2540. Minimum Common Value

/**
 * Problem: LeetCode 2540 - Minimum Common Value
 * Difficulty: Easy
 * Language: Java
 * Author: CodeByTushu
 */
class Solution {
    public int getCommon(int[] nums1, int[] nums2) {
        int i = 0;
        int j = 0;
        int n = nums1.length;
        int m = nums2.length;

        // Two-pointer linear traversal over both sorted arrays
        while (i < n && j < m) {
            if (nums1[i] == nums2[j]) {
                return nums1[i]; // Smallest common element found
            } else if (nums1[i] < nums2[j]) {
                i++;
            } else {
                j++;
            }
        }

        return -1;
    }
}

"""
Problem: LeetCode 2540 - Minimum Common Value
Difficulty: Easy
Language: Python
Author: CodeByTushu
"""
from typing import List

class Solution:
    def getCommon(self, nums1: List[int], nums2: List[int]) -> int:
        i, j = 0, 0
        n, m = len(nums1), len(nums2)

        # Two-pointer traversal over both sorted arrays
        while i < n and j < m:
            if nums1[i] == nums2[j]:
                return nums1[i]
            elif nums1[i] < nums2[j]:
                i += 1
            else:
                j += 1

        return -1

/**
 * Problem: LeetCode 2540 - Minimum Common Value
 * Difficulty: Easy
 * Language: C++
 * Author: CodeByTushu
 */
#include <vector>

using namespace std;

class Solution {
public:
    int getCommon(vector<int>& nums1, vector<int>& nums2) {
        int i = 0;
        int j = 0;
        int n = nums1.size();
        int m = nums2.size();

        // Two-pointer scan through sorted vectors
        while (i < n && j < m) {
            if (nums1[i] == nums2[j]) {
                return nums1[i];
            } else if (nums1[i] < nums2[j]) {
                i++;
            } else {
                j++;
            }
        }

        return -1;
    }
};

/**
 * Problem: LeetCode 2540 - Minimum Common Value
 * Difficulty: Easy
 * Language: JavaScript
 * Author: CodeByTushu
 */
/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number}
 */
var getCommon = function(nums1, nums2) {
    let i = 0;
    let j = 0;
    const n = nums1.length;
    const m = nums2.length;

    // Two-pointer traversal
    while (i < n && j < m) {
        if (nums1[i] === nums2[j]) {
            return nums1[i];
        } else if (nums1[i] < nums2[j]) {
            i++;
        } else {
            j++;
        }
    }

    return -1;
};