Check if two strings are k-anagrams or not

Given two strings of lowercase alphabets and a value k, the task is to find if two strings are K-anagrams of each other or not.
Two strings are called k-anagrams if following two conditions are true. 

1. Both have same number of characters.
2. Two strings can become anagram by changing at most k characters in a string.

Examples :

Input:  str1 = "anagram" , str2 = "grammar" , k = 3
Output:  Yes
Explanation: We can update maximum 3 values and 
it can be done in changing only 'r' to 'n' 
and 'm' to 'a' in str2.
Input:  str1 = "geeks", str2 = "eggkf", k = 1
Output:  No
Explanation: We can update or modify only 1 
value but there is a need of modifying 2 characters. 
i.e. g and f in str 2.

BRUTE METHOD:(USING MAP)

Intuition:

1. We create a Map<Character,Integer> to store the frequency-map of the string.
2. We push all the character of string 1 in the map.
3. Then we iterate through the string 2 and if the character if present ,we decrease it value by 1.
4. At-last, we iterate over the map and store the frequency of the remaining  characters in a variable count and check if its greater than K, we return false else return true. 

Implementation:

C++

#include <iostream>
#include <unordered_map>
using namespace std;

bool areKAnagrams(string s1, string s2, int k)
{
    if (s1.length() != s2.length()) {
        return false;
    }

    unordered_map<char, int> map;
    for (int i = 0; i < s1.length(); i++) {
        char ch = s1[i];
        map[ch]++;
    }

    for (int i = 0; i < s2.length(); i++) {
        char ch = s2[i];
        if (map[ch] > 0) {
            map[ch]--;
        }
    }

    int count = 0;
    for (auto it = map.begin(); it != map.end(); it++) {
        count += it->second;
    }

    if (count > k)
        return false;
    else
        return true;
}

int main()
{
    string str1 = "anagram";
    string str2 = "grammar";
    int k = 2;
    if (areKAnagrams(str1, str2, k))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;

    return 0;
}

// This code is contributed by Prajwal Kandekar

Java

// Java program to check if two strings are k anagram
// or not.

import java.io.*;
import java.util.*;

class GFG {
    public static boolean areKAnagrams(String s1, String s2, int k)
    {
        // code here
        if (s1.length() != s2.length()) {
            return false;
        }

        Map<Character, Integer> map = new HashMap<>();
        for (int i = 0; i < s1.length(); i++) {
            char ch = s1.charAt(i);
            map.put(ch, map.getOrDefault(ch, 0) + 1);
        }

        for (int i = 0; i < s2.length(); i++) {
            char ch = s2.charAt(i);
            if (map.getOrDefault(ch, 0) > 0) {
                map.put(ch, map.get(ch) - 1);
            }
        }

        int count = 0;
        for (char ch : map.keySet()) {
            count += map.get(ch);
        }

        if (count > k)
            return false;
        else
            return true;
    }
    public static void main(String[] args)
    {
        String str1 = "anagram";
        String str2 = "grammar";
        int k = 2;
        if (areKAnagrams(str1, str2, k))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
// This code is contributed by Raunak Singh

Python3

# Python program to check if two strings are k anagram
# or not.

from collections import defaultdict

def are_k_anagrams(s1, s2, k):
    if len(s1) != len(s2):
        return False

    count = defaultdict(int)
    for ch in s1:
        count[ch] += 1

    for ch in s2:
        if count[ch] > 0:
            count[ch] -= 1

    diff_count = sum(count.values())

    if diff_count > k:
        return False
    else:
        return True

str1 = "anagram"
str2 = "grammar"
k = 2

if are_k_anagrams(str1, str2, k):
    print("Yes")
else:
    print("No")


# This code is contrubuted by Vishal Dhaygude

C#

using System;
using System.Collections.Generic;

class GFG
{
    static bool AreKAnagrams(string s1, string s2, int k)
    {
        if (s1.Length != s2.Length)
        {
            return false;
        }

        Dictionary<char, int> map = new Dictionary<char, int>();
        foreach (char ch in s1)
        {
            if (map.ContainsKey(ch))
            {
                map[ch]++;
            }
            else
            {
                map[ch] = 1;
            }
        }

        foreach (char ch in s2)
        {
            if (map.ContainsKey(ch) && map[ch] > 0)
            {
                map[ch]--;
            }
        }

        int count = 0;
        foreach (var pair in map)
        {
            count += pair.Value;
        }

        return count <= k;
    }

    static void Main()
    {
        string str1 = "anagram";
        string str2 = "grammar";
        int k = 2;
        if (AreKAnagrams(str1, str2, k))
        {
            Console.WriteLine("Yes");
        }
        else
        {
            Console.WriteLine("No");
        }
    }
}

