Solution Explanation for 1121. Divide Array Into Increasing Sequences

This problem asks whether a sorted array nums can be partitioned into one or more increasing subsequences, each of length at least k. The solution leverages a crucial observation to avoid complex dynamic programming or greedy approaches.

Core Idea:

The most efficient way to solve this problem relies on analyzing the frequency of the most frequent element in nums. Let's denote this frequency as maxFreq. The key insight is:

• Minimum Subsequences: To satisfy the condition, we need at least maxFreq increasing subsequences because each occurrence of the most frequent element must belong to a separate subsequence.

• Sufficient Condition: If maxFreq * k <= nums.length, then we can create enough subsequences of length at least k. Each subsequence can accommodate at least k elements from nums.

• Insufficient Condition: If maxFreq * k > nums.length, it's impossible to create the required number of subsequences with sufficient length. There simply aren't enough elements.

Algorithm:

1. Find the Most Frequent Element: Iterate through nums to find the element with the highest frequency (maxFreq). This can be efficiently done using a hash map (dictionary in Python) or by iterating and tracking the current and maximum frequencies.

2. Check the Condition: If maxFreq * k <= nums.length, return true; otherwise, return false.

Time and Space Complexity:

• Time Complexity: O(N) - The algorithm iterates through nums once to find the most frequent element. The time spent is directly proportional to the input size.

• Space Complexity: O(1) or O(M) - If we use a hash map to count frequencies, the space complexity depends on the number of unique elements (M) in nums. In the best case (few unique elements), it's close to O(1). A simple iteration to find maxFreq without a hashmap achieves O(1) space complexity.

Code Examples:

Python:

from itertools import groupby
 
class Solution:
    def canDivideIntoSubsequences(self, nums: List[int], k: int) -> bool:
        max_freq = max(len(list(g)) for _, g in groupby(nums))  #Efficiently finds max frequency using itertools.groupby
        return max_freq * k <= len(nums)
 

Java:

import java.util.HashMap;
import java.util.Map;
 
class Solution {
    public boolean canDivideIntoSubsequences(int[] nums, int k) {
        Map<Integer, Integer> counts = new HashMap<>();
        int maxFreq = 0;
        for (int num : nums) {
            counts.put(num, counts.getOrDefault(num, 0) + 1);
            maxFreq = Math.max(maxFreq, counts.get(num));
        }
        return maxFreq * k <= nums.length;
    }
}

C++:

class Solution {
public:
    bool canDivideIntoSubsequences(vector<int>& nums, int k) {
        map<int, int> counts;
        int maxFreq = 0;
        for (int num : nums) {
            counts[num]++;
            maxFreq = max(maxFreq, counts[num]);
        }
        return (long long)maxFreq * k <= nums.size(); //Avoid potential integer overflow
    }
};

Go:

func canDivideIntoSubsequences(nums []int, k int) bool {
	counts := make(map[int]int)
	maxFreq := 0
	for _, num := range nums {
		counts[num]++
		if counts[num] > maxFreq {
			maxFreq = counts[num]
		}
	}
	return int64(maxFreq)*int64(k) <= int64(len(nums)) //Avoid potential integer overflow
 
}

Note: The C++ and Go examples include explicit type casting to long long (C++) and int64 (Go) to prevent potential integer overflow issues if maxFreq and k are large. This is crucial for robustness.