LeetCode #983 — MEDIUM

Minimum Cost For Tickets

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

The Problem

Problem Statement

You have planned some train traveling one year in advance. The days of the year in which you will travel are given as an integer array days. Each day is an integer from 1 to 365.

Train tickets are sold in three different ways:

a 1-day pass is sold for costs[0] dollars,
a 7-day pass is sold for costs[1] dollars, and
a 30-day pass is sold for costs[2] dollars.

The passes allow that many days of consecutive travel.

For example, if we get a 7-day pass on day 2, then we can travel for 7 days: 2, 3, 4, 5, 6, 7, and 8.

Return the minimum number of dollars you need to travel every day in the given list of days.

Example 1:

Input: days = [1,4,6,7,8,20], costs = [2,7,15]
Output: 11
Explanation: For example, here is one way to buy passes that lets you travel your travel plan:
On day 1, you bought a 1-day pass for costs[0] = $2, which covered day 1.
On day 3, you bought a 7-day pass for costs[1] = $7, which covered days 3, 4, ..., 9.
On day 20, you bought a 1-day pass for costs[0] = $2, which covered day 20.
In total, you spent $11 and covered all the days of your travel.

Example 2:

Input: days = [1,2,3,4,5,6,7,8,9,10,30,31], costs = [2,7,15]
Output: 17
Explanation: For example, here is one way to buy passes that lets you travel your travel plan:
On day 1, you bought a 30-day pass for costs[2] = $15 which covered days 1, 2, ..., 30.
On day 31, you bought a 1-day pass for costs[0] = $2 which covered day 31.
In total, you spent $17 and covered all the days of your travel.

Constraints:

1 <= days.length <= 365
1 <= days[i] <= 365
days is in strictly increasing order.
costs.length == 3
1 <= costs[i] <= 1000

Step 01

Brute Force Baseline

Problem summary: You have planned some train traveling one year in advance. The days of the year in which you will travel are given as an integer array days. Each day is an integer from 1 to 365. Train tickets are sold in three different ways: a 1-day pass is sold for costs[0] dollars, a 7-day pass is sold for costs[1] dollars, and a 30-day pass is sold for costs[2] dollars. The passes allow that many days of consecutive travel. For example, if we get a 7-day pass on day 2, then we can travel for 7 days: 2, 3, 4, 5, 6, 7, and 8. Return the minimum number of dollars you need to travel every day in the given list of days.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming

Example 1

[1,4,6,7,8,20]
[2,7,15]

Example 2

[1,2,3,4,5,6,7,8,9,10,30,31]
[2,7,15]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #983: Minimum Cost For Tickets
class Solution {
    private final int[] valid = {1, 7, 30};
    private int[] days;
    private int[] costs;
    private Integer[] f;
    private int n;

    public int mincostTickets(int[] days, int[] costs) {
        n = days.length;
        f = new Integer[n];
        this.days = days;
        this.costs = costs;
        return dfs(0);
    }

    private int dfs(int i) {
        if (i >= n) {
            return 0;
        }
        if (f[i] != null) {
            return f[i];
        }
        f[i] = Integer.MAX_VALUE;
        for (int k = 0; k < 3; ++k) {
            int j = Arrays.binarySearch(days, days[i] + valid[k]);
            j = j < 0 ? -j - 1 : j;
            f[i] = Math.min(f[i], dfs(j) + costs[k]);
        }
        return f[i];
    }
}

// Accepted solution for LeetCode #983: Minimum Cost For Tickets
func mincostTickets(days []int, costs []int) int {
	valid := [3]int{1, 7, 30}
	n := len(days)
	f := make([]int, n)
	var dfs func(int) int
	dfs = func(i int) int {
		if i >= n {
			return 0
		}
		if f[i] > 0 {
			return f[i]
		}
		f[i] = 1 << 30
		for k := 0; k < 3; k++ {
			j := sort.SearchInts(days, days[i]+valid[k])
			f[i] = min(f[i], dfs(j)+costs[k])
		}
		return f[i]
	}
	return dfs(0)
}

# Accepted solution for LeetCode #983: Minimum Cost For Tickets
class Solution:
    def mincostTickets(self, days: List[int], costs: List[int]) -> int:
        @cache
        def dfs(i: int) -> int:
            if i >= n:
                return 0
            ans = inf
            for c, v in zip(costs, valid):
                j = bisect_left(days, days[i] + v)
                ans = min(ans, c + dfs(j))
            return ans

        n = len(days)
        valid = [1, 7, 30]
        return dfs(0)

// Accepted solution for LeetCode #983: Minimum Cost For Tickets
struct Solution;

impl Solution {
    fn mincost_tickets(days: Vec<i32>, costs: Vec<i32>) -> i32 {
        let n = days.len();
        let mut dp: Vec<i32> = vec![];
        let pass = vec![1, 7, 30];
        for i in 0..n {
            let mut mins = costs.clone();
            for k in 0..3 {
                for j in (0..i).rev() {
                    if days[i] - days[j] >= pass[k] {
                        mins[k] += dp[j];
                        break;
                    }
                }
            }
            dp.push(*mins.iter().min().unwrap());
        }
        dp[n - 1]
    }
}

#[test]
fn test() {
    let days = vec![1, 4, 6, 7, 8, 20];
    let costs = vec![2, 7, 15];
    let res = 11;
    assert_eq!(Solution::mincost_tickets(days, costs), res);
    let days = vec![1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 30, 31];
    let costs = vec![2, 7, 15];
    let res = 17;
    assert_eq!(Solution::mincost_tickets(days, costs), res);
}

// Accepted solution for LeetCode #983: Minimum Cost For Tickets
function mincostTickets(days: number[], costs: number[]): number {
    const n = days.length;
    const f: number[] = Array(n).fill(0);
    const valid: number[] = [1, 7, 30];
    const search = (x: number): number => {
        let [l, r] = [0, n];
        while (l < r) {
            const mid = (l + r) >> 1;
            if (days[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
    const dfs = (i: number): number => {
        if (i >= n) {
            return 0;
        }
        if (f[i]) {
            return f[i];
        }
        f[i] = Infinity;
        for (let k = 0; k < 3; ++k) {
            const j = search(days[i] + valid[k]);
            f[i] = Math.min(f[i], dfs(j) + costs[k]);
        }
        return f[i];
    };
    return dfs(0);
}

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 × log n)

Space

O(n)

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.