How do you solve Count Numbers with Non-Decreasing Digits on LeetCode?

Count Numbers with Non-Decreasing Digits (LeetCode #3519) 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: Non-decreasing sequences can be counted using combinatorial methods.

What is the time complexity of Count Numbers with Non-Decreasing Digits ?

The optimal solution for Count Numbers with Non-Decreasing Digits runs in O(n) time and uses O(n) space. The DP approach efficiently counts valid numbers without generating them, leading to linear complexity relative to the number of digits.

What is the brute force solution for Count Numbers with Non-Decreasing Digits ?

The brute force approach for Count Numbers with Non-Decreasing Digits is Brute Force, with O(n²) time complexity and O(1) space complexity. Check each number in the range [l, r] and convert it to base b. Then verify if its digits are non-decreasing. This method is straightforward but inefficient for large ranges. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Numbers with Non-Decreasing Digits ?

Count Numbers with Non-Decreasing Digits uses the following concepts: Math, String, Dynamic Programming. The recommended patterns to study are: Digit DP, Combinatorial Counting.

What are the interview tips for Count Numbers with Non-Decreasing Digits ?

Explain your thought process clearly. Discuss edge cases and constraints. Practice similar problems to build confidence.

What are common mistakes when solving Count Numbers with Non-Decreasing Digits ?

Forgetting to handle leading zeros in base conversions. Not considering the tight condition in DP.

#3519

Count Numbers with Non-Decreasing Digits

Hard

MathStringDynamic ProgrammingDigit DPCombinatorial Counting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Use digit dynamic programming to count valid numbers directly without generating them. This approach leverages the properties of non-decreasing sequences.

⚙️

Algorithm

3 steps

1Step 1: Define a DP function that counts non-decreasing numbers up to a given limit in base b.
2Step 2: Use the DP function to calculate counts for r and l-1, then subtract to get the count in the range [l, r].
3Step 3: Handle the base conversion and non-decreasing condition within the DP logic.

solution.py15 lines

1def countNonDecreasing(l, r, b):
2    def dp(pos, last_digit, tight, digits):
3        if pos == len(digits): return 1
4        if (pos, last_digit, tight) in memo: return memo[(pos, last_digit, tight)]
5        limit = digits[pos] if tight else b - 1
6        count = 0
7        for digit in range(last_digit, limit + 1):
8            count += dp(pos + 1, digit, tight and (digit == limit), digits)
9        memo[(pos, last_digit, tight)] = count
10        return count
11    def count_up_to(x):
12        digits = list(map(int, str(int(x, b))))
13        return dp(0, 0, True, digits)
14    memo = {}
15    return (count_up_to(r) - count_up_to(l) + 10**9 + 7) % (10**9 + 7)

ℹ

Complexity note: The DP approach efficiently counts valid numbers without generating them, leading to linear complexity relative to the number of digits.

1Non-decreasing sequences can be counted using combinatorial methods.
2Dynamic programming can optimize counting by avoiding redundant calculations.

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