The O(n) Club: How Many Ways Can You Lose Count? (A Java Dev’s Guide to Combination Sum IV)

The O(n) Club: How Many Ways Can You Lose Count? (A Java Dev’s Guide to Combination Sum IV)

⚡ TL;DR

You want to count the number of ordered ways to reach a target using unlimited repeats of numbers from nums—think less chef’s choice, more fast-food “Any Combo, Any Order.” Stop brute-forcing (unless you enjoy debugging while yelling into your keyboard):

// Pure brute-force (so bad, it hurts):
int combo(int[] nums, int target) {
    if (target == 0) return 1;
    if (target < 0) return 0;
    int total = 0;
    for (int n : nums) total += combo(nums, target - n);
    return total;
}

🧠 How Devs Usually Mess This Up

Poor souls confuse “combination” with “set” and start deduping everything, or think you can use each number only once (nope). Some even code up the full permutation list in memory, at which point their laptop’s fans are legally allowed to file for overtime pay. It’s chaos, but not the fun kind. The real kicker? Forgetting that 1+2 is not the same as 2+1—surprise, your answer is less than half of what it should be.

🔍 What’s Actually Going On

Imagine trying to pay for overpriced coffee using any number of coins, in any order, as many times as you want. Each sequence—regardless of order—is a new way to hand over money (or mess up the barista). Here, we’re living that dream mathematically: For every value from 1 to target, consider each coin (number), and tally up all possible builds for the remaining amount. Use a one-dimensional DP array, dp[i], that records the number of ordered permutations that sum to i. dp[0]=1 (doing nothing is a valid strategy, at least here). Work your way up the barista’s bill—er, the target.

🛠️ PseudoCode

• Start with dp[0] = 1; because using nothing is the one thing we all do right.
• For each sum i from 1 to target:
• For every x in nums:
• If x ≤ i, add dp[i - x] to dp[i] (tack this number on the end of every smaller solution—you know, like adding one more shot to your espresso… again).
• Return dp[target] (AKA, how many times you nearly lose count while paying).

// Pseudo-Java style, for maximum clarity
int[] dp = new int[target + 1];
dp[0] = 1;
for (int i = 1; i <= target; i++) {
    for (int num : nums) {
        if (num <= i) {
            dp[i] += dp[i - num];
        }
    }
}
return dp[target];

💻 The Code

public int combinationSum4(int[] nums, int target) {
    int[] dp = new int[target + 1];
    dp[0] = 1;
    for (int i = 1; i <= target; i++) {
        for (int num : nums) {
            if (num <= i) {
                dp[i] += dp[i - num];
            }
        }
    }
    return dp[target];
}

⚠️ Pitfalls, Traps & Runtime Smacks

• Order matters! Miss this, and you’ll undercount hopelessly.
• Repetition is allowed. You can use each number a billion times (or until you run out of patience).
• Integer overflow? Try a long[] if your combos surge past Integer.MAX_VALUE (rare, but bugs always pick worst times).
• Negative targets are illegal—you’ll loop forever. Read the constraints, please.
• Complexity: It’s O(target × N) time and O(target) space. Respectable for something that counts so many ways to confuse an interviewer.

🧠 Interviewer Brain-Tattoo

“In permutation chaos, order isn’t just important—it’s the entire question.” Or: Always ask if order matters, or you’ll be ordering takeout instead of getting hired.