README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Hard	https://github.com/doocs/leetcode/edit/main/solution/1100-1199/1121.Divide%20Array%20Into%20Increasing%20Sequences/README_EN.md	1664	Biweekly Contest 4 Q4	Array Counting

1121. Divide Array Into Increasing Sequences 🔒

中文文档

Description

Given an integer array nums sorted in non-decreasing order and an integer k, return true if this array can be divided into one or more disjoint increasing subsequences of length at least k, or false otherwise.

 

Example 1:

Input: nums = [1,2,2,3,3,4,4], k = 3
Output: true
Explanation: The array can be divided into two subsequences [1,2,3,4] and [2,3,4] with lengths at least 3 each.

Example 2:

Input: nums = [5,6,6,7,8], k = 3
Output: false
Explanation: There is no way to divide the array using the conditions required.

 

Constraints:

• 1 <= k <= nums.length <= 105
• 1 <= nums[i] <= 105
• nums is sorted in non-decreasing order.

Solutions

Solution 1: Quick Thinking

We assume that the array can be divided into $m$ strictly increasing subsequences of length at least $k$. If the number of the most frequent number in the array is $cnt$, then these $cnt$ numbers must be in different subsequences, so $m \geq cnt$. Also, since the length of $m$ subsequences is at least $k$, the fewer the number of subsequences, the better, so $m = cnt$. Therefore, $cnt \times k \leq n$ must be satisfied. Hence, we only need to count the number of the most frequent number $cnt$ in the array, and then judge whether $cnt \times k \leq n$. If it is, return true, otherwise return false.

The time complexity is $O(n)$, and the space complexity is $O(1)$. Here, $n$ is the length of the array $nums$.

Python3

class Solution:
    def canDivideIntoSubsequences(self, nums: List[int], k: int) -> bool:
        mx = max(len(list(x)) for _, x in groupby(nums))
        return mx * k <= len(nums)

Java

class Solution {
    public boolean canDivideIntoSubsequences(int[] nums, int k) {
        Map<Integer, Integer> cnt = new HashMap<>();
        int mx = 0;
        for (int x : nums) {
            mx = Math.max(mx, cnt.merge(x, 1, Integer::sum));
        }
        return mx * k <= nums.length;
    }
}

C++

class Solution {
public:
    bool canDivideIntoSubsequences(vector<int>& nums, int k) {
        int cnt = 0;
        int a = 0;
        for (int& b : nums) {
            cnt = a == b ? cnt + 1 : 1;
            if (cnt * k > nums.size()) {
                return false;
            }
            a = b;
        }
        return true;
    }
};

Go

func canDivideIntoSubsequences(nums []int, k int) bool {
	cnt, a := 0, 0
	for _, b := range nums {
		cnt++
		if a != b {
			cnt = 1
		}
		if cnt*k > len(nums) {
			return false
		}
		a = b
	}
	return true
}

Solution 2

Java

class Solution {
    public boolean canDivideIntoSubsequences(int[] nums, int k) {
        int cnt = 0;
        int a = 0;
        for (int b : nums) {
            cnt = a == b ? cnt + 1 : 1;
            if (cnt * k > nums.length) {
                return false;
            }
            a = b;
        }
        return true;
    }
}