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Hard
2381
Biweekly Contest 82 Q4
Stack
Union Find
Array
Monotonic Stack

中文文档

Description

You are given an integer array nums and an integer threshold.

Find any subarray of nums of length k such that every element in the subarray is greater than threshold / k.

Return the size of any such subarray. If there is no such subarray, return -1.

A subarray is a contiguous non-empty sequence of elements within an array.

 

Example 1:

Input: nums = [1,3,4,3,1], threshold = 6
Output: 3
Explanation: The subarray [3,4,3] has a size of 3, and every element is greater than 6 / 3 = 2.
Note that this is the only valid subarray.

Example 2:

Input: nums = [6,5,6,5,8], threshold = 7
Output: 1
Explanation: The subarray [8] has a size of 1, and 8 > 7 / 1 = 7. So 1 is returned.
Note that the subarray [6,5] has a size of 2, and every element is greater than 7 / 2 = 3.5. 
Similarly, the subarrays [6,5,6], [6,5,6,5], [6,5,6,5,8] also satisfy the given conditions.
Therefore, 2, 3, 4, or 5 may also be returned.

 

Constraints:

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

Solutions

Solution 1

Python3

class Solution:
    def validSubarraySize(self, nums: List[int], threshold: int) -> int:
        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        def merge(a, b):
            pa, pb = find(a), find(b)
            if pa == pb:
                return
            p[pa] = pb
            size[pb] += size[pa]

        n = len(nums)
        p = list(range(n))
        size = [1] * n
        arr = sorted(zip(nums, range(n)), reverse=True)
        vis = [False] * n
        for v, i in arr:
            if i and vis[i - 1]:
                merge(i, i - 1)
            if i < n - 1 and vis[i + 1]:
                merge(i, i + 1)
            if v > threshold // size[find(i)]:
                return size[find(i)]
            vis[i] = True
        return -1

Java

class Solution {
    private int[] p;
    private int[] size;

    public int validSubarraySize(int[] nums, int threshold) {
        int n = nums.length;
        p = new int[n];
        size = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
            size[i] = 1;
        }
        int[][] arr = new int[n][2];
        for (int i = 0; i < n; ++i) {
            arr[i][0] = nums[i];
            arr[i][1] = i;
        }
        Arrays.sort(arr, (a, b) -> b[0] - a[0]);
        boolean[] vis = new boolean[n];
        for (int[] e : arr) {
            int v = e[0], i = e[1];
            if (i > 0 && vis[i - 1]) {
                merge(i, i - 1);
            }
            if (i < n - 1 && vis[i + 1]) {
                merge(i, i + 1);
            }
            if (v > threshold / size[find(i)]) {
                return size[find(i)];
            }
            vis[i] = true;
        }
        return -1;
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }

    private void merge(int a, int b) {
        int pa = find(a), pb = find(b);
        if (pa == pb) {
            return;
        }
        p[pa] = pb;
        size[pb] += size[pa];
    }
}

C++

using pii = pair<int, int>;

class Solution {
public:
    vector<int> p;
    vector<int> size;

    int validSubarraySize(vector<int>& nums, int threshold) {
        int n = nums.size();
        p.resize(n);
        for (int i = 0; i < n; ++i) p[i] = i;
        size.assign(n, 1);
        vector<pii> arr(n);
        for (int i = 0; i < n; ++i) arr[i] = {nums[i], i};
        sort(arr.begin(), arr.end());
        vector<bool> vis(n);
        for (int j = n - 1; ~j; --j) {
            int v = arr[j].first, i = arr[j].second;
            if (i && vis[i - 1]) merge(i, i - 1);
            if (j < n - 1 && vis[i + 1]) merge(i, i + 1);
            if (v > threshold / size[find(i)]) return size[find(i)];
            vis[i] = true;
        }
        return -1;
    }

    int find(int x) {
        if (p[x] != x) p[x] = find(p[x]);
        return p[x];
    }

    void merge(int a, int b) {
        int pa = find(a), pb = find(b);
        if (pa == pb) return;
        p[pa] = pb;
        size[pb] += size[pa];
    }
};

