Non decreasing numbers with n digits
Detailed guide and Python implementation for the 'Non decreasing numbers with n digits' problem.
1. Concept Overview
The 'Non decreasing numbers with n digits' problem is a key challenge in the Dynamic Programming section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Non decreasing numbers with n digits.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Non decreasing numbers with n digits carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
Write a function count_non_decreasing(n) that returns the count of non-decreasing numbers with n digits. A number is non-decreasing if every digit is greater than or equal to the digit to its left. Leading zeros are allowed (e.g. 012 is non-decreasing).
- •1 <= n <= 20
Examples
count_non_decreasing(1)
10
All single digit numbers (0 to 9) are non-decreasing.
count_non_decreasing(2)
55
There are 55 non-decreasing numbers of 2 digits (like 00, 01, ..., 11, 12, ..., 99).
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
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Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
def count_non_decreasing_opt(n):
dp = [[0] * 10 for _ in range(n + 1)]
for i in range(10): dp[1][i] = 1
for i in range(2, n + 1):
for j in range(10):
for k in range(j + 1): dp[i][j] += dp[i - 1][k]
return sum(dp[n])Brute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def count_non_decreasing_brute(n):
def solve(n, last):
if n == 0: return 1
res = 0
for i in range(last, 10): res += solve(n - 1, i)
return res
return solve(n, 0)Algorithm Pattern Checklist
When dealing with Dynamic Programming data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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