Leetcode #2605: Form Smallest Number From Two Digit Arrays

In this guide, we solve Leetcode #2605 Form Smallest Number From Two Digit Arrays in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

Given two arrays of unique digits nums1 and nums2, return the smallest number that contains at least one digit from each array. Example 1: Input: nums1 = [4,1,3], nums2 = [5,7] Output: 15 Explanation: The number 15 contains the digit 1 from nums1 and the digit 5 from nums2.

Quick Facts

Difficulty: Easy
Premium: No
Tags: Array, Hash Table, Enumeration

Intuition

Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.

By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.

Approach

Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.

This keeps the solution linear while remaining easy to explain in an interview setting.

Steps:

Initialize a hash map for seen items or counts.
Iterate through the input, querying/updating the map.
Return the first valid result or the final computed value.

Example

Input: nums1 = [4,1,3], nums2 = [5,7]
Output: 15
Explanation: The number 15 contains the digit 1 from nums1 and the digit 5 from nums2. It can be proven that 15 is the smallest number we can have.

Python Solution

class Solution:
    def minNumber(self, nums1: List[int], nums2: List[int]) -> int:
        ans = 100
        for a in nums1:
            for b in nums2:
                if a == b:
                    ans = min(ans, a)
                else:
                    ans = min(ans, 10 * a + b, 10 * b + a)
        return ans

Complexity

The time complexity is $O(m \times n)$ , and the space complexity is $O(1)$ , where $m$ and $n$ are the lengths of the arrays $nums1$ and $nums2$ . The space complexity is $O(1)$ , where $m$ and $n$ are the lengths of the arrays $nums1$ and $nums2$ .

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.

Approach

Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.

This keeps the solution linear while remaining easy to explain in an interview setting.

Steps:

Initialize a hash map for seen items or counts.

Iterate through the input, querying/updating the map.

Return the first valid result or the final computed value.

Python Solution

class Solution:
    def minNumber(self, nums1: List[int], nums2: List[int]) -> int:
        ans = 100
        for a in nums1:
            for b in nums2:
                if a == b:
                    ans = min(ans, a)
                else:
                    ans = min(ans, 10 * a + b, 10 * b + a)
        return ans

Complexity

The time complexity is

O(m \times n)

, and the space complexity is

O(1)

, where

m

and

n

are the lengths of the arrays

nums1

and

nums2

. The space complexity is

O(1)

, where

m

and

n

are the lengths of the arrays

nums1

and

nums2

Leetcode #2605: Form Smallest Number From Two Digit Arrays

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #2605: Form Smallest Number From Two Digit Arrays

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview