defcount_substring(string,sub_string):count=0;foriinrange(len(sub_string),len(string)+1):if(string[i-len(sub_string):i]==sub_string):count+=1returncount count_substring():=0foriinrange(0,len(string)):if(string[i:i+len(sub_string)]==sub_string):count+=1print(count)s=input()s...
1a =raw_input().strip()2b =raw_input().strip()3cnt =0;4foriinxrange(len(a)):5cnt += 1ifa.find(b, i, i+len(b))!= -1else06printcnt
Output Format Number of digits in that number exactly divides that number. 题解: 1importjava.io.*;2importjava.util.*;345publicclassSolution {67publicstaticvoidmain(String[] args) {8Scanner in =newScanner(System.in);9intt =in.nextInt();10for(inti = 0; i < t; i++){11Long num =i...
Here’s a quick example. Let’s say you wanted to concatenate a string in C#, but you have no idea how. All you would have to do is search the query and hit enter. You’ll immediately get the accurate solution, as well as the opportunity to edit the code right in your search resu...
yashpalsinghdeo1 4 years ago here is problem solution in python 2 and python 3 programming. https://programs.programmingoneonone.com/2021/01/hackerrank-find-a-string-solution-python.html -8|ParentPermalink tamylv_pb 4 years ago bro is this your website? 0|ParentPermalinkBlog...
defcount_substring(string,sub_string):counter=0sub_len=len(sub_string)foriinrange(0,len(string)):ifstring[i]==sub_string[0]:ifstring[i:(i+sub_len)]==sub_string:counter=counter+1returncounter pphelan 8 years ago Very simple and elegant... like it. ...
题解:久闻Partition函数可以用来找到乱序数组的中位数,今天终于实现了一把。 设置一个变量need为数组长度的一半,另一个变量hasFound为当前比找到的比中位数小的数的个数,当hasFound=need的时候,我们就找到了中位数。 在每次Partition后,看比pivot小的那部分数组有多少个元素,如果hasFound加上这部分元素正好等于nee...
classSolution {public:/** * @param A: An integer matrix * @return: The index of the peak*/vector<int> findPeakII(vector<vector<int> >A) {intn =A.size();intm = A[0].size();inti =1, j =1;while(i < m -1&& j < n -1) ...
Find a string check this def count_substring(string, sub_string): count =0 for i in range(0,len(string)-len(sub_string)+1): try: if string[i:i+len(sub_string)]==sub_string: count+=1 except IndexError: break return count
classSolution {public:intfindDuplicate(vector<int>&nums) {intnp =nums.size();intn = np -1;ints =1, e =n;while(s<e) {intmid = (s + e) /2;intcnt =0;for(auto v : nums)if(v <= mid) cnt++;if(cnt >mid) { e=mid; ...