Minimum sum formed by digits
Detailed guide and Python implementation for the 'Minimum sum formed by digits' problem.
1. Concept Overview
The 'Minimum sum formed by digits' problem is a key challenge in the Greedy 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 Minimum sum formed by digits.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Minimum sum formed by 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 min_sum_digits(arr) that takes a list of single-digit integers arr (0-9) and returns the minimum sum of two numbers formed by using all the digits in the array.
- •2 <= len(arr) <= 30
- •0 <= arr[i] <= 9
Examples
min_sum_digits([6, 8, 4, 5, 2, 3])
604
The minimum sum is obtained by forming the numbers 246 and 358: 246 + 358 = 604.
min_sum_digits([5, 3, 0, 7, 4])
82
The minimum sum is obtained by forming the numbers 35 and 47 (or 047): 35 + 47 = 82.
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
import heapq
def min_sum_opt(arr):
heapq.heapify(arr); n1 = n2 = 0
while arr:
n1 = n1 * 10 + heapq.heappop(arr)
if arr: n2 = n2 * 10 + heapq.heappop(arr)
return n1 + n2Brute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def min_sum_brute(arr):
arr.sort(); n1 = n2 = 0
for i in range(len(arr)):
if i % 2 == 0: n1 = n1 * 10 + arr[i]
else: n2 = n2 * 10 + arr[i]
return n1 + n2Algorithm Pattern Checklist
When dealing with Greedy data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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