Leetcode 2540: Minimum Common Value Solution

Here, we will see how to solve Minimum Common Value Solution of leet code 2540 problem.

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

Note: 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.

Approach:

• Iterate both the arrays from left to right.
• If the number is equal in nums1 and nums2 then return that number
• If the number is not equal and nums1’s number is less than nums2’s number in iterator then increment iterator of nums1.
• If the number is not equal and nums1’s number is greater than nums2’s number in iterator then increment iterator of nums2.

Minimum Common Value Solution in C++ and go:

Here, we will be solving problem in multiple ways with code.

C++ code 1:

class Solution {
public:
    int getCommon(vector<int>& nums1, vector<int>& nums2) {
        int len1 = nums1.size() - 1, i = 0;
        int len2 = nums2.size() - 1, j = 0;
        
        while(i <= len1 && j <= len2) {
            if(nums1[i] == nums2[j]) {
                return nums1[i];
            } else if(nums1[i] < nums2[j]) {
                i++;
            } else {
                j++;
            }
        }
        return -1;
    }
};

Go code 1:

func getCommon(nums1 []int, nums2 []int) int {
    var len1, i = len(nums1) - 1, 0
    var len2, j = len(nums2) - 1, 0
        
    for i <= len1 && j <= len2 {
        if nums1[i] == nums2[j] {
            return nums1[i]
        } else if nums1[i] < nums2[j] {
            i++
        } else {
            j++
        }
    }
    return -1;
}

Output:

Input: nums1 = [1,2,3,6], nums2 = [2,3,4,5]
Output: 2

Time complexity: O(n)

Space complexity: O(1)

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