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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,29 @@ | ||
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| def homework(n, arr): | ||
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| pos = dict() | ||
| for i, num in enumerate(arr): | ||
| pos[num] = i | ||
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| cnt = 0 | ||
| sorted_arr = sorted(arr) | ||
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| for i in range(len(arr)): | ||
| if arr[i] != sorted_arr[i]: | ||
| cnt += 1 | ||
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| new_idx = pos[sorted_arr[i]] | ||
| pos[arr[i]] = new_idx | ||
| arr[i], arr[new_idx] = arr[new_idx], arr[i] | ||
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| return cnt | ||
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| # input the number of elements of the array | ||
| n = int(input().strip()) | ||
| # input elements of array | ||
| arr = list(map(int, input().strip().split())) | ||
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| asc = homework(n, arr) | ||
| desc = homework(n, list(reversed(arr))) | ||
| # print the minimum number of swaps needed to make the array beautiful | ||
| print(min(asc, desc)) |
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,31 @@ | ||
| /* | ||
| We could use a greedy approach and starting from the | ||
| left everytime we see a 010 replace the last 0 with a 1 | ||
| and continue | ||
| */ | ||
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| import java.io.*; | ||
| import java.util.*; | ||
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| public class Solution { | ||
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| public static void main(String[] args) { | ||
| // Read input from STDIN and Print output to STDOUT | ||
| Scanner input = new Scanner(System.in); | ||
| int n = input.nextInt(); // length of binary string | ||
| input.nextLine(); | ||
| String s = input.nextLine(); // single binary string of length n | ||
| int switches = 0; | ||
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| for(int i = 0; i < s.length()-2; i++) // left to right character of binary string | ||
| { | ||
| if(s.charAt(i) == '0' && s.charAt(i+1) == '1' && s.charAt(i+2) == '0') | ||
| { | ||
| switches++; | ||
| i += 2; | ||
| } | ||
| } | ||
| // minimum number of steps needed to make the string beautiful | ||
| System.out.println(switches); | ||
| } | ||
| } |
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,35 @@ | ||
| # input the number of digits and the maximum number of changes allowed | ||
| n,k=map(int,input().split(' ')) | ||
| # input the n-digit string of numbers | ||
| num=input() | ||
| ss=list(num) | ||
| ch=[False for i in range(n)] | ||
| offset=0 | ||
| cc=0 | ||
| # print a string representation of the highest value achievable or -1 | ||
| for i in range(n>>1): | ||
| if ord(ss[i])>ord(ss[n-1-i]): | ||
| ss[n-1-i]=ss[i] | ||
| ch[n-1-i]=True | ||
| cc+=1 | ||
| elif ord(ss[i])<ord(ss[n-1-i]): | ||
| ss[i] = ss[n-1-i] | ||
| ch[i]=True | ||
| cc+=1 | ||
| if cc>k: | ||
| print(-1) | ||
| else: | ||
| k-=cc | ||
| i=0 | ||
| while k and i<(n>>1): | ||
| if ss[i]!='9': | ||
| if ch[i] or ch[n-i-1]: | ||
| ss[i]=ss[n-1-i]='9' | ||
| k-=1 | ||
| elif k>=2: | ||
| ss[i] = ss[n - 1 - i] = '9' | ||
| k -= 2 | ||
| i+=1 | ||
| if n&1 and k: | ||
| ss[(n>>1)]='9' | ||
| print(''.join(ss)) |
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,49 @@ | ||
| import java.io.*; | ||
| import java.math.*; | ||
| import java.security.*; | ||
| import java.text.*; | ||
| import java.util.*; | ||
| import java.util.concurrent.*; | ||
| import java.util.regex.*; | ||
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| public class Solution { | ||
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| // makingAnagrams function | ||
| static int makingAnagrams(String s1, String s2) { | ||
| int[] freqs1 = new int[26]; | ||
| int[] freqs2 = new int[26]; | ||
| int deletion = 0; | ||
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| for(int i = 0; i < s1.length(); i++) | ||
| freqs1[s1.charAt(i)-'a'] = freqs1[s1.charAt(i)-'a'] + 1; //Increment frequency of char at i | ||
| for(int i = 0; i < s2.length(); i++) | ||
| freqs2[s2.charAt(i)-'a'] = freqs2[s2.charAt(i)-'a'] + 1; //Increment frequency of char at i | ||
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| for(int i = 0; i < 26; i++) | ||
| deletion += Math.abs(freqs1[i] - freqs2[i]); //Track the total deletions needed | ||
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| // minimum number of characters which must be deleted to make the two strings anagrams of each other | ||
| return deletion; | ||
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| } | ||
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| private static final Scanner scanner = new Scanner(System.in); | ||
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| // main function | ||
| public static void main(String[] args) throws IOException { | ||
| BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH"))); | ||
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| String s1 = scanner.nextLine(); // first string | ||
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| String s2 = scanner.nextLine(); // second string | ||
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| int result = makingAnagrams(s1, s2); // function call | ||
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| bufferedWriter.write(String.valueOf(result)); | ||
| bufferedWriter.newLine(); | ||
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| bufferedWriter.close(); | ||
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| scanner.close(); | ||
