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Commit 9a3378c

Browse files
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1 parent b2b3704 commit 9a3378c

17 files changed

+543
-21
lines changed

‎Decode_The_Message.cpp‎

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/* https://leetcode.com/problems/decode-the-message/ */
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class Solution {
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public:
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string decodeMessage(string key, string msg) {
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char c='a';
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unordered_map<char,char>m;
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for(int i=0;i<key.length();i++)
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{
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if(key[i]==' ' || m.find(key[i])!=m.end())
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{ continue;
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}
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else{
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m[key[i]]=c;
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c++;
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}
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}
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string ans;
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for(auto i:msg)
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{
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if(m.find(i)!=m.end() && i!=' ')
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{
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ans+= m[i];
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}
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else
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{
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ans+=" ";
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}
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}
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for(auto i:m)
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{
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cout<<i.first<<" " << i.second<<endl;
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}
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return ans;
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}
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};

‎Longest_common_prefix.cpp‎

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/*https://leetcode.com/problems/longest-common-prefix/description/*/
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/*optimal*/
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class Solution {
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public:
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string longestCommonPrefix(vector<string>& s) {
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sort(s.begin(),s.end());
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string s1=s[0];
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string s2=s[s.size()-1];
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string ans="";
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for(int i=0;i<s1.length();i++)
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{
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if(s1[i]==s2[i])
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{
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ans+=s1[i];
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}
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else
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{
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return ans;
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}
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}
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return ans;
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}
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};
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/*greedy */
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class Solution {
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public:
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string longestCommonPrefix(vector<string>& s) {
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if(s.empty()) return "";
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string temp=s[0];
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int mini=INT_MAX;
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string ans="";
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for(int i=1;i<s.size();i++)
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{ int count=0;
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for(int j=0;j<s[i].size();j++)
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{
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if(s[0][j]==s[i][j])
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{
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count++;
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}
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}
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mini=min(mini,count);
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}
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for(int i=0;i<mini;i++)
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{
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ans+=s[0][i];
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}
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return ans;
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}
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};
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‎Palindrome_String.cpp‎

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@@ -32,3 +32,33 @@ int isPalindrome(string s)
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}else
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return 1;
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}
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//**************************************** Using Recursion********************
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#include <iostream>
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using namespace std;
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bool f(int i,int n,string &s)
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{
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if(i>=n/2)
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{
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return true;
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}
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if(s[i]!=s[s.size()-i-1])
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{
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return false;
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}
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return f(i+1,n,s);
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}
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int main()
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{
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string s="ababa";
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int n=5;
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int i=0;
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cout<<f(i,n,s);
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return 0;
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}

‎Shuffle_string.cpp‎

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/* https://leetcode.com/problems/shuffle-string/ */
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class Solution {
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public:
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string restoreString(string s, vector<int>& indices) {
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int n=s.length();
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char ans[n];
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for(int i=0;i<indices.size();i++)
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{
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int occ=indices[i];
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ans[occ]=s[i];
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}
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string t;
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for(int i=0;i<s.length();i++)
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{
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t=t+ans[i];
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}
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return t;
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}
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};
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//t[index[i]]=s[i]; one step

‎aa.cpp‎

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/******************************************************************************
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Online C++ Compiler.
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Code, Compile, Run and Debug C++ program online.
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Write your code in this editor and press "Run" button to compile and execute it.
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*******************************************************************************/
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#include <iostream>
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using namespace std;
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int main()
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{ long long int num;
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int count1=0,count2=0;
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cin >> num;
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while(num!=1)
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{ if(num%2!=0)
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{
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num=(num*3) + 1;
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cout<<num<<" ";
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count1++;
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}
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else{
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num=num/2;
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cout<<num<<" ";
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count2++;
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}
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}
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cout<<" count total "<< count1+count2<< endl;
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return 0;
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}

‎aa.exe‎

51.2 KB
Binary file not shown.

‎anagram_string_or_not.cpp‎

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/*
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link:- https://practice.geeksforgeeks.org/problems/anagram-1587115620/1
4-
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https://leetcode.com/problems/valid-anagram/
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problem: Given two strings a and b consisting of lowercase characters.
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The task is to check whether two given strings are an anagram of each other or not.
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An anagram of a string is another string that contains the same characters,
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return a==b;
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}
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};
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/* using map*/
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class Solution {
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public:
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bool isAnagram(string s, string t) {
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map<char,int>m1;
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map<char,int>m2;
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for(int i=0;i<s.length();i++)
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{
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m1[s[i]]++;
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}
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for(int i=0;i<t.length();i++)
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{
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m2[t[i]]++;
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}
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return(m1==m2);
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}
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};
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/*https://leetcode.com/problems/check-if-all-characters-have-equal-number-of-occurrences/*/
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class Solution {
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public:
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bool areOccurrencesEqual(string s) {
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unordered_map<int,int>m;
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for(int i=0;i<s.length();i++)
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{
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m[s[i]]++;
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}
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vector<int>v;
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for(auto i=m.begin();i!=m.end();i++)
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{
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v.push_back(i->second);
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}
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for(int i=0;i<v.size();i++)
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{
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if(v[i]!=v[0])
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{
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return false;
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}
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}
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return true;
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}
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};
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/*
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link:- https://practice.geeksforgeeks.org/problems/check-if-strings-are-rotations-of-each-other-or-not-1587115620/1
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Given two strings s1 and s2. The task is to check if s2 is a rotated version of the string s1.
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The characters in the strings are in lowercase.
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Input:
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geeksforgeeks
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forgeeksgeeks
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Output:
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1
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Explanation: s1 is geeksforgeeks, s2 is
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forgeeksgeeks. Clearly, s2 is a rotated
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version of s1 as s2 can be obtained by
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left-rotating s1 by 5 units.
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Example 2:
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Input:
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mightandmagic
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andmagicmigth
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Output:
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0
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Explanation: Here with any amount of
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rotation s2 can't be obtained by s1.
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*/
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// approch:- do s1+s1 and find window of s2 in s1 and check if s2 is there or not.
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class Solution
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{
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public:
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//Function to check if two strings are rotations of each other or not.
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bool areRotations(string s1,string s2)
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{
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if(s1.length()!=s2.length())
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{
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return 0;
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}
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else
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{
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string s=s1+s1;
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if(s.find(s2)!=string::npos)
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return 1;
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else
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return 0;
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}
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}
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};

‎delete_column.cpp‎

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/*
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https://leetcode.com/problems/delete-columns-to-make-sorted/description/
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*/
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class Solution {
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public:
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int minDeletionSize(vector<string>& str) {
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int row=str.size();
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int column=str[0].size();
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int count=0;
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for (int j = 0; j < column; j++) {
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for (int i = 0; i < row-1; i++) {
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if(str[i][j]>str[i+1][j])
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{
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count++;
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break;
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}
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}
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}
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return count;
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}
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};

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