Check If All 1's Are at Least Length K Places Away

class Solution {
    public boolean checkDistance(int[] array, int distance) {
        int concatenated = 0;
        for (int element : array) {
            concatenated = (concatenated << 1) | element;
        }
        
        if (concatenated == 0 || distance == 0) {
            return true;
        }
        
        while ((concatenated & 1) == 0) {
            concatenated >>= 1;    
        }
        
        while (concatenated != 1) {
            concatenated >>= 1;
            
            int zeroCount = 0;
            while ((concatenated & 1) == 0) {
                concatenated >>= 1;
                zeroCount++;
            }
            
            if (zeroCount < distance) {
                return false;
            }
        }
        return true;
    }
}

This Java solution is designed to address the problem of determining if all the 1 values in a given binary array are located at least k places away from each other. Here's a breakdown of how the solution achieves this:

• Initialization of Variables:

  • An integer concatenated is initialized to combine all the elements of the input array into a single binary number. This transformation is achieved using a bitwise operation inside a loop.

• Edge Case Handling:

  • If the concatenated equals 0 (indicating there are no 1s in the array) or the distance distance is 0, the function returns true, as the condition is trivially satisfied.

• Main Loop to Check Distance:

  1. The function removes all trailing zeros from the binary representation using a bitwise right shift until a 1 is found.
  2. The concatenated value is then iteratively right-shifted. For each shift, a count of consecutive zeros (zeroCount) is maintained. This count helps track the number of places between consecutive 1s.
  3. During each zero-counting operation, if zeroCount is found to be less than the required distance, the function returns false, indicating the 1s are too close to each other.
  4. The loop continues until all bits in concatenated are processed.

• Conclusion:

  • If the function exits the loop without finding any inadequately spaced 1s, then it returns true, confirming all 1's are spaced at least distance places apart.

By leveraging bitwise operations, this method efficiently manipulates and checks the spacings in the array dynamically, ensuring optimal performance and logical clarity.

class Solution:
    def sufficientDistance(self, binary_nums: List[int], gap: int) -> bool:
        # Transform the list of binary numbers into a single integer value
        combined_num = 0
        for bit in binary_nums:
            combined_num = (combined_num << 1) | bit

        # handle cases where no separation is required or no binary numbers
        if combined_num == 0 or gap == 0:
            return True

        # Strip trailing zeroes from combined_num
        while combined_num & 1 == 0:
            combined_num >>= 1

        # Check the distances between consecutive '1' bits
        while combined_num != 1:
            # Shift right to skip a '1' bit
            combined_num >>= 1

            zeros_count = 0
            while combined_num & 1 == 0:
                combined_num >>= 1
                zeros_count += 1
            
            # Verify if the gap between 1s is sufficient
            if zeros_count < gap:
                return False

        return True

This Python solution checks if all '1's in an array of binary numbers are separated by at least a given minimum distance (gap). Here's how it functions:

• Start by converting the array of binary numbers into a single integer. This transformation involves iterating through the array and using bitwise operations to shift and set bits.

• If the resulting combined number is zero (i.e., no '1's present) or the gap requirement is zero, the function immediately returns True.

• Remove any trailing zeros from the right of the combined binary number to simplify further checks.

• Using a loop, travel through the number and count zeros between consecutive '1' bits. The count resets after encountering a '1' and starts again until the next '1'.

• For each group of zeros found between '1' bits, compare against the required gap. If any group is found to have fewer zeros than required, return False.

• If the function iterates through all the binary digits without finding gaps less than required, return True.

This logic ensures efficient checking by utilizing bitwise operations and shifts, making the approach both intuitive and optimal in terms of execution speed.