The O(n) Club: Minimum Cost for Tickets (aka, The ‘Why Can’t I Just Be Greedy’ Problem)

The O(n) Club: Minimum Cost for Tickets (aka, The ‘Why Can’t I Just Be Greedy’ Problem)

⚡ TL;DR

Given a random set of travel days, what’s the cheapest blend of 1-day, 7-day, and 30-day passes to cover them? If you try every combo, you’ll have more recursive calls than actual travel days.

// Brute force: Recursively try every pass option for each trip day
int minCost(int[] days, int[] costs, int i) {
  if (i >= days.length) return 0;
  int use1 = costs[0] + minCost(days, costs, i + 1);
  int j = i;
  while (j < days.length && days[j] < days[i] + 7) j++;
  int use7 = costs[1] + minCost(days, costs, j);
  j = i;
  while (j < days.length && days[j] < days[i] + 30) j++;
  int use30 = costs[2] + minCost(days, costs, j);
  return Math.min(use1, Math.min(use7, use30));
}

🧠 How Devs Usually Mess This Up

Raise your hand if you ever just grabbed the cheapest pass every time you saw a travel day. Yeah—put your hand down, we’ve all done it. Trouble is: passes overlap. Greedy buying means you’ll end up paying more when a big bundle would’ve done the trick. Some folks also try to process every boring day of the year, even the ones spent binge-watching Java tutorials (O(365) facepalm). And honestly, if you don’t skip days already paid for? You’re double-paying like your landlord asked extra for April. Don’t.

🔍 What’s Actually Going On

Imagine you’re a commuter in a city where transit passes come in flavors: 1-day (cheap, expires fast), 7-day (a splurge, covers a week), 30-day (the ticket for grownups). The trick: For each trip day, decide—do you splurge today to avoid repeat spending, or pinch pennies and buy short-term? You want minimum cash left at the end and zero repeats. That’s dynamic programming: at each decision, remember what the cheapest plan from here looks like. Memoization means you stop recalculating the same bad week over and over—your CPU thanks you with fewer existential crises.

🛠️ PseudoCode

• Start with a function dp(i): min cost needed to cover days[i...].
• If i is at the end, cost is 0—finally, a vacation.
• For each pass:
  • 1-day: Add its cost, solve dp(i + 1).
  • 7-day: Find the next day not covered by 7 days, call dp(next).
  • 30-day: Same—jump ahead to the first uncovered day after 30.
• Memoize results. No one likes reruns, especially your RAM.
• Return the cheapest route. Congrats, you’re now a ticket optimization machine.

💻 The Code

// Java—Dynamic Programming with Memoization
class Solution {
  public int mincostTickets(int[] days, int[] costs) {
    Integer[] memo = new Integer[days.length];
    return dp(0, days, costs, memo);
  }
  private int dp(int i, int[] days, int[] costs, Integer[] memo) {
    if (i >= days.length) return 0;
    if (memo[i] != null) return memo[i];
    int cost1 = costs[0] + dp(i + 1, days, costs, memo);
    int j = i;
    while (j < days.length && days[j] < days[i] + 7) j++;
    int cost7 = costs[1] + dp(j, days, costs, memo);
    j = i;
    while (j < days.length && days[j] < days[i] + 30) j++;
    int cost30 = costs[2] + dp(j, days, costs, memo);
    return memo[i] = Math.min(cost1, Math.min(cost7, cost30));
  }
}

⚠️ Pitfalls, Traps & Runtime Smacks

• Cluster bomb days vs. scattered days: All together, all apart, or a mish-mash—test all combos.
• Forgetting to leap over covered days: Don’t crawl day-by-day. After a weekly or monthly splurge, jump to the next day not covered or you’ll bleed recursion and cash.
• Complexity reality check: O(# travel days). Memo array matches days.length. Faster than your morning caffeine hitting your brain.

🧠 Interviewer Brain-Tattoo

“Always buying the cheapest ticket is a great way to go broke AND get wet. Sometimes you need to splurge once—your wallet (and code) will thank you later.”