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2100. Find Good Days to Rob the Bank

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.

 

Constraints:

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

Solutions

Solution 1

Thinking

Checking each candidate day \(i\) by scanning \(\textit{time}\) days to the left and right costs \(O(\textit{time})\) per index. With \(n\) and \(\textit{time}\) both up to \(10^5\), that approach does not pass.

The required non-increasing / non-decreasing runs can be accumulated along the array: if \(\textit{security}[i]\le \textit{security}[i-1]\), the leftward non-increasing length at \(i\) is one more than at \(i-1\); otherwise it resets. The rightward non-decreasing length is symmetric. Thus both sides can be obtained in two linear passes.

We therefore keep arrays \(\textit{left}\) and \(\textit{right}\), then collect indices with \(\min(\textit{left}[i],\textit{right}[i])\ge \textit{time}\). If \(n\le 2\cdot\textit{time}\), no day can have \(\textit{time}\) neighbors on both sides, so the answer is empty.

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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])]

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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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
    }
}

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