Javascript

function areKAnagrams(s1, s2, k) {
    if (s1.length !== s2.length) {
        return false;
    }

    const map = new Map();
    for (let i = 0; i < s1.length; i++) {
        const ch = s1[i];
        if (map.has(ch)) {
            map.set(ch, map.get(ch) + 1);
        } else {
            map.set(ch, 1);
        }
    }

    for (let i = 0; i < s2.length; i++) {
        const ch = s2[i];
        if (map.has(ch) && map.get(ch) > 0) {
            map.set(ch, map.get(ch) - 1);
        }
    }

    let count = 0;
    for (const [ch, freq] of map.entries()) {
        count += freq;
    }

    return count <= k;
}

const str1 = "anagram";
const str2 = "grammar";
const k = 2;
if (areKAnagrams(str1, str2, k)) {
    console.log("Yes");
} else {
    console.log("No");
}

Output

Yes

• Time Complexity: O(N)
• Auxiliary Space: O(N), since we are using a frequency map.

Method 1:

Below is a solution to check if two strings are k-anagrams of each other or not. 

1. Stores occurrence of all characters of both strings in separate count arrays.
2. Count number of different characters in both strings (in this if a string has 4 a and second has 3 ‘a’ then it will be also counted.
3. If count of different characters is less than or equal to k, then return true else false.

Implementation:

C++

// C++ program to check if two strings are k anagram
// or not.
#include<bits/stdc++.h>
using namespace std;
const int MAX_CHAR = 26;

// Function to check that string is k-anagram or not
bool arekAnagrams(string str1, string str2, int k)
{
    // If both strings are not of equal
    // length then return false
    int n = str1.length();
    if (str2.length() != n)
        return false;

    int count1[MAX_CHAR] = {0};
    int count2[MAX_CHAR] = {0};

    // Store the occurrence of all characters
    // in a hash_array
    for (int i = 0; i < n; i++)
        count1[str1[i]-'a']++;
    for (int i = 0; i < n; i++)
        count2[str2[i]-'a']++;

    int count = 0;

    // Count number of characters that are
    // different in both strings
    for (int i = 0; i < MAX_CHAR; i++)
        if (count1[i] > count2[i])
            count = count + abs(count1[i]-count2[i]);

    // Return true if count is less than or
    // equal to k
    return (count <= k);
}

// Driver code
int main()
{
    string str1 = "anagram";
    string str2 = "grammar";
    int k = 2;
    if (arekAnagrams(str1, str2, k))
        cout << "Yes";
    else
        cout<< "No";
    return 0;
}

C

#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

const int MAX_CHAR = 26;

// Function to check if str1 and str2 are k-anagrams or not
bool arekAnagrams(char str1[], char str2[], int k)
{
    // If both strings are not of equal length, then return
    // false
    int n = strlen(str1);
    if (strlen(str2) != n)
        return false;

    int count1[MAX_CHAR];
    int count2[MAX_CHAR];

    // Initialize count1 and count2 arrays with zero
    for (int i = 0; i < MAX_CHAR; i++) {
        count1[i] = 0;
        count2[i] = 0;
    }

    // Store the occurrence of all characters in count1 and
    // count2 arrays
    for (int i = 0; i < n; i++)
        count1[str1[i] - 'a']++;
    for (int i = 0; i < n; i++)
        count2[str2[i] - 'a']++;

    int count = 0;

    // Count the number of characters that are different in
    // both strings
    for (int i = 0; i < MAX_CHAR; i++)
        if (count1[i] > count2[i])
            count = count + abs(count1[i] - count2[i]);

    // Return true if count is less than or equal to k
    return (count <= k);
}

// Driver code
int main()
{
    char str1[] = "anagram";
    char str2[] = "grammar";
    int k = 2;

    if (arekAnagrams(str1, str2, k))
        printf("Yes\n");
    else
        printf("No\n");

    return 0;
}

Java

// Java program to check if two strings are k anagram
// or not.
public class GFG {

    static final int MAX_CHAR = 26;

    // Function to check that string is k-anagram or not
    static boolean arekAnagrams(String str1, String str2,
                                                 int k)
    {
        // If both strings are not of equal
        // length then return false
        int n = str1.length();
        if (str2.length() != n)
            return false;

        int[] count1 = new int[MAX_CHAR];
        int[] count2 = new int[MAX_CHAR];
        int count = 0;

        // Store the occurrence of all characters
        // in a hash_array
        for (int i = 0; i < n; i++)
            count1[str1.charAt(i) - 'a']++;
        for (int i = 0; i < n; i++)
            count2[str2.charAt(i) - 'a']++;