| } | ||
| } |
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,11 @@ | ||
| #!/bin/python3 | ||
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| import sys | ||
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| # input number of strings | ||
| n = int(input().strip()) | ||
| for a0 in range(n): | ||
| # input the strings | ||
| s = input().strip() | ||
| # print minimum cost of constructing new string | ||
| print(len(set(s))) |
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,52 @@ | ||
| import java.io.*; | ||
| import java.math.*; | ||
| import java.security.*; | ||
| import java.text.*; | ||
| import java.util.*; | ||
| import java.util.concurrent.*; | ||
| import java.util.regex.*; | ||
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| public class Solution { | ||
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| // theLoveLetterMystery function | ||
| static int theLoveLetterMystery(String s) { | ||
| int count = 0; | ||
| int i = 0; // pointer at starting of string s | ||
| int j = s.length() - 1; // pointer at end of string s | ||
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| // check for palindromic string | ||
| while (i <= j) { | ||
| if (s.charAt(i) != s.charAt(j)) { | ||
| count += Math.abs(((int) s.charAt(i)) - ((int) s.charAt(j))); | ||
| } | ||
| i++; j--; | ||
| } | ||
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| // minimum number of operations corresponding to each test case | ||
| return count; | ||
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| } | ||
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| private static final Scanner scanner = new Scanner(System.in); | ||
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| // main function | ||
| public static void main(String[] args) throws IOException { | ||
| BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH"))); | ||
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| int q = scanner.nextInt(); // number of queries | ||
| scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?"); | ||
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| for (int qItr = 0; qItr < q; qItr++) { | ||
| String s = scanner.nextLine(); // string | ||
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| int result = theLoveLetterMystery(s); | ||
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| bufferedWriter.write(String.valueOf(result)); | ||
| bufferedWriter.newLine(); | ||
| } | ||
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| bufferedWriter.close(); | ||
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| scanner.close(); | ||
| } | ||
| } |
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,37 @@ | ||
| #include <iostream> | ||
| #include <set> | ||
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| using namespace std ; | ||
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| int main () | ||
| { | ||
| // input the number of options to perform | ||
| int n ; | ||
| cin >> n ; | ||
| set < string > s ; | ||
| set < string > :: iterator it ; | ||
| while ( n -- ) | ||
| { | ||
| // input the operations | ||
| string a , b ; | ||
| cin >> a >> b ; | ||
| if ( a == "add" ) | ||
| { | ||
| s.insert ( b ) ; | ||
| } | ||
| else | ||
| { | ||
| it = s.lower_bound ( b ) ; | ||
| int counter = 0 ; | ||
| int len = b.length () ; | ||
| while ( it != s.end () && b == (*it).substr ( 0 , len ) ) | ||
| { | ||
| counter ++ ; | ||
| it ++ ; | ||
| } | ||
| // print the number of contact names starting with input string | ||
| cout << counter << endl ; | ||
| } | ||
| } | ||
| return 0 ; | ||
| } |
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,73 @@ | ||
| #include <cmath> | ||
| #include <cstdio> | ||
| #include <vector> | ||
| #include <iostream> | ||
| #include <algorithm> | ||
| #include <string> | ||
| using namespace std; | ||
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| // Trie node | ||
| struct Node { | ||
| char data; | ||
| vector<Node*> nexts = vector<Node*> {}; | ||
| Node(char c){ | ||
| data = c; | ||
| } | ||
| }; | ||
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| // The trie class | ||
| class Trie { | ||
| public: | ||
| bool repeater = false; | ||
| Node* head = new Node('z'); | ||
| void insert(const string& s){ | ||
| bool working = false; | ||
| Node* tmp = head; | ||
| for(auto it=s.begin(); it < s.end(); ++it) { | ||
| vector<Node*>::iterator vit; | ||
| for(vit = tmp->nexts.begin(); vit < tmp->nexts.end(); ++vit) { | ||
| if((*vit)->data == *it){ | ||
| tmp = *vit; | ||
| break; | ||
| } | ||
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| } | ||
| if(tmp->nexts.size() == 0 && tmp != head && !working) { | ||
| repeater = true; | ||
| return; | ||
| } | ||
| if(vit == tmp->nexts.end()){ | ||
| working = true; | ||
| auto vv = new Node(*it); | ||
| tmp->nexts.push_back(vv); | ||
| tmp = vv; | ||
| } | ||
| if(it==s.end()-1 && tmp->nexts.size() > 0){ | ||
| repeater = true; | ||
| return; | ||
| } | ||
| } | ||
| } | ||
| }; | ||
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| // driver function to test the above functions | ||
| int main() { | ||
| // input number of strings in the set | ||
| int x; | ||
| cin >> x; | ||
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| auto trie = Trie(); | ||
| string s; | ||
| //print GOOD SET if the set is valid else, print BAD SET followed by the first string for which the condition fails | ||
| for(int i = 0; i < x ; ++i) { | ||
| // input the string | ||
| cin >> s; | ||
| trie.insert(s); | ||
| if(trie.repeater){ | ||
| cout << "BAD SET" << "\n" << s << endl; | ||
| return 0; | ||
| } | ||
| } | ||
| cout << "GOOD SET" << endl; | ||
| return 0; | ||
| } |