你和一群强盗准备打劫银行。给你一个下标从 0 开始的整数数组 security ，其中 security[i] 是第 i 天执勤警卫的数量。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打劫银行的日子：
- 第
i天前和后都分别至少有time天。 - 第
i天前连续time天警卫数目都是非递增的。 - 第
i天后连续time天警卫数目都是非递减的。
更正式的，第 i 天是一个合适打劫银行的日子当且仅当：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 适合打劫银行的日子（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 是适合打劫银行的日子。
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打劫银行的日子，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 天警卫数目是非递增的。 所以没有适合打劫银行的日子，返回空数组。
提示：
1 <= security.length <= 1050 <= security[i], time <= 105
left, right 分别记录左右符合要求的天数。
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
}
}