3046. Split the Array

Description

You are given an integer array nums of even length. You have to split the array into two parts nums1 and nums2 such that:

• nums1.length == nums2.length == nums.length / 2.
• nums1 should contain distinct elements.
• nums2 should also contain distinct elements.

Return true if it is possible to split the array, and false otherwise.

Example 1:

Input: nums = [1,1,2,2,3,4]
Output: true
Explanation: One of the possible ways to split nums is nums1 = [1,2,3] and nums2 = [1,2,4].


Example 2:

Input: nums = [1,1,1,1]
Output: false
Explanation: The only possible way to split nums is nums1 = [1,1] and nums2 = [1,1]. Both nums1 and nums2 do not contain distinct elements. Therefore, we return false.


Constraints:

• 1 <= nums.length <= 100
• nums.length % 2 == 0
• 1 <= nums[i] <= 100

Solutions

Solution 1: Counting

According to the problem, we need to divide the array into two parts, and the elements in each part are all distinct. Therefore, we can count the occurrence of each element in the array. If an element appears three or more times, it cannot satisfy the problem’s requirements. Otherwise, we can divide the array into two parts.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array.

• class Solution {
public boolean isPossibleToSplit(int[] nums) {
int[] cnt = new int[101];
for (int x : nums) {
if (++cnt[x] >= 3) {
return false;
}
}
return true;
}
}

• class Solution {
public:
bool isPossibleToSplit(vector<int>& nums) {
int cnt[101]{};
for (int x : nums) {
if (++cnt[x] >= 3) {
return false;
}
}
return true;
}
};

• class Solution:
def isPossibleToSplit(self, nums: List[int]) -> bool:
return max(Counter(nums).values()) < 3


• func isPossibleToSplit(nums []int) bool {
cnt := [101]int{}
for _, x := range nums {
cnt[x]++
if cnt[x] >= 3 {
return false
}
}
return true
}

• function isPossibleToSplit(nums: number[]): boolean {
const cnt: number[] = Array(101).fill(0);
for (const x of nums) {
if (++cnt[x] >= 3) {
return false;
}
}
return true;
}