# 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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