Is Count Integers With Even Digit Sum easy or hard?

Count Integers With Even Digit Sum is classified as an Easy problem on LeetCode. The brute-force solution is straightforward, while the optimal solution requires recognizing a simple mathematical parity pattern.

Count Integers With Even Digit Sum Python/Java solution

Python and Java implementations usually follow either a brute-force loop that calculates digit sums or the optimized math formula. The optimized version computes the digit sum of num and returns num / 2 if it is even, otherwise (num - 1) / 2.

How to solve Count Integers With Even Digit Sum in O(n)?

Iterate from 1 to num and compute the digit sum of each number using repeated modulo and division operations. If the sum is even, increment a counter. This simulation-based method runs in O(n * d) time where d is the number of digits in each number.

What is the best approach for Count Integers With Even Digit Sum?

The mathematical pattern approach is the best solution. Instead of checking every number, compute the digit sum of num and use the parity rule: if the digit sum is even, the result is num / 2; otherwise it is (num - 1) / 2. This runs in O(d) time where d is the number of digits.

Is Count Integers With Even Digit Sum asked at Google/Amazon/Meta?

Problems involving digit manipulation and mathematical patterns appear frequently in coding interviews at companies like Google, Amazon, and Meta. While this exact problem is categorized as easy, the reasoning pattern shows up in more complex math and digit DP questions.

What data structure is used in Count Integers With Even Digit Sum?

No specialized data structure is required. The problem mainly uses arithmetic operations on integers and simple iteration. The optimized solution relies on mathematical reasoning rather than additional storage.

What is the time complexity of Count Integers With Even Digit Sum?

The brute-force method takes O(n * d) time because each number from 1 to n requires computing its digit sum. The optimized mathematical solution only calculates the digit sum of n once, giving O(d) time and O(1) space.

Description Hints 2 Solution Video 5:30

2180. Count Integers With Even Digit Sum

Easy71.1% acceptance1 company

Given a positive integer num, return the number of positive integers less than or equal to num whose digit sums are even.

The digit sum of a positive integer is the sum of all its digits.

Example 1:

Input: num = 4
Output: 2
Explanation:
The only integers less than or equal to 4 whose digit sums are even are 2 and 4.

Example 2:

Input: num = 30
Output: 14
Explanation:
The 14 integers less than or equal to 30 whose digit sums are even are
2, 4, 6, 8, 11, 13, 15, 17, 19, 20, 22, 24, 26, and 28.

Constraints:

1 <= num <= 1000

Asked at 1 company

Mindtree

Topics

Math Simulation

Hints open one at a time

Hint 1

Iterate through all integers from 1 to num.

Hint 2

For any integer, extract the individual digits to compute their sum and check if it is even.

5:30

Still stuck?Read 3 approaches, or watch a walkthrough.

Open solution