Question

2100. 适合野炊的日子

English Version

题目描述

你和朋友们准备去野炊。给你一个下标从 0 开始的整数数组 security ，其中 security[i] 是第 i 天的建议出行指数。日子从 0 开始编号。同时给你一个整数 time 。

如果第 i 天满足以下所有条件，我们称它为一个适合野炊的日子：

• 第 i 天前和后都分别至少有 time 天。
• 第 i 天前连续 time 天建议出行指数都是非递增的。
• 第 i 天后连续 time 天建议出行指数都是非递减的。

更正式的，第 i 天是一个适合野炊的日子当且仅当：security[i - time] >= security[i - time + 1] >= ... >= security[i] <= ... <= security[i + time - 1] <= security[i + time].

请你返回一个数组，包含 所有 适合野炊的日子（下标从 0 开始）。返回的日子可以 任意 顺序排列。

 

示例 1：

输入：security = [5,3,3,3,5,6,2], time = 2
输出：[2,3]
解释：
第 2 天，我们有 security[0] >= security[1] >= security[2] <= security[3] <= security[4] 。
第 3 天，我们有 security[1] >= security[2] >= security[3] <= security[4] <= security[5] 。
没有其他日子符合这个条件，所以日子 2 和 3 是适合野炊的日子。

示例 2：

输入：security = [1,1,1,1,1], time = 0
输出：[0,1,2,3,4]
解释：
因为 time 等于 0 ，所以每一天都是适合野炊的日子，所以返回每一天。

示例 3：

输入：security = [1,2,3,4,5,6], time = 2
输出：[]
解释：
没有任何一天的前 2 天建议出行指数是非递增的。
所以没有适合野炊的日子，返回空数组。

 

提示：

• 1 <= security.length <= 105
• 0 <= security[i], time <= 105

解法

方法一

class Solution:
  def goodDaysToRobBank(self, security: List[int], time: int) -> List[int]:
    n = len(security)
    if n <= time * 2:
      return []
    left, right = [0] * n, [0] * n
    for i in range(1, n):
      if security[i] <= security[i - 1]:
        left[i] = left[i - 1] + 1
    for i in range(n - 2, -1, -1):
      if security[i] <= security[i + 1]:
        right[i] = right[i + 1] + 1
    return [i for i in range(n) if time <= min(left[i], right[i])]

class Solution {
  public List<Integer> goodDaysToRobBank(int[] security, int time) {
    int n = security.length;
    if (n <= time * 2) {
      return Collections.emptyList();
    }
    int[] left = new int[n];
    int[] right = new int[n];
    for (int i = 1; i < n; ++i) {
      if (security[i] <= security[i - 1]) {
        left[i] = left[i - 1] + 1;
      }
    }
    for (int i = n - 2; i >= 0; --i) {
      if (security[i] <= security[i + 1]) {
        right[i] = right[i + 1] + 1;
      }
    }
    List<Integer> ans = new ArrayList<>();
    for (int i = time; i < n - time; ++i) {
      if (time <= Math.min(left[i], right[i])) {
        ans.add(i);
      }
    }
    return ans;
  }
}

class Solution {
public:
  vector<int> goodDaysToRobBank(vector<int>& security, int time) {
    int n = security.size();
    if (n <= time * 2) return {};
    vector<int> left(n);
    vector<int> right(n);
    for (int i = 1; i < n; ++i)
      if (security[i] <= security[i - 1])
        left[i] = left[i - 1] + 1;
    for (int i = n - 2; i >= 0; --i)
      if (security[i] <= security[i + 1])
        right[i] = right[i + 1] + 1;
    vector<int> ans;
    for (int i = time; i < n - time; ++i)
      if (time <= min(left[i], right[i]))
        ans.push_back(i);
    return ans;
  }
};

func goodDaysToRobBank(security []int, time int) []int {
  n := len(security)
  if n <= time*2 {
    return []int{}
  }
  left := make([]int, n)
  right := make([]int, n)
  for i := 1; i < n; i++ {
    if security[i] <= security[i-1] {
      left[i] = left[i-1] + 1
    }
  }
  for i := n - 2; i >= 0; i-- {
    if security[i] <= security[i+1] {
      right[i] = right[i+1] + 1
    }
  }
  var ans []int
  for i := time; i < n-time; i++ {
    if time <= left[i] && time <= right[i] {
      ans = append(ans, i)
    }
  }
  return ans
}

function goodDaysToRobBank(security: number[], time: number): number[] {
  const n = security.length;
  if (n <= time * 2) {
    return [];
  }
  const l = new Array(n).fill(0);
  const r = new Array(n).fill(0);
  for (let i = 1; i < n; i++) {
    if (security[i] <= security[i - 1]) {
      l[i] = l[i - 1] + 1;
    }
    if (security[n - i - 1] <= security[n - i]) {
      r[n - i - 1] = r[n - i] + 1;
    }
  }
  const res = [];
  for (let i = time; i < n - time; i++) {
    if (time <= Math.min(l[i], r[i])) {
      res.push(i);
    }
  }
  return res;
}

use std::cmp::Ordering;

impl Solution {
  pub fn good_days_to_rob_bank(security: Vec<i32>, time: i32) -> Vec<i32> {
    let time = time as usize;
    let n = security.len();
    if time * 2 >= n {
      return vec![];
    }
    let mut g = vec![0; n];
    for i in 1..n {
      g[i] = match security[i].cmp(&security[i - 1]) {
        Ordering::Less => -1,
        Ordering::Greater => 1,
        Ordering::Equal => 0,
      };
    }
    let (mut a, mut b) = (vec![0; n + 1], vec![0; n + 1]);
    for i in 1..=n {
      a[i] = a[i - 1] + (if g[i - 1] == 1 { 1 } else { 0 });
      b[i] = b[i - 1] + (if g[i - 1] == -1 { 1 } else { 0 });
    }
    let mut res = vec![];
    for i in time..n - time {
      if a[i + 1] - a[i + 1 - time] == 0 && b[i + 1 + time] - b[i + 1] == 0 {
        res.push(i as i32);
      }
    }
    res
  }
}