LeetCode #2540 — EASY

Minimum Common Value

Build confidence with an intuition-first walkthrough focused on array fundamentals.

Solve on LeetCode

The Problem

Problem Statement

Given two integer arrays nums1 and nums2, sorted in non-decreasing order, return the minimum integer common to both arrays. If there is no common integer amongst nums1 and nums2, return -1.

Note that an integer is said to be common to nums1 and nums2 if both arrays have at least one occurrence of that integer.

Example 1:

Input: nums1 = [1,2,3], nums2 = [2,4]
Output: 2
Explanation: The smallest element common to both arrays is 2, so we return 2.

Example 2:

Input: nums1 = [1,2,3,6], nums2 = [2,3,4,5]
Output: 2
Explanation: There are two common elements in the array 2 and 3 out of which 2 is the smallest, so 2 is returned.

Constraints:

1 <= nums1.length, nums2.length <= 10⁵
1 <= nums1[i], nums2[j] <= 10⁹
Both nums1 and nums2 are sorted in non-decreasing order.

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: Given two integer arrays nums1 and nums2, sorted in non-decreasing order, return the minimum integer common to both arrays. If there is no common integer amongst nums1 and nums2, return -1. Note that an integer is said to be common to nums1 and nums2 if both arrays have at least one occurrence of that integer.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Two Pointers · Binary Search

Example 1

[1,2,3]
[2,4]

Example 2

[1,2,3,6]
[2,3,4,5]

Related Problems

Intersection of Two Arrays (intersection-of-two-arrays)
Intersection of Two Arrays II (intersection-of-two-arrays-ii)

Step 02

Core Insight

What unlocks the optimal approach

Try to use a set.
Otherwise, try to use a two-pointer approach.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2540: Minimum Common Value
class Solution {
    public int getCommon(int[] nums1, int[] nums2) {
        int m = nums1.length, n = nums2.length;
        for (int i = 0, j = 0; i < m && j < n;) {
            if (nums1[i] == nums2[j]) {
                return nums1[i];
            }
            if (nums1[i] < nums2[j]) {
                ++i;
            } else {
                ++j;
            }
        }
        return -1;
    }
}

// Accepted solution for LeetCode #2540: Minimum Common Value
func getCommon(nums1 []int, nums2 []int) int {
	m, n := len(nums1), len(nums2)
	for i, j := 0, 0; i < m && j < n; {
		if nums1[i] == nums2[j] {
			return nums1[i]
		}
		if nums1[i] < nums2[j] {
			i++
		} else {
			j++
		}
	}
	return -1
}

# Accepted solution for LeetCode #2540: Minimum Common Value
class Solution:
    def getCommon(self, nums1: List[int], nums2: List[int]) -> int:
        i = j = 0
        m, n = len(nums1), len(nums2)
        while i < m and j < n:
            if nums1[i] == nums2[j]:
                return nums1[i]
            if nums1[i] < nums2[j]:
                i += 1
            else:
                j += 1
        return -1

// Accepted solution for LeetCode #2540: Minimum Common Value
impl Solution {
    pub fn get_common(nums1: Vec<i32>, nums2: Vec<i32>) -> i32 {
        let m = nums1.len();
        let n = nums2.len();
        let mut i = 0;
        let mut j = 0;
        while i < m && j < n {
            if nums1[i] == nums2[j] {
                return nums1[i];
            }
            if nums1[i] < nums2[j] {
                i += 1;
            } else {
                j += 1;
            }
        }
        -1
    }
}

// Accepted solution for LeetCode #2540: Minimum Common Value
function getCommon(nums1: number[], nums2: number[]): number {
    const m = nums1.length;
    const n = nums2.length;
    let i = 0;
    let j = 0;
    while (i < m && j < n) {
        if (nums1[i] === nums2[j]) {
            return nums1[i];
        }
        if (nums1[i] < nums2[j]) {
            i++;
        } else {
            j++;
        }
    }
    return -1;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.