README_EN.md

2100. Find Good Days to Rob the Bank

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Description

You and a gang of thieves are planning on robbing a bank. You are given a 0-indexed integer array security, where security[i] is the number of guards on duty on the ith day. The days are numbered starting from 0. You are also given an integer time.

The ith day is a good day to rob the bank if:

• There are at least time days before and after the ith day,
• The number of guards at the bank for the time days before i are non-increasing, and
• The number of guards at the bank for the time days after i are non-decreasing.

More formally, this means day i is a good day to rob the bank if and only if security[i - time] >= security[i - time + 1] >= ... >= security[i] <= ... <= security[i + time - 1] <= security[i + time].

Return a list of all days (0-indexed) that are good days to rob the bank. The order that the days are returned in does not matter.

 

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 satisfy this condition, so days 2 and 3 are the only good days to rob the bank.

Example 2:

Input: security = [1,1,1,1,1], time = 0
Output: [0,1,2,3,4]
Explanation:
Since time equals 0, every day is a good day to rob the bank, so return every day.

Example 3:

Input: security = [1,2,3,4,5,6], time = 2
Output: []
Explanation:
No day has 2 days before it that have a non-increasing number of guards.
Thus, no day is a good day to rob the bank, so return an empty list.

Example 4:

Input: security = [1], time = 5
Output: []
Explanation:
No day has 5 days before and after it.
Thus, no day is a good day to rob the bank, so return an empty list.

 

Constraints:

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

Solutions

Python3

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])]

Java

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 = 0; i < n; ++i) {
            if (time <= Math.min(left[i], right[i])) {
                ans.add(i);
            }
        }
        return ans;
    }
}

C++

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 = 0; i < n; ++i)
            if (time <= min(left[i], right[i]))
                ans.push_back(i);
        return ans;
    }
};

Go

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 := 0; i < n; i++ {
		if time <= left[i] && time <= right[i] {
			ans = append(ans, i)
		}
	}
	return ans
}

TypeScript

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