        // Count number of characters that are
        // different in both strings
        for (int i = 0; i < MAX_CHAR; i++)
            if (count1[i] > count2[i])
                count = count + Math.abs(count1[i] -
                                          count2[i]);

        // Return true if count is less than or
        // equal to k
        return (count <= k);
    }

    // Driver code
    public static void main(String args[])
    {
        String str1 = "anagram";
        String str2 = "grammar";
        int k = 2;
        if (arekAnagrams(str1, str2, k))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
// This code is contributed by Sumit Ghosh

Python3

# Python3 program to check if two
# strings are k anagram or not.
MAX_CHAR = 26

# Function to check that is
# k-anagram or not
def arekAnagrams(str1, str2, k) :

    # If both strings are not of equal
    # length then return false
    n = len(str1)
    if (len(str2)!= n) :
        return False

    count1 = [0] * MAX_CHAR
    count2 = [0] * MAX_CHAR

    # Store the occurrence of all
    # characters in a hash_array
    for i in range(n):
        count1[ord(str1[i]) -
               ord('a')] += 1
    for i in range(n):
        count2[ord(str2[i]) -
               ord('a')] += 1

    count = 0

    # Count number of characters that
    # are different in both strings
    for i in range(MAX_CHAR):
        if (count1[i] > count2[i]) :
            count = count + abs(count1[i] -
                                count2[i])

    # Return true if count is less
    # than or equal to k
    return (count <= k)

# Driver Code
if __name__ == '__main__':
    str1 = "anagram"
    str2 = "grammar"
    k = 2
    if (arekAnagrams(str1, str2, k)):
        print("Yes")
    else:
        print("No")

# This code is contributed
# by SHUBHAMSINGH10

C#

// C# program to check if two
// strings are k anagram or not.
using System;
class GFG {

    static int MAX_CHAR = 26;

    // Function to check that
    // string is k-anagram or not
    static bool arekAnagrams(string str1,
                             string str2,
                                      int k)
    {

        // If both strings are not of equal
        // length then return false
        int n = str1.Length;
        if (str2.Length != n)
            return false;

        int[] count1 = new int[MAX_CHAR];
        int[] count2 = new int[MAX_CHAR];
        int count = 0;

        // Store the occurrence
        // of all characters
        // in a hash_array
        for (int i = 0; i < n; i++)
            count1[str1[i] - 'a']++;
        for (int i = 0; i < n; i++)
            count2[str2[i] - 'a']++;

        // Count number of characters that are
        // different in both strings
        for (int i = 0; i < MAX_CHAR; i++)
            if (count1[i] > count2[i])
                count = count + Math.Abs(count1[i] -
                                         count2[i]);

        // Return true if count is
        // less than or equal to k
        return (count <= k);
    }

    // Driver code
    public static void Main()
    {
        string str1 = "anagram";
        string str2 = "grammar";
        int k = 2;
        if (arekAnagrams(str1, str2, k))
            Console.Write("Yes");
        else
            Console.Write("No");
    }
}
// This code is contributed by nitin mittal.

Javascript

<script>

// Javascript program to check if two
// strings are k anagram or not.
let MAX_CHAR = 26;

// Function to check that string is
// k-anagram or not
function arekAnagrams(str1, str2, k)
{

    // If both strings are not of equal
    // length then return false
    let n = str1.length;
    if (str2.length != n)
        return false;

    let count1 = new Array(MAX_CHAR);
    let count2 = new Array(MAX_CHAR);
    let count = 0;

    // Store the occurrence of all characters
    // in a hash_array
    for(let i = 0; i < n; i++)
        count1[str1[i].charCodeAt(0) -
                   'a'.charCodeAt(0)]++;
    for(let i = 0; i < n; i++)
        count2[str2[i].charCodeAt(0) -
                   'a'.charCodeAt(0)]++;

    // Count number of characters that are
    // different in both strings
    for(let i = 0; i < MAX_CHAR; i++)
        if (count1[i] > count2[i])
            count = count + Math.abs(count1[i] -
                                     count2[i]);

    // Return true if count is less than or
    // equal to k
    return (count <= k);
}

// Driver code
let str1 = "anagram";
let str2 = "grammar";
let k = 2;

if (arekAnagrams(str1, str2, k))
    document.write("Yes");
else
    document.write("No");

// This code is contributed by rag2127

</script>

PHP

<?php
// PHP program to check
// if two strings are
// k anagram or not.
$MAX_CHAR = 26;

// Function to check that
// string is k-anagram or not
function arekAnagrams($str1, $str2, $k)
{
    global $MAX_CHAR;
    // If both strings are not of
    // equal length then return false
    $n = strlen($str1);
    if (strlen($str2) != $n)
        return false;

    $count1 = (0);
    $count2 = (0);

