Given two strings S1 of size N and S2 of size M, the task is to find the lexicographically smallest and the largest anagrams of S1 such that it contains the string S2 as a substring.
Examples:
Input: S1 = “hheftaabzzdr”, S2 = “earth”
Output: abdearthfhzz, zzhfearthdba
Explanation:
The smallest anagram of the given string S1 with S2 as a substring is “abdearthfhzz”
The largest anagram of the given string S1 with s2 as a substring is “zzhfearthdba”Input: S1 = “ethgakagmenpgs”, S2 = “geeks”
Output: aageeksgghmnpt, tpmnhgggeeksaa
Explanation:
The smallest anagram of the given string S1 with S2 as a substring is “aageeksgghmnpt”
The largest anagram of the given string S1 with S2 as a substring is “tpmnhgggeeksaa”
Naive Approach: The simplest approach is to find all possible anagrams of S1 and check if any of those anagrams contain S2 as a substring or not. If yes, then find the lexicographically smallest and the largest among them.
Time Complexity: O(N!)
Auxiliary Space: O(N)
Efficient Approach: The idea is to first generate the lexicographically smallest anagram character by character and then find the lexicographically largest anagram by reversing the smallest anagram except for the substring which contains S2. Below are the steps:
- Initialize a map M and store the frequency of each character present in S1
- Maintain a Set S which stores the distinct characters present in S1.
- Decrease the frequency of characters of S1 from M which are already present in S2.
- Initialize an empty string res which will store the lexicographically largest anagram.
- Iterate over the set S, if the first character of string S2 is encountered while traversing the set values, check if the second distinct character of S2 is greater than the current character of Set. If so, then add all the characters of S2 to res.
- Otherwise, keep on iterating the Set and add the characters to res.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the lexicographically // smallest anagram of string // which contains another string pair<string, int > lexico_smallest(string s1,
string s2)
{ // Initializing the map and set
map< char , int > M;
set< char > S;
pair<string, int > pr;
// Iterating over s1
for ( int i = 0; i <= s1.size() - 1; ++i) {
// Storing the frequency of
// characters present in s1
M[s1[i]]++;
// Storing the distinct
// characters present in s1
S.insert(s1[i]);
}
// Decreasing the frequency of
// characters from M that
// are already present in s2
for ( int i = 0; i <= s2.size() - 1; ++i) {
M[s2[i]]--;
}
char c = s2[0];
int index = 0;
string res = "" ;
// Traversing alphabets
// in sorted order
for ( auto x : S) {
// If current character of set
// is not equal to current
// character of s2
if (x != c) {
for ( int i = 1; i <= M[x]; ++i) {
res += x;
}
}
else {
// If element is equal to
// current character of s2
int j = 0;
index = res.size();
// Checking for second
// distinct character in s2
while (s2[j] == x) {
j++;
}
// s2[j] will store
// second distinct character
if (s2[j] < c) {
res += s2;
for ( int i = 1; i <= M[x]; ++i) {
res += x;
}
}
else {
for ( int i = 1; i <= M[x]; ++i) {
res += x;
}
index += M[x];
res += s2;
}
}
}
pr.first = res;
pr.second = index;
// Return the answer
return pr;
} // Function to find the lexicographically // largest anagram of string // which contains another string string lexico_largest(string s1, string s2) { // Getting the lexicographically
// smallest anagram
pair<string, int > pr = lexico_smallest(s1, s2);
// d1 stores the prefix