Go

func validSubarraySize(nums []int, threshold int) int {
	n := len(nums)
	p := make([]int, n)
	size := make([]int, n)
	for i := range p {
		p[i] = i
		size[i] = 1
	}
	var find func(int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	merge := func(a, b int) {
		pa, pb := find(a), find(b)
		if pa == pb {
			return
		}
		p[pa] = pb
		size[pb] += size[pa]
	}

	arr := make([][]int, n)
	for i, v := range nums {
		arr[i] = []int{v, i}
	}
	sort.Slice(arr, func(i, j int) bool {
		return arr[i][0] > arr[j][0]
	})
	vis := make([]bool, n)
	for _, e := range arr {
		v, i := e[0], e[1]
		if i > 0 && vis[i-1] {
			merge(i, i-1)
		}
		if i < n-1 && vis[i+1] {
			merge(i, i+1)
		}
		if v > threshold/size[find(i)] {
			return size[find(i)]
		}
		vis[i] = true
	}
	return -1
}

Solution 2

Python3

class Solution:
    def validSubarraySize(self, nums: List[int], threshold: int) -> int:
        n = len(nums)
        left = [-1] * n
        right = [n] * n
        stk = []
        for i, v in enumerate(nums):
            while stk and nums[stk[-1]] >= v:
                stk.pop()
            if stk:
                left[i] = stk[-1]
            stk.append(i)
        stk = []
        for i in range(n - 1, -1, -1):
            while stk and nums[stk[-1]] >= nums[i]:
                stk.pop()
            if stk:
                right[i] = stk[-1]
            stk.append(i)
        for i, v in enumerate(nums):
            k = right[i] - left[i] - 1
            if v > threshold // k:
                return k
        return -1

Java

class Solution {
    public int validSubarraySize(int[] nums, int threshold) {
        int n = nums.length;
        int[] left = new int[n];
        int[] right = new int[n];
        Arrays.fill(left, -1);
        Arrays.fill(right, n);
        Deque<Integer> stk = new ArrayDeque<>();
        for (int i = 0; i < n; ++i) {
            int v = nums[i];
            while (!stk.isEmpty() && nums[stk.peek()] >= v) {
                stk.pop();
            }
            if (!stk.isEmpty()) {
                left[i] = stk.peek();
            }
            stk.push(i);
        }
        stk.clear();
        for (int i = n - 1; i >= 0; --i) {
            int v = nums[i];
            while (!stk.isEmpty() && nums[stk.peek()] >= v) {
                stk.pop();
            }
            if (!stk.isEmpty()) {
                right[i] = stk.peek();
            }
            stk.push(i);
        }
        for (int i = 0; i < n; ++i) {
            int v = nums[i];
            int k = right[i] - left[i] - 1;
            if (v > threshold / k) {
                return k;
            }
        }
        return -1;
    }
}

C++

class Solution {
public:
    int validSubarraySize(vector<int>& nums, int threshold) {
        int n = nums.size();
        vector<int> left(n, -1);
        vector<int> right(n, n);
        stack<int> stk;
        for (int i = 0; i < n; ++i) {
            int v = nums[i];
            while (!stk.empty() && nums[stk.top()] >= v) stk.pop();
            if (!stk.empty()) left[i] = stk.top();
            stk.push(i);
        }
        stk = stack<int>();
        for (int i = n - 1; ~i; --i) {
            int v = nums[i];
            while (!stk.empty() && nums[stk.top()] >= v) stk.pop();
            if (!stk.empty()) right[i] = stk.top();
            stk.push(i);
        }
        for (int i = 0; i < n; ++i) {
            int v = nums[i];
            int k = right[i] - left[i] - 1;
            if (v > threshold / k) return k;
        }
        return -1;
    }
};

Go

func validSubarraySize(nums []int, threshold int) int {
	n := len(nums)
	left := make([]int, n)
	right := make([]int, n)
	for i := range left {
		left[i] = -1
		right[i] = n
	}
	var stk []int
	for i, v := range nums {
		for len(stk) > 0 && nums[stk[len(stk)-1]] >= v {
			stk = stk[:len(stk)-1]
		}
		if len(stk) > 0 {
			left[i] = stk[len(stk)-1]
		}
		stk = append(stk, i)
	}
	stk = []int{}
	for i := n - 1; i >= 0; i-- {
		v := nums[i]
		for len(stk) > 0 && nums[stk[len(stk)-1]] >= v {
			stk = stk[:len(stk)-1]
		}
		if len(stk) > 0 {
			right[i] = stk[len(stk)-1]
		}
		stk = append(stk, i)
	}
	for i, v := range nums {
		k := right[i] - left[i] - 1
		if v > threshold/k {
			return k
		}
	}
	return -1
}