Armstrong number in a given range
Detailed guide and Python implementation for the 'Armstrong number in a given range' problem.
1. Concept Overview
The 'Armstrong number in a given range' problem is a key challenge in the Basics 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 Armstrong number in a given range.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Armstrong number in a given range 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 armstrongs_in_range(a, b) that takes two integers a and b (where a <= b) and returns a list of all Armstrong numbers in the range [a, b] inclusive, in ascending order. An Armstrong number is a number that equals the sum of its digits each raised to the power of the number of digits.
- •1 <= a <= b <= 10^5
Examples
a = 100, b = 500
[153, 370, 371, 407]
The Armstrong numbers between 100 and 500 are 153, 370, 371, and 407.
a = 1, b = 9
[1, 2, 3, 4, 5, 6, 7, 8, 9]
All single-digit numbers are Armstrong numbers since d^1 = d.
a = 200, b = 300
[]
There are no Armstrong numbers between 200 and 300.
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 armstrong_in_range_opt(l: int, r: int) -> list:
res = []
for num in range(l, r + 1):
if num < 0: continue
digits, temp = [], num
while temp > 0:
digits.append(temp % 10)
temp //= 10
p = len(digits)
if num == sum(d ** p for d in digits):
res.append(num)
return resBrute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def armstrong_in_range_brute(l: int, r: int) -> list:
res = []
for num in range(l, r + 1):
s = str(num)
p = len(s)
if num == sum(int(d) ** p for d in s):
res.append(num)
return resAlgorithm Pattern Checklist
When dealing with Basics data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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