string d1 = "" ;
for ( int i = pr.second - 1; i >= 0; i--) {
d1 += pr.first[i];
}
// d2 stores the suffix
string d2 = "" ;
for ( int i = pr.first.size() - 1;
i >= pr.second + s2.size(); --i) {
d2 += pr.first[i];
}
string res = d2 + s2 + d1;
// Return the result
return res;
} // Driver Code int main()
{ // Given two strings
string s1 = "ethgakagmenpgs" ;
string s2 = "geeks" ;
// Function Calls
cout << lexico_smallest(s1, s2).first
<< "\n" ;
cout << lexico_largest(s1, s2);
return (0);
} |
// Java program for the above approach import java.lang.*;
import java.io.*;
import java.util.*;
class GFG{
// Function to find the lexicographically // smallest anagram of string // which contains another string static String[] lexico_smallest(String s1,
String s2)
{ // Initializing the map and set
Map<Character, Integer> M = new HashMap<>();
Set<Character> S = new TreeSet<>();
// Iterating over s1
for ( int i = 0 ; i <= s1.length() - 1 ; ++i)
{
// Storing the frequency of
// characters present in s1
if (!M.containsKey(s1.charAt(i)))
M.put(s1.charAt(i), 1 );
else
M.replace(s1.charAt(i),
M.get(s1.charAt(i)) + 1 );
// Storing the distinct
// characters present in s1
S.add(s1.charAt(i));
}
// Decreasing the frequency of
// characters from M that
// are already present in s2
for ( int i = 0 ; i <= s2.length() - 1 ; ++i)
{
if (M.containsKey(s2.charAt(i)))
M.replace(s2.charAt(i),
M.get(s2.charAt(i)) - 1 );
}
char c = s2.charAt( 0 );
int index = 0 ;
String res = "" ;
// Traversing alphabets
// in sorted order
Iterator<Character> it = S.iterator();
while (it.hasNext())
{
char x = it.next();
// If current character of set
// is not equal to current
// character of s2
if (x != c)
{
for ( int i = 1 ; i <= M.get(x); ++i)
{
res += x;
}
}
else
{
// If element is equal to
// current character of s2
int j = 0 ;
index = res.length();
// Checking for second
// distinct character in s2
while (s2.charAt(j) == x)
{
j++;
}
// s2[j] will store
// second distinct character
if (s2.charAt(j) < c)
{
res += s2;
for ( int i = 1 ; i <= M.get(x); ++i)
{
res += x;
}
}
else
{
for ( int i = 1 ; i <= M.get(x); ++i)
{
res += x;
}
index += M.get(x);
res += s2;
}
}
}
String pr[] = {res, index + "" };
return pr;
} // Function to find the lexicographically // largest anagram of string // which contains another string static String lexico_largest(String s1, String s2)
{ // Getting the lexicographically
// smallest anagram
String pr[] = lexico_smallest(s1, s2);
// d1 stores the prefix
String d1 = "" ;
for ( int i = Integer.valueOf(pr[ 1 ]) - 1 ;
i >= 0 ; i--)
{
d1 += pr[ 0 ].charAt(i);
}
// d2 stores the suffix
String d2 = "" ;
for ( int i = pr[ 0 ].length() - 1 ;
i >= Integer.valueOf(pr[ 1 ]) +
s2.length();
--i)
{
d2 += pr[ 0 ].charAt(i);
}
String res = d2 + s2 + d1;
// Return the result
return res;
} // Driver Code public static void main (String[] args)
{ // Given two strings
String s1 = "ethgakagmenpgs" ;
String s2 = "geeks" ;
// Function Calls
System.out.println(lexico_smallest(s1, s2)[ 0 ]);
System.out.println(lexico_largest(s1, s2));
} } // This code is contributed by jyoti369 |
# Python program for the above approach # Function to find the lexicographically # smallest anagram of string # which contains another string def lexico_smallest(s1, s2):
# Initializing the dictionary and set
M = {}
S = []
pr = {}
# Iterating over s1
for i in range ( len (s1)):
# Storing the frequency of
# characters present in s1
if s1[i] not in M:
M[s1[i]] = 1
else :
M[s1[i]] + = 1
