LeetCode 2100. A good day to rob a bank

【LetMeFly】2100. A good day to rob a bank

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Leetcode question link: https://leetcode.cn/problems/find-good-days-to-rob-the-bank/

You and a group of robbers are going to rob a bank. You are given an array of integers whose index starts from 0security , where security[i] is i the number of guards on duty on the day. Days 0 are numbered from the beginning. Also gives you an integer time .

If i day satisfies all of the following conditions, we call it a suitable day to rob a bank:

i There are at least one day before and after the first day time .
i The number of guards on consecutive days before the first day time is non-increasing.
The number of guards is non-decreasing on i consecutive days after the first day.time

More formally, day iis a suitable day to rob a bank if and only if: security[i - time] >= security[i - time + 1] >= ... >= security[i] <= ... <= security[i + time - 1] <= security[i + time].

Please return an array containing all days suitable for bank robbery (the subscript starts from 0 ). The days returned can be in any order.

Example 1:

Input: security = [5,3,3,3,5,6,2], time = 2
 Output: [2,3]
 Explanation: 
On day 2, we have security[0] >= security[1] >= security[2] <= security[3] <= security[4] . 
On day 3, we have security[1] >= security[2] >= security[3] <= security[4] <= security[5] . 
No other days fit this criteria, so days 2 and 3 are good days for bank robberies.

Example 2:

Input: security = [1,1,1,1,1], time = 0
 Output: [0,1,2,3,4]
 Explanation: 
Because time is equal to 0, every day is a suitable day to rob a bank. So return every day.

Example 3:

Input: security = [1,2,3,4,5,6], time = 2
 Output: []
 Explanation: 
There is no non-increasing number of guards in the first 2 days of any day. 
So there are no days suitable for bank robbery, and an empty array is returned.

hint:

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

Ideas

Method 1: Classification discussion

<small time complexity $O (n)$ , space complexity O (1), difficulty ※※** </smtps://leetcode-cn.com/problems/find-good-days-to-rob-the-bank/solution/letmefly- fen-lei-tao-lun-on-o1-by-letmef-1jgw/](https://img-blog.csdnimg.cn/img_convert/ca4a6ec46181f3756420620dbe776075.jpeg)

$t im e = In the case of 0,$ special consideration should be given to inserting links and pictures.

Therefore we use two variables $l ian X u X ia D a ys$ (representing non-increasing days) and $co u l d A s U p 4 B e g in$ (The day from now on when non-decreasing can start)

That is to say, in consecutive $l ian X u X ia D a ys$ after non-increasing days, if $l ian X u X ia D a ys \geq ti im e$ , then as long asthe continuoustoday $Tim e is$ not decreasing day by day, today is the "robbery day".

So we are in $lianXuXiaDays\geq time$ , you can change $co u l d A s U p 4 B e g in$ Mark it as today.

after $t im e$ days and above are not decreasing, then the recorded $co u l d A s U p 4 B e g in$ is a "robbery day."

Therefore, in the backward traversal, if it is still in the non-decreasing state, it can be judged whether there could $co u l d A s U p 4 B e g in$ , if there is ( $\neq -1$ ) Just judge today’s distance $co u l d A s U p 4 B e g in$ whether $\geq time$ days if $\geq time$ ， Employment $co u l d A s U p 4 B e g in$ the continuous $t im e$ days are non-decreasing, so $co u l d A s U p 4 B e g in$ is a robbery day.

For a more detailed description, please refer to the comments

AC code

C++

class Solution {
    
    
public:
    vector<int> goodDaysToRobBank(vector<int>& security, int time) {
    
    
        if (!time) {
    
      // time = 0，每天都是“打劫日”
            vector<int> ans(security.size());  // 答案共有security.size()天
            for (int i = 0; i < security.size(); i++) {
    
    
                ans[i] = i;  // 第i个答案是第i天
            }
            return ans;
        }
        vector<int> ans;
        int lianXuXiaDays = 0;  // 连续↓或→的天数
        int couldAsUp4Begin = -1;  // 最早可以认为是开始连续上升的那一天 | 如果couldAsUp4Begin=a≠-1，说明第a天之前至少有time天的非递增
        for (int i = 1; i < security.size(); i++) {
    
      // 从第二天开始遍历
            if (security[i] < security[i - 1]) {
    
      // ↓
                lianXuXiaDays++;  // 连续非递增天数++
                if (lianXuXiaDays >= time) {
    
      // 如果连续非递增天数≥time，那么今天之前就有≥time的非递减
                    couldAsUp4Begin = i;  // 从今天开始可以非递减了
                }
                else {
    
      // 还没有连续非递增time天
                    couldAsUp4Begin = -1;
                }
            }
            else if (security[i] == security[i - 1]) {
    
      // 今天和昨天相等，也就是说既符合非递增又符合非递减
                lianXuXiaDays++;  // 符合非递增，连续非递增天数++
                if (couldAsUp4Begin != -1) {
    
      // 前面有≥time的非递减，并且从那天起没有递增的一天 | Lable1
                    if (i - couldAsUp4Begin >= time) {
    
      // 如果今天距离那天≥time，那天就是抢劫日
                        ans.push_back(couldAsUp4Begin);  // 先把抢劫日添加到答案中去
                        if (security[couldAsUp4Begin + 1] <= security[couldAsUp4Begin]) {
    
      // 如果抢劫日的下一天仍然是非递增，那么下一天之前肯定有至少time天的非递增
                            couldAsUp4Begin++;  // 下一天也可以作为开始非递减的一天
                        }
                        else {
    
      // 否则
                            couldAsUp4Begin = -1;  // 下一天＞这个抢劫日，说明下一天必不满足“前面有至少time天的非递增”
                        }
                    }
                }
                else {
    
      // couldAsUp4Begin = -1
                    if (lianXuXiaDays >= time) {
    
      // 连续非递增天数≥time
                        couldAsUp4Begin = i;  // 从今天起可以开始非递减了
                    }
                }
            }
            else {
    
      // 今 > 昨
                lianXuXiaDays = 0;  // 连续非递减天数中断
                if (couldAsUp4Begin != -1) {
    
      // 这个同理于上面的“Lable1”处
                    if (i - couldAsUp4Begin >= time) {
    
    
                        ans.push_back(couldAsUp4Begin);
                        if (security[couldAsUp4Begin + 1] <= securmai ty[coul AsUp4Begin])   {
    
                         couldA Up4B gectin++;
                           }
 已完成          else {
    
    
                  }
 couldA     }
        }
Up4Begin = -1;
                       }
                    }
                }
            }
        return ans;  // 返回答案即可
    }
};

Method Two

Time complexity $O (n)$ , space complexity O(n), difficulty※

This method is easier to implement than the previous method, but the space complexity is higher than the previous method.

We can use The time complexity of $O$ $($ $n$ $) is to find the "number of consecutive non-increasing days before" and "the number of consecutive non-decreasing days after" for each day.$

$x ia [i]$ display No. $There are non-increasing days before i$ day, $s han g [i]$ display No.How many days before $i day are non-decreasing$

Specific method: traverse the array from front to back, if 今天≤昨天 , then xia[i] = xia[i - 1] + 1 ; otherwise, xia[i] = 0 . Initial value xia[0] = 0 Traverse the array from back to front, if 今天≤昨天 , then shang[i] = shang[i + 1] + 1 ; otherwise, shang[i] = 0 . initial value shang[security.size() - 1] = 0

Then we traverse each day, if a day satisfies $xia[i]\geq time$ 和 $\geq time$ , this day is robbery day.

AC code

C++

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

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