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Problem solution · C++

Lexicographically Maximum Mex Array

Lexicographically Maximum Mex Array: a C++ solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Binary search
Source
Kamyu LeetCode Solutions
Length
100 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Binary search

For Lexicographically Maximum Mex Array, the implementation exploits a monotonic condition to discard half of the remaining search space after every check.

  1. Identify the ordered answer range or sorted search domain.
  2. Write a predicate whose truth changes only once.
  3. Move the appropriate boundary after each midpoint check and return the final feasible position.

Code notes

  • 100 lines of C++ from the credited upstream file lexicographically-maximum-mex-array.cpp.
  • The implementation visibly relies on sequence storage.
  • 8 loop blocks detected.

Complexity

Multiply the logarithmic number of midpoint checks by the cost of one predicate evaluation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeLexicographically Maximum Mex Array · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(n)// Space: O(n) // hash table, prefix sum, greedyclass Solution {public:    vector<int> maximumMEX(vector<int>& nums) {        int ver = -1;        vector<int> lookup(size(nums), ver);        vector<int> suffix(size(nums));        ++ver;        int mex = 0;        for (int i = size(nums) - 1; i >= 0; --i) {            if (nums[i] < size(lookup)) {                lookup[nums[i]] = ver;            }            while (mex < size(lookup) && lookup[mex] == ver) {                ++mex;            }            suffix[i] = mex;        }        vector<int> result;        ++ver;        mex = 0;        int j = 0;        for (int i = 0; i < size(nums); ++i) {            if (!suffix[j]) {                break;            }            if (nums[i] < size(lookup)) {                lookup[nums[i]] = ver;            }            while (mex < size(lookup) && lookup[mex] == ver) {                ++mex;            }            if (mex != suffix[j]) {                continue;            }            result.emplace_back(mex);            ++ver;            mex = 0;            j = i + 1;        }        result.resize(size(result) + (size(nums) - j));        return result;    }}; // Time:  O(n)// Space: O(n)// hash table, freq table, greedyclass Solution2 {public:    vector<int> maximumMEX(vector<int>& nums) {        int ver = -1;        vector<int> lookup(size(nums), ver);        vector<int> cnt(size(nums));        ++ver;        int mex = 0;        for (int i = 0; i < size(nums); ++i) {            if (nums[i] < size(lookup)) {                lookup[nums[i]] = ver;                ++cnt[nums[i]];            }            while (mex < size(lookup) && lookup[mex] == ver) {                ++mex;            }        }        int suffix = mex, new_suffix = mex;        vector<int> result;        ++ver;        mex = 0;        int j = 0;        for (int i = 0; i < size(nums); ++i) {            if (!suffix) {                break;            }            if (nums[i] < size(lookup)) {                lookup[nums[i]] = ver;                if (!--cnt[nums[i]] && nums[i] < new_suffix) {                    new_suffix = nums[i];                }            }            while (mex < size(lookup) && lookup[mex] == ver) {                ++mex;            }            if (mex != suffix) {                continue;            }            result.emplace_back(mex);            ++ver;            mex = 0;            j = i + 1;            suffix = new_suffix;        }        result.resize(size(result) + (size(nums) - j));        return result;    }};

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