it's not possible.

Code

C++

class Solution {
public:
    int minKBitFlips(vector<int>& nums, int k) {
        int flag = 0;
        int res = 0;
        int n = nums.size();

        for (int i = 0; i < n; i++) {
            int cur = nums[i] ^ flag;
            if (cur == 0) {
                if (i + k > n) return -1;
                flag ^= 1;
                res++;
            }
            nums[i] = 1 - cur;
            if (i + 1 - k >= 0) flag ^= nums[i + 1 - k];
        }
        return res;
    }
};

Python

class Solution:
    def minKBitFlips(self, nums: List[int], k: int) -> int:
        flag = 0
        res = 0
        n = len(nums)

        for i in range(n):
            cur = nums[i] ^ flag
            if cur == 0:
                if i + k > n:
                    return -1
                flag ^= 1
                res += 1
            nums[i] = 1 - cur
            if i + 1 - k >= 0:
                flag ^= nums[i + 1 - k]
        
        return res

Java

class Solution {
    public int minKBitFlips(int[] nums, int k) {
        int flag = 0;
        int res = 0;
        int n = nums.length;

        for (int i = 0; i < n; i++) {
            int cur = nums[i] ^ flag;
            if (cur == 0) {
                if (i + k > n) return -1;
                flag ^= 1;
                res++;
            }
            nums[i] = 1 - cur;
            if (i + 1 - k >= 0) flag ^= nums[i + 1 - k];
        }
        return res;
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number}
 */
var minKBitFlips = function (nums, k) {
    var flag = 0;
    var res = 0;
    var n = nums.length;

    for (var i = 0; i < n; i++) {
        var cur = nums[i] ^ flag;
        if (cur === 0) {
            if (i + k > n) return -1;
            flag ^= 1;
            res++;
        }
        nums[i] = 1 - cur;
        if (i + 1 - k >= 0) flag ^= nums[i + 1 - k];
    }
    return res;
};