[Leetcode] 321.Create Maximum Number

[Leetcode] 321.Create Maximum Number

[Description]

321. Create Maximum Number

[Analysis]

To create max number of length k from two arrays, you need to create max number of length i from array one and max number of length k-i from array two, then combine them together. After trying all possible i, you will get the max number created from two arrays.

[Code]

#include <iostream>
#include <vector>

using namespace std;

class Solution {
private:
    vector<vector<int>> helper(vector<int>& nums, int k) {
        vector<vector<int>> numsTopK(k + 1, vector<int>{});
        vector<int> tmp;
        for (int i = 0; i < nums.size(); i++) {
            for (int j = k; j > 0; j--) {
                if (numsTopK[j].size() < j) {
                    numsTopK[j].push_back(nums[i]);
                } else {
                    tmp = numsTopK[j - 1];
                    tmp.push_back(nums[i]);
                    if (compare(tmp, numsTopK[j])) {
                        numsTopK[j] = tmp;
                    }
                }
            }
        }
        return numsTopK;
    }

    vector<int> merge(vector<int>& nums1, vector<int>& nums2) {
        vector<int> result;
        int i = 0, j = 0;
        while (i < nums1.size() && j < nums2.size()) {
            if (compare(nums1, nums2, i, j)) {
                result.push_back(nums1[i]);
                i++;
            } else {
                result.push_back(nums2[j]);
                j++;
            }
        }
        while (i < nums1.size()) {
            result.push_back(nums1[i]);
            i++;
        }

        while (j < nums2.size()) {
            result.push_back(nums2[j]);
            j++;
        }
        return result;
    }

    bool compare(vector<int>& nums1, vector<int>& nums2, int i, int j) {
        while (i < nums1.size() && j < nums2.size()) {
            if (nums1[i] > nums2[j]) {
                return true;
            } else if (nums1[i] < nums2[j]) {
                return false;
            } else {
                i++;
                j++;
            }
        }
        if (i == nums1.size()) {
            return false;
        }
        return true;
    }

    bool compare(vector<int>& nums1, vector<int>& nums2) {
        if (nums1.size() > nums2.size()) {
            return true;
        } else if (nums1.size() < nums2.size()) {
            return false;
        } else {
            for (int i = 0; i < nums1.size(); i++) {
                if (nums1[i] > nums2[i]) {
                    return true;
                } else if (nums1[i] < nums2[i]) {
                    return false;
                }
            }
        }
        return true;
    }
public:
    vector<int> maxNumber(vector<int>& nums1, vector<int>& nums2, int k) {
        vector<vector<int>> nums1TopK = helper(nums1, k);
        vector<vector<int>> nums2TopK = helper(nums2, k);
        vector<int> maxResult;
        vector<int> tmp;
        for (int i = 0; i <= k; i++) {
            tmp = merge(nums1TopK[i], nums2TopK[k - i]);
            if (compare(tmp, maxResult)) {
                maxResult = tmp;
            }
        }
        return maxResult;
    }
};
int main() {
    vector<int> nums1{2,5,6,4,4,0};
    vector<int> nums2{7, 3, 8, 0, 6, 5, 7, 6, 2};
    Solution so;
    so.maxNumber(nums1, nums2, 15);
    return 0;
}

[Complexity]

Time Complexity: O(n * k)
Space Complexity: O( k^2 )

The solution is pretty simple so I will try to optimize the solution sooner.