How do you solve Binary Prefix Divisible By 5 on LeetCode?

Binary Prefix Divisible By 5 (LeetCode #1018) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Divisibility by 5 can be checked using modulo operation.

What is the time complexity of Binary Prefix Divisible By 5?

The optimal solution for Binary Prefix Divisible By 5 runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only traverse the array once, and the space complexity is O(n) for the result array.

What is the brute force solution for Binary Prefix Divisible By 5?

The brute force approach for Binary Prefix Divisible By 5 is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves calculating the decimal value of the binary number formed by each prefix of the array and checking if it is divisible by 5. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Binary Prefix Divisible By 5?

Binary Prefix Divisible By 5 uses the following concepts: Array, Bit Manipulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Binary Prefix Divisible By 5?

Explain your thought process clearly. Discuss time and space complexity before coding. Consider edge cases and test your code with them.

What are common mistakes when solving Binary Prefix Divisible By 5?

Not using modulo to keep the number manageable. Recalculating the binary number from scratch instead of building it up.

#1018

Binary Prefix Divisible By 5

Easy

ArrayBit ManipulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

Instead of recalculating the decimal value for each prefix, we can build the number iteratively. We can keep a running total and use modulo 5 to check divisibility, which is more efficient.

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Algorithm

3 steps

1Step 1: Initialize an empty result array and a variable to hold the current number.
2Step 2: For each index i in the nums array, update the current number using the formula: current_number = (current_number * 2 + nums[i]) % 5.
3Step 3: Check if current_number is 0 (which means it's divisible by 5) and store the result in the result array.

solution.py7 lines

1# Full working Python code
2result = []
3current_number = 0
4for num in nums:
5    current_number = (current_number * 2 + num) % 5
6    result.append(current_number == 0)
7return result

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Complexity note: The time complexity is O(n) because we only traverse the array once, and the space complexity is O(n) for the result array.

1Divisibility by 5 can be checked using modulo operation.
2Building the number iteratively avoids redundant calculations.

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