コード例 #1
0
/** Find the Missing Number
 *
 * @reference   https://www.geeksforgeeks.org/find-the-missing-number/
 *
 * You are given a list of n-1 integers and these integers are in the range of 1 to n.
 * There are no duplicates in list. One of the integers is missing in the list.
 * Write an efficient code to find the missing integer.
 *
 * @complexity  O(n)
 */
auto FindTheMissingNumberSum(const ArrayType &integers) {
    assert(not integers.empty());

    const auto N = integers.size() + 1u;
    const auto sum = N * (N + 1) / 2;
    return std::accumulate(integers.cbegin(), integers.cend(), sum, std::minus<ArrayType::value_type> {});
}
コード例 #2
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auto FindTheMissingNumberXor(const ArrayType &integers) {
    assert(not integers.empty());

    const auto N = integers.size() + 1u;
    auto xor_of_all = integers[0];
    for (ArrayType::size_type i = 1ul; i < integers.size(); ++i) {
        xor_of_all ^= (i ^ integers[i]);
    }

    return xor_of_all ^ N ^ integers.size();
}
コード例 #3
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std::pair<int, int> FindSubarrayWithGivenSum(const ArrayType &integers,
        const ArrayType::value_type SUM) {
    assert(not integers.empty());

    auto start = integers.cbegin();
    auto current_sum = *start;
    for (auto i = start + 1; i != integers.cend(); ++i) {
        while (current_sum > SUM and start < i - 1) {
            current_sum -= *start++;
        }

        if (current_sum == SUM) {
            return std::make_pair(start - integers.cbegin(), i - integers.cbegin() - 1);
        }

        current_sum += *i;
    }

    return NOT_FOUND;
}