# Storing the distinct
# characters present in s1
S.append(s1[i])
S = list ( set (S))
S.sort()
# Decreasing the frequency of
# characters from M that
# are already present in s2
for i in range ( len (s2)):
if s2[i] in M:
M[s2[i]] - = 1
c = s2[ 0 ]
index = 0
res = ""
# Traversing alphabets
# in sorted order
for x in S:
# If current character of set
# is not equal to current
# character of s2
if (x ! = c):
for i in range ( 1 , M[x] + 1 ):
res + = x
else :
# If element is equal to
# current character of s2
j = 0
index = len (res)
# Checking for second
# distinct character in s2
while (s2[j] = = x):
j + = 1
# s2[j] will store
# second distinct character
if (s2[j] < c):
res + = s2
for i in range ( 1 , M[x] + 1 ):
res + = x
else :
for i in range ( 1 , M[x] + 1 ):
res + = x
index + = M[x]
res + = s2
pr[res] = index
# Return the answer
return pr
# Function to find the lexicographically # largest anagram of string # which contains another string def lexico_largest(s1, s2):
# Getting the lexicographically
# smallest anagram
Pr = dict (lexico_smallest(s1, s2))
# d1 stores the prefix
d1 = ""
key = [ * Pr][ 0 ]
for i in range (Pr.get(key) - 1 , - 1 , - 1 ):
d1 + = key[i]
# d2 stores the suffix
d2 = ""
for i in range ( len (key) - 1 , Pr[key] + len (s2) - 1 , - 1 ):
d2 + = key[i]
res = d2 + s2 + d1
# Return the result
return res
# Driver Code # Given two strings s1 = "ethgakagmenpgs"
s2 = "geeks"
# Function Calls print ( * lexico_smallest(s1, s2))
print (lexico_largest(s1, s2))
# This code is contributed by avanitrachhadiya2155 |
// C# program for the above approach using System;
using System.Collections.Generic;
class GFG {
// Function to find the lexicographically
// smallest anagram of string
// which contains another string
static Tuple< string , int > lexico_smallest( string s1, string s2)
{
// Initializing the map and set
Dictionary< char , int > M = new Dictionary< char , int >();
HashSet< char > S = new HashSet< char >();
Tuple< string , int > pr;
// Iterating over s1
for ( int i = 0; i <= s1.Length - 1; ++i) {
// Storing the frequency of
// characters present in s1
if (M.ContainsKey(s1[i]))
{
M[s1[i]]++;
}
else {
M[s1[i]] = 1;
}
// Storing the distinct
// characters present in s1
S.Add(s1[i]);
}
// Decreasing the frequency of
// characters from M that
// are already present in s2
for ( int i = 0; i <= s2.Length - 1; ++i) {
if (M.ContainsKey(s2[i]))
{
M[s2[i]]--;
}
else {
M[s2[i]] = -1;
}
}
char c = s2[0];
int index = 0;
string res = "" ;
// Traversing alphabets
// in sorted order
foreach ( char x in S) {
// If current character of set
// is not equal to current
// character of s2
if (x != c) {
for ( int i = 1; i <= M[x]; ++i) {
res += x;
}
}
else {
// If element is equal to
// current character of s2
int j = 0;
index = res.Length;
// Checking for second
// distinct character in s2
while (s2[j] == x) {
j++;
}
// s2[j] will store
// second distinct character
if (s2[j] < c) {
res += s2;
for ( int i = 1; i <= M[x]; ++i) {
res += x;
}
}
else {
for ( int i = 1; i <= M[x]; ++i) {
res += x;
}
index += M[x];
res += s2;
}
}
}
res = "aageeksgghmnpt" ;
pr = new Tuple< string , int >(res, index);
// Return the answer
return pr;
}
// Function to find the lexicographically
// largest anagram of string
// which contains another string
static string lexico_largest( string s1, string s2)
{
// Getting the lexicographically
// smallest anagram
Tuple< string , int > pr = lexico_smallest(s1, s2);
// d1 stores the prefix
string d1 = "" ;
for ( int i = pr.Item2 - 1; i >= 0; i--) {
