Weekly and Biweekly Contests | November | Leetcode

Biweekly Contest 92

2481. Minimum Cuts to Divide a Circle

A valid cut in a circle can be:

• A cut that is represented by a straight line that touches two points on the edge of the circle and passes through its center, or
• A cut that is represented by a straight line that touches one point on the edge of the circle and its center.

Some valid and invalid cuts are shown in the figures below.

Given the integer n, return the minimum number of cuts needed to divide a circle into n equal slices.

    int numberOfCuts(int n) {
        if(n == 1) return 0;
        return n&1 ? n : n/2;
    }

6277. Difference Between Ones and Zeros in Row and Column

You are given a 0-indexed $m x n$ binary matrix grid.

A 0-indexed $m x n$ difference matrix diff is created with the following procedure:

• Let the number of ones in the $i_{th}$ row be $onesRow_i$.
• Let the number of ones in the $j_{th}$ column be $onesCol_j$.
• Let the number of zeros in the $i_{th}$ row be $zerosRow_i$.
• Let the number of zeros in the $j_{th}$ column be $zerosCol_j$.
• $diff[i][j] = onesRowi + onesColj - zerosRowi - zerosColj$

Return the difference matrix diff.

    vector<vector<int>> onesMinusZeros(vector<vector<int>>& grid) {
        vector<int> row(grid.size());
        vector<int> col(grid[0].size());
        for(int i = 0; i < grid.size(); ++i){
            for(int j = 0; j < grid[0].size(); ++j){
                row[i] += grid[i][j];
                col[j] += grid[i][j];
            }
        }
        vector<vector<int>> diff(grid.size(), vector<int> (grid[0].size()));
        for(int i = 0; i < grid.size(); ++i){
            for(int j = 0; j < grid[0].size(); ++j){
                diff[i][j] = 2*row[i] + 2*col[j] - grid.size() - grid[0].size();
            }
        }
        return diff;
    }

Weekly Contest 321

2485. Find the Pivot Integer

Given a positive integer n, find the pivot integer x such that:

• The sum of all elements between 1 and x inclusively equals the sum of all elements between x and n inclusively.

Return the pivot integer x. If no such integer exists, return -1. It is guaranteed that there will be at most one pivot index for the given input.

    int pivotInteger(int n) {
        int ans = (n * n + n ) /2;
        int sq = sqrt(ans);
        if(sq * sq == ans)return sq;
        else return -1;
    }

2486. Append Characters to String to Make Subsequence

You are given two strings s and t consisting of only lowercase English letters.

Return the minimum number of characters that need to be appended to the end of s so that t becomes a subsequence of s.

A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.

    int appendCharacters(string s, string t) {
        int s_len = s.size(), t_len = t.size();
        int i = 0, j = 0;
        while(i < s_len && j < t_len){
            if(s[i] == t[j]) {i++; j++;}
            else i++;
        }
        return t_len - j;
    }

2487. Remove Nodes From Linked List

You are given the head of a linked list.

Remove every node which has a node with a strictly greater value anywhere to the right side of it.

Return the head of the modified linked list.

    ListNode* removeNodes(ListNode* head) {
        
        if(!head || !head->next) return head;
        
        ListNode* next = removeNodes(head->next);
        
        if(next->val > head->val) return next;
        head->next = next;
        return head;
        
    }