Composing the machines

Combining machines is where exam questions live. The recipe: decompose the prompt into named machines, then wire them in one pass.

"Return the longest String in words; if several tie for longest, return the earliest. Empty array returns null."

public static String longest(String[] words) {
    if (words.length == 0) { return null; }
    String best = words[0];
    for (int i = 1; i < words.length; i++) {
        if (words[i].length() > best.length()) {
            best = words[i];
        }
    }
    return best;
}

Decomposition: guard the vacuous case → seed from element 0 → scan with strict > so ties keep the earlier winner (Trap 1 + tie-breaking in one). Compare with >= and the answer silently becomes the last tie — a one-character difference the exam adores.

Second combination — "count how many elements are strictly between the first and last occurrence of key":

public static int between(int[] arr, int key) {
    int first = -1, last = -1;
    for (int i = 0; i < arr.length; i++) {
        if (arr[i] == key) {
            if (first == -1) { first = i; }
            last = i;
        }
    }
    if (first == -1) { return 0; }
    return last - first - 1;
}

Two sentinel scans in one loop (first stays, last overwrites), then a guard that also rejects a single occurrence (last == first means "nothing in between"), then index algebra. No nested loops, no second pass. When you can name the machines, "hard" problems assemble from parts you already own.