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Commit 4f40051

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‎Binary_String.cpp‎

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/*
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link:- https://practice.geeksforgeeks.org/problems/binary-string-1587115620/1
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problem:
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Given a binary string S. The task is to count the number of substrings that start and end with 1.
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For example, if the input string is "00100101", then there are three substrings "1001", "100101" and "101".
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Input:
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N = 4
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S = 1111
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Output: 6
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Explanation: There are 6 substrings from
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the given string. They are 11, 11, 11,
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111, 111, 1111.
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Input:
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N = 5
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S = 01101
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Output: 3
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Explanation: There 3 substrings from the
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given string. They are 11, 101, 1101.
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*/
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class Solution
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{
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public:
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//Function to count the number of substrings that start and end with 1.
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long binarySubstring(int n, string a){
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// n = length of string , a = binary string
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int ans=0;
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sort(a.begin(),a.end());
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for(int i=0;i<n;i++)
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{
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if(a[i]!='0')
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{
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ans++;
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}
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//ans= count(a.begin(), a.end(), '1');
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}
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// return count;
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return (ans*(ans-1))/2;
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}
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};
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/*
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approch:-
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1.count the total number of 1's in the string ( using sort string and for loop)
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2. observe pattern if
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number of 1 is 1 then ans 1
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number of 1 is 2 then ans 2
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number of 1 is 3 then ans 3
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number of 1 is 4 then ans 6
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number of 1 is 5 then ans 10
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.
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.
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number of 1 is n then ans n*(n-1)/2
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*/
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‎Integer_To_Roman.cpp‎

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/*
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link:- https://practice.geeksforgeeks.org/problems/convert-to-roman-no/1
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problem:- covert integer to roman number
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Input:
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n = 5
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Output: V
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Input:
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n = 3
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Output: III
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*/
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class Solution{
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public:
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string convertToRoman(int n) {
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// code here
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int num[]={1,4,5,9,10,40,50,90,100,400,500,900,1000};
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string s[]={"I","IV","V","IX","X","XL","L","XC","C","CD","D","CM","M"};
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string ans;
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int i=12;
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while(n>0)
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{
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int div=n/num[i];
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n=n%num[i];
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while(div--)
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{
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ans=ans+s[i];
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}
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i--;
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}
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return ans;
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}
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};

‎anagram_make.cpp‎

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/*
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problem link:- https://practice.geeksforgeeks.org/problems/anagram-of-string/1
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problem statement:- Given two strings S1 and S2 in lowercase, the task is to make them anagram.
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The only allowed operation is to remove a character from any string.
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Find the minimum number of characters to be deleted to make both the strings anagram.
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Two strings are called anagram of each other if one of them can be converted into another by rearranging its letters.
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Example 1:
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Input:
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S1 = bcadeh
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S2 = hea
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Output: 3
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Explanation: We need to remove b, c
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and d from S1.
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Example 2:
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Input:
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S1 = cddgk
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S2 = gcd
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Output: 2
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Explanation: We need to remove d and
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k from S1.
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*/
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int remAnagram(string str1, string str2)
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{
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int m1[26]={0},m2[26]={0},res=0;
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for(int i=0;i<str1.length();i++)
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{
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m1[str1[i]-'a']++;
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}
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for(int i=0;i<str2.length();i++)
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{
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m2[str2[i]-'a']++;
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}
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for(int i=0;i<26;i++)
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{
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res=res+abs(m1[i]-m2[i]);
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}
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return res;
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}
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//Dry run
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/*
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s1=cddgk
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s2=gcd
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s1 map | s2 map | diff | res
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c-1 | c-1 | 0 | 0
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d-2 | d-1 | 1 | 1
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g-1 | g-1 | 0 | 0
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k-1 | | 1 | 1+1=2
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res=output=2
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*/

‎anagram_string_or_not.cpp‎

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/*
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link:- https://practice.geeksforgeeks.org/problems/anagram-1587115620/1
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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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only the order of characters can be different.
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For example, act and tac are an anagram of each other.
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Input: a = geeksforgeeks, b = forgeeksgeeks
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Output: YES
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Explanation: Both the string have same characters with
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same frequency. So, both are anagrams.
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*/
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class Solution
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{
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public:
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bool isAnagram(string a, string b){
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sort(a.begin(),a.end());
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sort(b.begin(),b.end());
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return a==b;
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}
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};

‎array_subset_of_another_sum.cpp‎

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/*
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link:- https://practice.geeksforgeeks.org/problems/array-subset-of-another-array2317/1
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problem:-
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Given two arrays: a1[0..n-1] of size n and a2[0..m-1] of size m. Task is to
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check whether a2[] is a subset of a1[] or not.
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Both the arrays can be sorted or unsorted.
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Input:
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a1[] = {11, 1, 13, 21, 3, 7}
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a2[] = {11, 3, 7, 1}
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Output:
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Yes
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Explanation:
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a2[] is a subset of a1[]
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Input:
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a1[] = {10, 5, 2, 23, 19}
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a2[] = {19, 5, 3}
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Output:
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No
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Explanation:
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a2[] is not a subset of a1[]
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*/
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string isSubset(int a1[], int a2[], int n, int m)
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{
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unordered_set<int> s(a1, a1 + n);
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for (int i = 0; i < n; i++)
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{
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s.insert(a1[i]);
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}
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int cnt = 0;
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// cout<< s.size();
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for (int i = 0; i < m; i++)
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{
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if (s.count(a2[i]))
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{
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cnt++;
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}
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if (cnt == m)
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{
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return "Yes";
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}
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}
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return "No";
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}
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/*
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OTHER APPROCH:-
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string isSubset(int a1[], int a2[], int n, int m) {
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vector<int>v1(a1,a1+n);
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vector<int>v2(a2,a2+m);
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int f=0;
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vector<int>::iterator it;
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int i=0;
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while(i!=v2.size())
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{
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it=find(v1.begin(),v1.end(),v2[i]);
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i++;
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if(it!=v1.end())
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{
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}
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else
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{
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f=1;
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}
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}
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if(f==1)
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{
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return "No";
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}
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else
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{
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return "Yes";
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}
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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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};

‎find_occurence_of_s2_in_s1.cpp‎

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/*
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link:- https://practice.geeksforgeeks.org/problems/implement-strstr/1
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Input:
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s = GeeksForGeeks, x = Fr
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Output: -1
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Explanation: Fr is not present in the
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string GeeksForGeeks as substring.
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*/
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//Function to locate the occurrence of the string x in the string s.
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int strstr(string s, string x)
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{
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//Your code here
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// char p=strstr(s,x);
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if(s.find(x)!=string::npos)
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{
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return s.find(x); //returns the index of the first occurence of x in s
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}
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else
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{
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return -1;
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}
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}

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