LeetCode #2100 — MEDIUM

Find Good Days to Rob the Bank

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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 i^th day. The days are numbered starting from 0. You are also given an integer time.

The i^th day is a good day to rob the bank if:

There are at least time days before and after the i^th 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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

[5,3,3,3,5,6,2]
2

Example 2

[1,1,1,1,1]
0

Example 3

[1,2,3,4,5,6]
2

Core Insight

What unlocks the optimal approach

The trivial solution is to check the time days before and after each day. There are a lot of repeated operations using this solution. How could we optimize this solution?
We can use precomputation to make the solution faster.
Use an array to store the number of days before the i<sup>th</sup> day that is non-increasing, and another array to store the number of days after the i<sup>th</sup> day that is non-decreasing.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2100: Find Good Days to Rob the Bank
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;
    }
}

// Accepted solution for LeetCode #2100: Find Good Days to Rob the Bank
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
}

# Accepted solution for LeetCode #2100: Find Good Days to Rob the Bank
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])]

// Accepted solution for LeetCode #2100: Find Good Days to Rob the Bank
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
    }
}

// Accepted solution for LeetCode #2100: Find Good Days to Rob the Bank
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.