Can you solve this real interview question? Count Good Numbers - A digit string is good if the digits (0-indexed) at even indices are even and the digits at odd indices are prime (2, 3, 5, or 7). * For example, "2582" is good because the digits (2 and
Input:nums = [5,2,6,1]Output:[2,1,1,0]Explanation:To the right of 5 there are2smaller elements (2 and 1). To the right of 2 there is only1smaller element (1). To the right of 6 there is1smaller element (1). To the right of 1 there is0smaller element. Example 2: Input...
In this, i will post my solutions of POTD's of Leetcode - Leetcode/2057-count-salary-categories/2057-count-salary-categories.sql at main · raghav161/Leetcode
A set of practice note, solution, complexity analysis and test bench to leetcode problem set - leetcode/CountSubArrayFixBound.drawio at b58bcceb0ea27d0756ad72fb6a64b3b547fae221 · brianchiang-tw/leetcode
* are smaller that the given integer*/vector<int> countOfSmallerNumber(vector<int> &A, vector<int> &queries) {intmaxNo =0;if(A.size() >0) { maxNo= *max_element(A.begin(), A.end()); } ft.assign(maxNo+1,0);//memory complexity may not be goodfor(auto v : A) ...
的一个逆序对,也称作逆序数。 例题1 - LeetCode剑指 Offer 51. 数组中的逆序对 题目 在数组中的两个数字,如果前面一个数字大于后面的数字,则这两个数字组成一个逆序对。输入一个数组,求出这个数组中的逆序对的总数。 输入: [7,5,6,4] 输出: 5 ...
0-indexedinteger arraynumsof lengthnand an integerk, returnthenumber of pairs(i, j)such that: 0 <= i < j <= n - 1and nums[i] * nums[j]is divisible byk. Example 1: Input:nums = [1,2,3,4,5], k = 2Output:7Explanation:The 7 pairs of indices whose corresponding products are...