The O(n) Club: Burst Balloons: When DP Pops and Padding Saves Your Sanity

The O(n) Club: Burst Balloons—When DP Pops and Padding Saves Your Sanity

⚡ TL;DR

Maximum coins by popping balloons—but if you think popping the fattest (or skinniest) ones first is clever, re-read the problem. Brute force is O(n!), interval DP is O(n³), and here’s both, side by side:

// Please don't brute force:
int maxCoins(int[] nums) {
    // tries every explosion order—bring snacks
    return dfs(new ArrayList<>(), nums);
}
// Interval DP saves careers:
// dp[i][j]: max coins from balloons[i]..balloons[j]

🧠 How Devs Usually Mess This Up

Every newbie’s gremlin says, “Just pop the biggest balloon!” Or, “Just sweep from left to right!” That gremlin also formats production servers. The adjacent balloons change on every pop, so greedy fails, and hardcoding the order is just asking for LeetCode to slap you.
One tragic classic: forgetting to pad the array with 1s at both ends. Those little sentinels keep your edge cases from turning into existential interviews.

🔍 What’s Actually Going On

Picture balloons as dominoes: every burst ripples through its neighbors. For each subinterval, ask, “Which balloon should I pop last for the fattest profit?” This splits the subproblem and cleverly lets your left and right intervals just do their thing, then get stitched up. You basically host startup demos in each subsegment—and the boundaries are guarded by the bouncer 1s.
Welcome to interval DP, starring you, a pile of balloons, and enough padding to survive any recursion cliff.

🛠️ PseudoCode

• Pad with 1s: Add 1 to both ends, so edge pops don’t go haywire.

  // Example: nums = [3,1,5,8] → balloons = [1,3,1,5,8,1]

• Build dp[i][j]: Max coins from index i to j (exclusive).
• Loop for lengths 2 to n:
  • For every (left, right) interval:
  • Try every position k as “last to burst” in this interval
  • Update:

    for (int len = 2; len < n; len++) {
        for (int left = 0; left < n - len; left++) {
            int right = left + len;
            for (int k = left + 1; k < right; k++) {
                dp[left][right] = Math.max(dp[left][right],
                    balloons[left] * balloons[k] * balloons[right] +
                    dp[left][k] + dp[k][right]);
            }
        }
    }

  • Translation: Pop k last, scoop up coins, join with both sides’ best results, and repeat. Yes, O(n³) means you have time to refill your coffee.
• Profit: dp[0][n-1] is the answer. Interviewer sighs in impressed defeat.

💻 The Code

public int maxCoins(int[] nums) {
    int n = nums.length + 2;
    int[] balloons = new int[n];
    balloons[0] = balloons[n - 1] = 1;
    for (int i = 0; i < nums.length; i++) balloons[i + 1] = nums[i];
    int[][] dp = new int[n][n];
    for (int len = 2; len < n; len++) {
        for (int left = 0; left < n - len; left++) {
            int right = left + len;
            for (int k = left + 1; k < right; k++) {
                dp[left][right] = Math.max(dp[left][right],
                  balloons[left] * balloons[k] * balloons[right]
                    + dp[left][k] + dp[k][right]);
            }
        }
    }
    return dp[0][n - 1];
}

⚠️ Pitfalls, Traps & Runtime Smacks

• Forget the 1s at both ends? Your DP table will cry and so will you.
• Off-by-one errors in intervals? Welcome to debugging limbo.
• Trying to outsmart O(n³) time? Unless you have a supercomputer or a time machine, just relax and let DP cook.
• Space is O(n²), so pray you don’t get a LeetCode case with a billion balloons.

🧠 Interviewer Brain-Tattoo

“If popping order matters, let each subarray answer: Who gets popped last? And always bring your extra 1s—they’re the coffee for tired intervals.”