How do you solve Count Integers With Even Digit Sum on LeetCode?

Count Integers With Even Digit Sum (LeetCode #2180) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The digit sums of consecutive integers alternate between even and odd.

What is the brute force solution for Count Integers With Even Digit Sum?

The brute force approach for Count Integers With Even Digit Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each integer from 1 to num, calculating its digit sum, and determining if that sum is even. This method is straightforward but inefficient for larger values of num. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Integers With Even Digit Sum?

Count Integers With Even Digit Sum uses the following concepts: Math, Simulation. The recommended patterns to study are: Math, Simulation.

What are the interview tips for Count Integers With Even Digit Sum?

Always start with a brute-force solution to understand the problem better. Look for patterns in the data that can help optimize your solution. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Count Integers With Even Digit Sum?

Not handling the digit extraction correctly, leading to incorrect sums. Overlooking the efficiency of the solution, leading to unnecessary complexity.

#2180

Count Integers With Even Digit Sum

Easy

MathSimulationMathSimulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal approach leverages the fact that the digit sums of consecutive integers alternate between even and odd. Therefore, we can calculate the count of even digit sums directly based on the parity of num.

⚙️

Algorithm

5 steps

1Step 1: Initialize a counter to zero.
2Step 2: Loop through each integer from 1 to num.
3Step 3: Calculate the digit sum for each integer.
4Step 4: If the digit sum is even, increment the counter.
5Step 5: Return the counter after the loop.

solution.py9 lines

1# Full working Python code
2
3def count_even_digit_sum(num):
4    count = 0
5    for i in range(1, num + 1):
6        if sum(int(d) for d in str(i)) % 2 == 0:
7            count += 1
8    return count
9

ℹ

Complexity note: The time complexity is O(n) because we only loop through the integers up to num once, and the space complexity is O(1) since we are using a fixed amount of space regardless of input size.

1The digit sums of consecutive integers alternate between even and odd.
2Counting can be done efficiently without recalculating for every integer.

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