LeetCode weekly contest 139

Rank	Name	Score	Finish Time	Q1 (4)	Q2 (5)	Q3 (6)	Q4 (8)
855 / 3985	YoungForest	10	1:03:50	0:53:00	1:03:50	solved after the contest	null

When I got up on Sunday, it was already past 11, so I joined the contest 40 minutes late. I solved the first two problems smoothly. I took some detours on the third problem and only solved it after the contest. If there had been enough time, solving the third problem should not have been an issue.

1071. Greatest Common Divisor of Strings

Intuition:
This problem is equivalent to finding the greatest common divisor of two numbers.
The length of Greatest Common Divisor must be equal to the greatest common divisor or 0.
A simple proof is as follows:
Suppose the answer length is x. Since str1 is composed of x, x must be a divisor of str1.length. Similarly, it is also a divisor of str2.length.
If x is not the greatest common divisor but can form both str1 and str2, then the greatest common divisor must also be able to form both str1 and str2.

Time complexity: O(log N)
Space complexity: O(N)

class Solution {
    int gcd(int a, int b) {
        if (b > a) {
            swap(a, b);
        }
        while (b > 0) {
            auto temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }
public:
    string gcdOfStrings(string str1, string str2) {
        int len1 = str1.size();
        int len2 = str2.size();
        auto common_len = gcd(len1, len2);
        auto t1 = str1.substr(0, common_len);
        auto t2 = str2.substr(0, common_len);
        return t1 == t2 ? t1 : "";
    }
};

1072. Flip Columns For Maximum Number of Equal Rows

Intuition:
Find the maximum number of complementary rows.

Time complexity: O(matrix.length * matrix[0].length)
Space complexity: O(matrix.length * matrix[0].length)

class Solution {
public:
    int maxEqualRowsAfterFlips(vector<vector<int>>& matrix) {
        unordered_map<string, int> count;
        int ret = 0;
        for (const auto& row : matrix) {
            string right, complete;
            for (int item : row) {
                right.push_back(item + '0');
                complete.push_back(((~item) & 1) + '0');
            }
            ++count[right];
            ++count[complete];
            ret = max({ret, count[right], count[complete]});
        }
        return ret;
    }
};

1073. Adding Two Negabinary Numbers

Intuition:
Imitate binary addition. The difference is that the carry may be -1, 1, or 0.

Time complexity: O(max(arr.size(), arr.size()))
Space complexity: O(max(arr.size(), arr.size()))

class Solution {
    void update(int& digit, int& carry) {
        if (digit == -1) {
            digit = 1;
            carry = 1;
        } else if (digit == 2) {
            carry = -1;
            digit = 0;
        } else if (digit == 3) {
            carry = -1;
            digit = 1;
        } else {
            carry = 0;
        }
    }
public:
    vector<int> addNegabinary(vector<int>& arr1, vector<int>& arr2) {
        int carry = 0;
        vector<int> ret;
        int i, j;
        for (i = arr1.size() - 1, j = arr2.size() - 1; i >= 0 && j >= 0; --i, --j) {
            int digit = arr1[i] + arr2[j] + carry;
            update(digit, carry);
            ret.push_back(digit);
        }
        for (; i >= 0; --i) {
            int digit = arr1[i] + carry;
            update(digit, carry);
            ret.push_back(digit);
        }
        for (; j >= 0; --j) {
            int digit = arr2[j] + carry;
            update(digit, carry);
            ret.push_back(digit);
        }
        if (carry == 1)
            ret.push_back(1);
        else if (carry == -1) {
            ret.push_back(1);
            ret.push_back(1);
        }
        while(ret.size() > 1 && ret.back() == 0)
            ret.pop_back();
        reverse(ret.begin(), ret.end());
        
        return ret;
    }
};

1074. Number of Submatrices That Sum to Target