LeetCode 2540: Minimum Common Value ·

Table of Contents

Meta Data
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Difficulty: easy First Attempt: 2026-01-31

Submissions: Accepted (first try)

Approach
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Both arrays are sorted. Keep one pointer in each and compare from the smallest:

Equal: this is the smallest common value; return it.
Not equal: the smaller value cannot appear later in the other array (later values are only larger), so advance the smaller side.

My Solution
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class Solution:
    def getCommon(self, nums1: List[int], nums2: List[int]) -> int:
        i = j = 0
        while i < len(nums1) and j < len(nums2):
            if nums1[i] == nums2[j]:
                return nums1[i]
            if nums1[i] < nums2[j]:
                i += 1
            else:
                j += 1
        return -1

My original checked “is this already the last element” before advancing and broke out if so. That is unnecessary: once a pointer reaches the end, the while condition stops the loop.

Complexity
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Time O(n + m), space O(1).

Mistakes to Avoid
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Advancing the larger side: it skips the answer.
Intersecting sets and taking the minimum: it works, but uses O(n) space and ignores the sorting.

Interview Explanation
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Both arrays are sorted, so I compare from the front with two pointers. Equal values give the smallest common value; otherwise the smaller value cannot be found in the other array anymore, so I advance its side. If either side runs out first, I return -1. Time O(n + m), space O(1).