How do you solve Sum of K-Digit Numbers in a Range on LeetCode?

Sum of K-Digit Numbers in a Range (LeetCode #3855) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(k) time complexity and O(1) space complexity. Key insight: Each digit position contributes independently to the total sum.

What is the time complexity of Sum of K-Digit Numbers in a Range?

The optimal solution for Sum of K-Digit Numbers in a Range runs in O(k) time and uses O(1) space. This approach computes contributions directly, leading to linear complexity with respect to k, making it efficient.

What is the brute force solution for Sum of K-Digit Numbers in a Range?

The brute force approach for Sum of K-Digit Numbers in a Range is Brute Force, with O(n²) time complexity and O(1) space complexity. Generate all possible k-digit numbers using digits from the range [l, r] and sum them up directly. This approach is straightforward but inefficient for larger ranges. The optimal Optimal Solution approach improves this to O(k).

What algorithm or data structure is used to solve Sum of K-Digit Numbers in a Range?

Sum of K-Digit Numbers in a Range uses the following concepts: Math, Divide and Conquer, Combinatorics, Number Theory. The recommended patterns to study are: Combinatorics, Dynamic Programming.

What are the interview tips for Sum of K-Digit Numbers in a Range?

Explain your thought process clearly. Discuss edge cases like single-digit ranges. Practice combinatorial problems to build intuition.

What are common mistakes when solving Sum of K-Digit Numbers in a Range?

Forgetting to apply modulo operations. Incorrectly calculating the sum of the range.

#3855

Sum of K-Digit Numbers in a Range

Hard

MathDivide and ConquerCombinatoricsNumber TheoryCombinatoricsDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(k)Space O(1)

Use combinatorial mathematics to calculate the contribution of each digit position directly, avoiding the need to generate all combinations.

⚙️

Algorithm

3 steps

1Step 1: Calculate the sum of digits from l to r: sum_digits = (r - l + 1) * (l + r) / 2.
2Step 2: For each digit position, compute its contribution as: contribution = sum_digits * (10^p) * (r - l + 1)^(k-1).
3Step 3: Sum contributions for all k positions and return the result modulo 10^9 + 7.

solution.py7 lines

1def sum_k_digit_numbers_optimal(l, r, k):
2    MOD = 10**9 + 7
3    sum_digits = (r - l + 1) * (l + r) // 2 % MOD
4    total_sum = 0
5    for p in range(k):
6        total_sum = (total_sum + sum_digits * pow(10, p, MOD) * pow(r - l + 1, k - 1, MOD)) % MOD
7    return total_sum

ℹ

Complexity note: This approach computes contributions directly, leading to linear complexity with respect to k, making it efficient.

1Each digit position contributes independently to the total sum.
2Using combinatorial counting avoids generating all combinations.

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