intstringReduction(intn,conststring&s){vector<int>occurrence(3);for(charc:s)occurrence[c-'a']++;if(occurrence[0]==noroccurrence[1]==noroccurrence[2]==n)returnn;if(occurrence[0]%2==occurrence[1]%2andoccurrence[1]%2==occurrence[2]%2)return2;elsereturn1;}intmain(){intt;strings;cin>...
HackerRank - "String Modification" An ACM-level problem because it involves "advanced maths". It should not be marked as "Moderate". Other than that, it is a medium level DP one. *Math modeling in your mind: can you see through the problem statement and figure out that it isa DP-...
I checked Editorials, understand it and then rewrite their Jave code into my solution in C++ (it looks almost the same). #include <cmath>#include<cstdio>#include<vector>#include#include<set>#include<unordered_set>#include<string>#include<climits>#include<iostream>#include<algorithm>#include<un...
My working solution is here: def parse_input(): string = raw_input() return string def check_string(string): check_funs = [str.isalnum, str.isalpha, str.isdigit, str.islower, str.isupper, ] return [any(fun(char) for char in string) for fun in check_funs] def print_output...
};classSolution {public:stringrearrangeString(stringstr,intk) {if(k <2)returnstr;intn =str.length();//cntintmaxcnt =0; vector<int> cnt(26);for(auto c : str) { cnt[c-'a'] ++; maxcnt= max(maxcnt, cnt[c -'a']);
To trigger a specific action, I need to identify the last character of a string and input it intoif. Essentially, I need to determine if the last character is "A," "B," or "C." How can I accomplish this? Solution 1: Apply the string technique labeled asendswith. ...
怎么在添加背景图片string description = "Basic Vertex Lighting with a Texture"; //--- float4x4 viewProjection : View 分享1赞 山斯野人吧 albo6494650 【纯水】hackerrank记录最近摸鱼不敢瞎摸,只能写代码,所以玩玩hackerrank 随便镇楼 分享20赞 异尘浩劫吧 lv_kyo1 异尘浩劫秘籍戏中,按f11 , 输入下列密码, ...
classSolution {public:boolhasSameLetters(strings1,strings2) { std::sort(s1.begin(), s1.end()); std::sort(s2.begin(), s2.end());returns1.compare(s2) ==0; }boolisScramble(strings1,strings2) { size_t len1=s1.length(); size_t len2=s2.length();if(len1 != len2)returnfalse;...
Fun one. A matter of string generation by given rules. I know it can be much shorter.. but i'm lazy to do that. classSolution {public:intmagicalString(intn) {if(!n)return0;if(n <2)return1;intret =1;strings ="1";boolisOne =false;inti =0;//num inxwhile(s.length() <n) ...
classSolution {public:boolisInterleave(strings1,strings2,strings3) { size_t len1=s1.length(); size_t len2=s2.length(); size_t len3=s3.length();if((len1 + len2) != len3)returnfalse;//Initvector<bool> ret; ret.resize(len3 +1); ...