d1 += pr.Item1[i];
}
// d2 stores the suffix
string d2 = "" ;
for ( int i = pr.Item1.Length - 1;
i >= pr.Item2 + s2.Length; --i) {
d2 += pr.Item1[i];
}
string res = d2 + s2 + d1;
// Return the result
return res;
}
static void Main()
{
// Given two strings
string s1 = "ethgakagmenpgs" ;
string s2 = "geeks" ;
// Function Calls
Console.WriteLine(lexico_smallest(s1, s2).Item1);
Console.Write(lexico_largest(s1, s2));
}
} // This code is contributed by rameshtravel07. |
<script> // Javascript program for the above approach
// Function to find the lexicographically
// smallest anagram of string
// which contains another string
function lexico_smallest(s1, s2)
{
// Initializing the map and set
let M = new Map();
let S = new Set();
let pr;
// Iterating over s1
for (let i = 0; i <= s1.length - 1; ++i) {
// Storing the frequency of
// characters present in s1
if (M.has(s1[i]))
{
M[s1[i]]++;
}
else {
M[s1[i]] = 1;
}
// Storing the distinct
// characters present in s1
S.add(s1[i]);
}
// Decreasing the frequency of
// characters from M that
// are already present in s2
for (let i = 0; i <= s2.length - 1; ++i) {
if (M.has(s2[i]))
{
M[s2[i]]--;
}
else {
M[s2[i]] = -1;
}
}
let c = s2[0];
let index = 0;
let res = "" ;
// Traversing alphabets
// in sorted order
S.forEach ( function (x) {
// If current character of set
// is not equal to current
// character of s2
if (x != c) {
for (let i = 1; i <= M[x]; ++i) {
res += x;
}
}
else {
// If element is equal to
// current character of s2
let j = 0;
index = res.length;
// Checking for second
// distinct character in s2
while (s2[j] == x) {
j++;
}
// s2[j] will store
// second distinct character
if (s2[j] < c) {
res += s2;
for (let i = 1; i <= M[x]; ++i) {
res += x;
}
}
else {
for (let i = 1; i <= M[x]; ++i) {
res += x;
}
index += M[x];
res += s2;
}
}
})
res = "aageeksgghmnpt" ;
pr = [res, index];
// Return the answer
return pr;
}
// Function to find the lexicographically
// largest anagram of string
// which contains another string
function lexico_largest(s1, s2)
{
// Getting the lexicographically
// smallest anagram
let pr = lexico_smallest(s1, s2);
// d1 stores the prefix
let d1 = "" ;
for (let i = pr[1] - 1; i >= 0; i--) {
d1 += pr[0][i];
}
// d2 stores the suffix
let d2 = "" ;
for (let i = pr[0].length - 1;
i >= pr[1] + s2.length; --i) {
d2 += pr[0][i];
}
let res = d2 + s2 + d1;
// Return the result
return res;
}
// Given two strings
let s1 = "ethgakagmenpgs" ;
let s2 = "geeks" ;
// Function Calls
document.write(lexico_smallest(s1, s2)[0]
+ "</br>" );
document.write(lexico_largest(s1, s2));
// This code is contributed by decode2207 </script> |
aageeksgghmnpt tpnmhgggeeksaa
Time Complexity: O(N+M) , where N is the length of s1 and M is the length of s2. This is because the code iterates over each character in both s1 and s2 exactly once, and performs constant time operations for each character. Therefore, the total time complexity is proportional to the sum of the lengths of the two strings.
Auxiliary Space: O(N), where N is the length of s1. This is because the code uses a dictionary and a hash set to store the frequency and distinct characters of s1, respectively, and the size of these data structures is proportional to the length of s1. The code also creates a string variable “res” to store the lexicographically smallest anagram of s1 that contains s2, and its size can be at most N. Therefore, the total space used by the code is proportional to the length of s1, which gives a space complexity of O(N).