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Description

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.

Solutions

Python3

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

Java

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;
    }
}

C++

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

Go

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
}

TypeScript

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;
}

Rust

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
    }
}
impl Solution {
    pub fn get_common(nums1: Vec<i32>, nums2: Vec<i32>) -> i32 {
        let mut iter1 = nums1.iter();
        let mut iter2 = nums2.iter();
        let mut num1 = iter1.next();
        let mut num2 = iter2.next();

        while let (Some(n1), Some(n2)) = (num1, num2) {
            if n1 == n2 {
                return *n1;
            } else if n1 < n2 {
                num1 = iter1.next();
            } else {
                num2 = iter2.next();
            }
        }

        -1
    }
}

C

int getCommon(int* nums1, int nums1Size, int* nums2, int nums2Size) {
    int i = 0;
    int j = 0;
    while (i < nums1Size && j < nums2Size) {
        if (nums1[i] == nums2[j]) {
            return nums1[i];
        }
        if (nums1[i] < nums2[j]) {
            i++;
        } else {
            j++;
        }
    }
    return -1;
}

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