Computing XOR Over an Arithmetic Sequence

Given two integers n and start, an array is constructed using the rule arr[i] = start + 2 * i for zero-based indices, where n equals the length. The task is to return the bitwise XOR over all the elements.

A straightforward solution iterates through the range and accumulates the XOR result. No array alloctaion is required.

function computeXorSequence(n, start) {
    let result = 0;
    for (let idx = 0; idx < n; idx++) {
        result ^= start + 2 * idx;
    }
    return result;
}

The algorithm runs in O(n) time and uses O(1) extra space. Since constraints place n at most 1000, the linear approach is well within limits. Alternative implementations might precompute a filled array and reduce with XOR, but the explicit loop avoids unnecessary memory overhead.

Generating Distinct File Names with Minimal Suffixes

A list of requested folder names is provided, and each must be assigned an actual name that is unique at the moment of creation. If a name is already taken, the smallest positive integer k is appended in parentheses to form name(k); this check recurses if the suffixed variant also exists.

Context

Consider folders = ["alpha", "alpha", "beta"]. The outputs would be ["alpha", "alpha(1)", "beta"]. More complex cases involve existing "alpha(1)" entries causing later "alpha" to become "alpha(2)", or even "alpha(1)(1)" when the base plus suffix collides.

Implementation Strategy

A dictionary tracks the next candidate suffix for each base name, plus a set of all assigned names for quick existence checks. We process the list sequentially:

• If the requested name is not in the assigned set, use it directly and mark it.
• Otherwise, retrieve the stored counter for that base name. Increment the counter until a combination not present in the set is found. Assign the new name, update the counter for the base, and mark the suffixed name as used.

Be careful with names that already contain parentheses; they are treated as literal strings. The base for suffxiing is always the original requested name, not a stripped version.

function assignUniqueNames(names) {
    const used = new Set();
    const suffixCount = new Map();
    const result = [];

    for (let name of names) {
        if (!used.has(name)) {
            used.add(name);
            suffixCount.set(name, 1);
            result.push(name);
        } else {
            let count = suffixCount.get(name) ?? 1;
            let candidate;
            do {
                candidate = `${name}(${count})`;
                count++;
            } while (used.has(candidate));

            used.add(candidate);
            suffixCount.set(name, count);
            result.push(candidate);
        }
    }

    return result;
}

This method processes each name in O(k) amortized time, where k is the number of collisions for a given base. With the input length capped at 5 * 10^4 and short strings, it performs well. Using a Map and Set ensures fast lookups and updates.

Key Observations

• For the XOR problem, remembering that XOR with zero is identity allows safe initialization of the accumulator to 0.
• The file naming algorithm relies on tracking the smallest unused integer per base name and repeatedly probing untill a free variant is found. Memoizing the last assigned number avoids restarting from 1 on every collision.