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2863. Maximum Length of Semi-Decreasing Subarrays

Description

You are given an integer array nums.

Return the length of the longest semi-decreasing subarray of nums, and 0 if there are no such subarrays.

  • A subarray is a contiguous non-empty sequence of elements within an array.
  • A non-empty array is semi-decreasing if its first element is strictly greater than its last element.

 

Example 1:

Input: nums = [7,6,5,4,3,2,1,6,10,11]
Output: 8
Explanation: Take the subarray [7,6,5,4,3,2,1,6].
The first element is 7 and the last one is 6 so the condition is met.
Hence, the answer would be the length of the subarray or 8.
It can be shown that there aren't any subarrays with the given condition with a length greater than 8.

Example 2:

Input: nums = [57,55,50,60,61,58,63,59,64,60,63]
Output: 6
Explanation: Take the subarray [61,58,63,59,64,60].
The first element is 61 and the last one is 60 so the condition is met.
Hence, the answer would be the length of the subarray or 6.
It can be shown that there aren't any subarrays with the given condition with a length greater than 6.

Example 3:

Input: nums = [1,2,3,4]
Output: 0
Explanation: Since there are no semi-decreasing subarrays in the given array, the answer is 0.

 

Constraints:

  • 1 <= nums.length <= 105
  • -109 <= nums[i] <= 109

Solutions

Solution 1: Hash Table + Sorting

The problem is essentially finding the maximum length of the inverse pairs. We can use a hash table $d$ to record the index $i$ corresponding to each number $x$ in the array.

Next, we traverse the keys of the hash table in descending order of the numbers. We maintain a number $k$ to keep track of the smallest index that has appeared so far. For the current number $x$, we can get a maximum inverse pair length of $d[x][ d[x] -1]-k + 1$, where $ d[x] $ represents the length of the array $d[x]$, i.e., the number of times the number $x$ appears in the original array. We update the answer accordingly. Then, we update $k$ to $d[x][0]$, which is the index where the number $x$ first appears in the original array. We continue to traverse the keys of the hash table until all keys are traversed.

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

  • class Solution {
    public int maxSubarrayLength(int[] nums) {
        TreeMap<Integer, List<Integer>> d = new TreeMap<>(Comparator.reverseOrder());
        for (int i = 0; i < nums.length; ++i) {
            d.computeIfAbsent(nums[i], k -> new ArrayList<>()).add(i);
        }
        int ans = 0, k = 1 << 30;
        for (List<Integer> idx : d.values()) {
            ans = Math.max(ans, idx.get(idx.size() - 1) - k + 1);
            k = Math.min(k, idx.get(0));
        }
        return ans;
    }
}

  • class Solution {
public:
    int maxSubarrayLength(vector<int>& nums) {
        map<int, vector<int>, greater<int>> d;
        for (int i = 0; i < nums.size(); ++i) {
            d[nums[i]].push_back(i);
        }
        int ans = 0, k = 1 << 30;
        for (auto& [_, idx] : d) {
            ans = max(ans, idx.back() - k + 1);
            k = min(k, idx[0]);
        }
        return ans;
    }
};

  • class Solution:
    def maxSubarrayLength(self, nums: List[int]) -> int:
        d = defaultdict(list)
        for i, x in enumerate(nums):
            d[x].append(i)
        ans, k = 0, inf
        for x in sorted(d, reverse=True):
            ans = max(ans, d[x][-1] - k + 1)
            k = min(k, d[x][0])
        return ans


  • func maxSubarrayLength(nums []int) (ans int) {
	d := map[int][]int{}
	for i, x := range nums {
		d[x] = append(d[x], i)
	}
	keys := []int{}
	for x := range d {
		keys = append(keys, x)
	}
	sort.Slice(keys, func(i, j int) bool { return keys[i] > keys[j] })
	k := 1 << 30
	for _, x := range keys {
		idx := d[x]
		ans = max(ans, idx[len(idx)-1]-k+1)
		k = min(k, idx[0])
	}
	return ans
}

  • function maxSubarrayLength(nums: number[]): number {
    const d: Map<number, number[]> = new Map();
    for (let i = 0; i < nums.length; ++i) {
        if (!d.has(nums[i])) {
            d.set(nums[i], []);
        }
        d.get(nums[i])!.push(i);
    }
    const keys = Array.from(d.keys()).sort((a, b) => b - a);
    let ans = 0;
    let k = Infinity;
    for (const x of keys) {
        const idx = d.get(x)!;
        ans = Math.max(ans, idx.at(-1) - k + 1);
        k = Math.min(k, idx[0]);
    }
    return ans;
}

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