Find All Good Indices

Find All Good Indices

You are given a 0-indexed integer array nums of size $n$ and a positive integer $k$.

We call an index $i$ in the range $k \leq i < n - k$ good if the following conditions are satisfied:

• The $k$ elements that are just before the index $i$ are in non-increasing order.
• The $k$ elements that are just after the index $i$ are in non-decreasing order.

Return an array of all good indices sorted in increasing order.

Example 1:

Input: nums = [2,1,1,1,3,4,1], k = 2
Output: [2,3]
Explanation: There are two good indices in the array:
- Index 2. The subarray [2,1] is in non-increasing order, and the subarray [1,3] is in non-decreasing order.
- Index 3. The subarray [1,1] is in non-increasing order, and the subarray [3,4] is in non-decreasing order.
Note that the index 4 is not good because [4,1] is not non-decreasing.

Example 2:

Input: nums = [2,1,1,2], k = 2
Output: []
Explanation: There are no good indices in this array.

Constraints:

$n == nums.length$
$3 \leq n \leq {10}^{5}$
$1 \leq nums[i] \leq {10}^{6}$
$1 \leq k \leq n / 2$

 

解题思路

　　做这题的时候很容易想到对于每一个$i$，可以直接每次暴力枚举区间$[i-k,i-1]$和$[i+1,i+k]$看看是否满足要求，但这样做会超时，然后又发现这相当于询问一个区间是否满足非递增或非递减，因此比赛的时候直接用线段树去实现了。

　　AC代码如下：

const int N = 1e5 + 10;

class Solution {
public:
    struct Node {
        int l, r;
        bool f1, f2;
    }tr[N * 4];
    vector<int> a;
    
    void build(int u, int l, int r) {
        if (l == r) {
            tr[u] = {l, r, true, true};
        }
        else {
            int mid = l + r >> 1;
            build(u << 1, l, mid);
            build(u << 1 | 1, mid + 1, r);
            tr[u] = {l, r, tr[u << 1].f1 && tr[u << 1 | 1].f1 && a[mid] >= a[mid + 1], tr[u << 1].f2 && tr[u << 1 | 1].f2 && a[mid] <= a[mid + 1]};
        }
    }
    
    Node query(int u, int l, int r) {
        if (tr[u].l >= l && tr[u].r <= r) return tr[u];
        int mid = tr[u].l + tr[u].r >> 1;
        if (r <= mid) {
            return query(u << 1, l, r);
        }
        else if (l >= mid + 1) {
            return query(u << 1 | 1, l, r);
        }
        else {
            Node t1 = query(u << 1, l, r);
            Node t2 = query(u << 1 | 1, l, r);
            Node ret;
            ret.f1 = t1.f1 && t2.f1 && a[mid] >= a[mid + 1];
            ret.f2 = t1.f2 && t2.f2 && a[mid] <= a[mid + 1];
            return ret;
        }
    }
    
    vector<int> goodIndices(vector<int>& nums, int k) {
        int n = nums.size();
        a = nums;
        build(1, 0, n - 1);
        vector<int> ans;
        for (int i = k; i < n - k; i++) {
            if (query(1, i - k, i - 1).f1 && query(1, i + 1, i + k).f2) ans.push_back(i);
        }
        return ans;
    }
};

　　下面给出正解，反正我是想不到的。

　　先预处理一个$f(i)$，表示以$i$为右端点，且单调递减的序列的最大长度。再预处理一个$g(i)$表示以$i$为左端点，且单调递增的序列的最大长度。

　　对于$f(i)$，如果有$a_{i-1} \geq a_{i}$，那么$f(i) = f(i-1)+1$；否则如果有$a_{i-1} < a_{i}$，那么$f(i) = 1$。

　　同理对于$g(i)$，如果有$a_{i} \leq a_{i+1}$，那么$g(i) = g(i+1)+1$；否则如果有$a_{i} > a_{i+1}$，那么$g(i) = 1$。

　　因此当我们想要判断第$i$个位置是否满足要求，就可以看是否同时满足$f(i-1) \geq k$和$g(i+1) \geq k$。

　　AC代码如下：

class Solution {
public:
    vector<int> goodIndices(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> f(n), g(n);
        for (int i = 0; i < n; i++) {
            f[i] = 1;
            if (i && nums[i] <= nums[i - 1]) f[i] = f[i - 1] + 1;
        }
        for (int i = n - 1; i >= 0; i--) {
            g[i] = 1;
            if (i + 1 < n && nums[i] <= nums[i + 1]) g[i] = g[i + 1] + 1;
        }
        
        vector<int> ans;
        for (int i = k; i < n - k; i++) {
            if (f[i - 1] >= k && g[i + 1] >= k) ans.push_back(i);
        }
        return ans;
    }
};

 

参考资料

　　力扣第312场周赛：https://www.bilibili.com/video/BV13Y4y1K7eU