How do you solve Count of Integers on LeetCode?

Count of Integers (LeetCode #2719) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * max_sum) time complexity and O(n * max_sum) space complexity. Key insight: Using dynamic programming can significantly reduce time complexity.

What is the brute force solution for Count of Integers?

The brute force approach for Count of Integers is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check each integer from num1 to num2 and calculate its digit sum. If the digit sum falls within the specified range, we count it as a good integer. This approach is straightforward but inefficient for large ranges. The optimal Optimal Solution approach improves this to O(n * max_sum).

What algorithm or data structure is used to solve Count of Integers?

Count of Integers uses the following concepts: Math, String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Digit DP.

What are the interview tips for Count of Integers?

Explain your thought process clearly while coding. Discuss edge cases, especially with large inputs. Practice dynamic programming problems to get comfortable with state transitions.

What are common mistakes when solving Count of Integers?

Not handling large numbers correctly due to overflow. Forgetting to apply modulo operation in final results.

#2719

Count of Integers

Hard

MathStringDynamic ProgrammingDynamic ProgrammingDigit DP

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n * max_sum)Space O(n * max_sum)

Instead of checking each integer, we can use a dynamic programming approach to count the number of integers with a digit sum within the specified range. This is more efficient as it avoids redundant calculations.

⚙️

Algorithm

5 steps

1Step 1: Define a function f(n, l, r) to count integers from 1 to n with digit sums between l and r.
2Step 2: Use dynamic programming to build a table that keeps track of the number of ways to form digit sums.
3Step 3: Calculate f(num2, min_sum, max_sum) and f(num1 - 1, min_sum, max_sum).
4Step 4: The result is f(num2, min_sum, max_sum) - f(num1 - 1, min_sum, max_sum).
5Step 5: Return the result modulo 10^9 + 7.

solution.py15 lines

1# Full working Python code
2
3def count_digit_sum(n, min_sum, max_sum):
4    dp = [[0] * (max_sum + 1) for _ in range(len(n) + 1)]
5    dp[0][0] = 1
6    for i in range(1, len(n) + 1):
7        for j in range(10):
8            for k in range(max_sum + 1):
9                if k + j <= max_sum:
10                    dp[i][k + j] = (dp[i][k + j] + dp[i - 1][k]) % (10**9 + 7)
11    return sum(dp[len(n)][min_sum:max_sum + 1]) % (10**9 + 7)
12
13
14def count_good_integers(num1, num2, min_sum, max_sum):
15    return (count_digit_sum(num2, min_sum, max_sum) - count_digit_sum(str(int(num1) - 1), min_sum, max_sum)) % (10**9 + 7)

ℹ

Complexity note: The time complexity is O(n * max_sum) because we are filling a DP table of size (length of n) x (max_sum). This is efficient compared to the brute force approach.

1Using dynamic programming can significantly reduce time complexity.
2Understanding digit sums and their properties is key to solving this problem.

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