How do you solve Minimum Common Value on LeetCode?

Minimum Common Value (LeetCode #2540) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Both arrays are sorted, which allows for efficient searching techniques.

What is the brute force solution for Minimum Common Value?

The brute force approach for Minimum Common Value is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each element in the first array against every element in the second array. This is straightforward but inefficient, especially for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Common Value?

Minimum Common Value uses the following concepts: Array, Hash Table, Two Pointers, Binary Search. The recommended patterns to study are: Two Pointers, Array.

What are the interview tips for Minimum Common Value?

Always consider the properties of the input data (like sorted arrays) to optimize your solution. Explain your thought process clearly, especially when transitioning from a brute force to an optimal solution. Practice writing clean and concise code, as readability is important in interviews.

What are common mistakes when solving Minimum Common Value?

Not leveraging the sorted property of the arrays, leading to inefficient solutions. Failing to handle edge cases, such as when there are no common elements.

#2540

Minimum Common Value

Easy

ArrayHash TableTwo PointersBinary SearchTwo PointersArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

Using a two-pointer technique leverages the fact that both arrays are sorted. This allows us to efficiently find the smallest common value without unnecessary comparisons.

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Algorithm

5 steps

1Step 1: Initialize two pointers, one for each array, starting at the beginning.
2Step 2: While both pointers are within the bounds of their respective arrays, compare the elements at the pointers.
3Step 3: If the elements are equal, return the element as it is the minimum common value.
4Step 4: If the element in nums1 is smaller, increment the pointer for nums1; otherwise, increment the pointer for nums2.
5Step 5: If the loop ends without finding a common element, return -1.

solution.py12 lines

1# Full working Python code
2
3def min_common_value(nums1, nums2):
4    i, j = 0, 0
5    while i < len(nums1) and j < len(nums2):
6        if nums1[i] == nums2[j]:
7            return nums1[i]
8        elif nums1[i] < nums2[j]:
9            i += 1
10        else:
11            j += 1
12    return -1

ℹ

Complexity note: The time complexity is O(n) because we are traversing each array at most once. The space complexity is O(1) since we are using a constant amount of extra space.

1Both arrays are sorted, which allows for efficient searching techniques.
2Using two pointers minimizes unnecessary comparisons.

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