    // Store the occurrence of all
    // characters in a hash_array
    $count = 0;

    // Count number of characters that
    // are different in both strings
    for ($i = 0; $i < $MAX_CHAR; $i++)
        if ($count1[$i] > $count2[$i])
            $count = $count + abs($count1[$i] -
                                  $count2[$i]);

    // Return true if count is
    // less than or equal to k
    return ($count <= $k);
}

// Driver Code
$str1 = "anagram";
$str2 = "grammar";
$k = 2;
if (arekAnagrams($str1, $str2, $k))
    echo "Yes";
else
    echo "No";

// This code is contributed by m_kit
?>

Output

Yes

Time complexity: O(n)
Auxiliary Space: O(1)

Method 2: We can optimize above solution. Here we use only one count array to store counts of characters in str1. We traverse str2 and decrement occurrence of every character in count array that is present in str2. If we find a character that is not there in str1, we increment count of different characters. If count of different character become more than k, we return false.

Implementation:

C++

// Optimized C++ program to check if two strings
// are k anagram or not.
#include<bits/stdc++.h>
using namespace std;

const int MAX_CHAR = 26;

// Function to check if str1 and str2 are k-anagram
// or not
bool areKAnagrams(string str1, string str2, int k)
{
    // If both strings are not of equal
    // length then return false
    int n = str1.length();
    if (str2.length() != n)
        return false;

    int hash_str1[MAX_CHAR] = {0};

    // Store the occurrence of all characters
    // in a hash_array
    for (int i = 0; i < n ; i++)
        hash_str1[str1[i]-'a']++;

    // Store the occurrence of all characters
    // in a hash_array
    int count = 0;
    for (int i = 0; i < n ; i++)
    {
        if (hash_str1[str2[i]-'a'] > 0)
            hash_str1[str2[i]-'a']--;
        else
            count++;

        if (count > k)
            return false;
    }

    // Return true if count is less than or
    // equal to k
    return true;
}

// Driver code
int main()
{
    string str1 = "fodr";
    string str2 = "gork";
    int k = 2;
    if (areKAnagrams(str1, str2, k) == true)
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

C

#include <stdbool.h>
#include <stdio.h>
#include <string.h>

#define MAX_CHAR 26

// Function to check if str1 and str2 are k-anagram or not
bool areKAnagrams(char* str1, char* str2, int k)
{
    // If both strings are not of equal length, then return
    // false
    int n = strlen(str1);
    if (strlen(str2) != n)
        return false;

    int hash_str1[MAX_CHAR] = { 0 };

    // Store the occurrence of all characters in a hash
    // array
    for (int i = 0; i < n; i++)
        hash_str1[str1[i] - 'a']++;

    int count = 0;

    // Check the occurrence of characters in str2
    for (int i = 0; i < n; i++) {
        if (hash_str1[str2[i] - 'a'] > 0)
            hash_str1[str2[i] - 'a']--;
        else
            count++;

        if (count > k)
            return false;
    }

    // Return true if count is less than or equal to k
    return true;
}

// Driver code
int main()
{
    char str1[] = "fodr";
    char str2[] = "gork";
    int k = 2;

    if (areKAnagrams(str1, str2, k))
        printf("Yes\n");
    else
        printf("No\n");

    return 0;
}

Java

// Optimized Java program to check if two strings
// are k anagram or not.
public class GFG {

    static final int MAX_CHAR = 26;

    // Function to check if str1 and str2 are k-anagram
    // or not
    static boolean areKAnagrams(String str1, String str2,
                                                  int k)
    {
        // If both strings are not of equal
        // length then return false
        int n = str1.length();
        if (str2.length() != n)
            return false;

        int[] hash_str1 = new int[MAX_CHAR];

        // Store the occurrence of all characters
        // in a hash_array
        for (int i = 0; i < n ; i++)
            hash_str1[str1.charAt(i)-'a']++;

        // Store the occurrence of all characters
        // in a hash_array
        int count = 0;
        for (int i = 0; i < n ; i++)
        {
            if (hash_str1[str2.charAt(i)-'a'] > 0)
                hash_str1[str2.charAt(i)-'a']--;
            else
                count++;

            if (count > k)
                return false;
        }

        // Return true if count is less than or
        // equal to k
        return true;
    }

    // Driver code
    public static void main(String args[])
    {
        String str1 = "fodr";
        String str2 = "gork";
        int k = 2;
        if (areKAnagrams(str1, str2, k) == true)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
// This code is contributed by Sumit Ghosh

Python3

# Optimized Python3 program
# to check if two strings
# are k anagram or not.
MAX_CHAR = 26;

# Function to check if str1
# and str2 are k-anagram or not
def areKAnagrams(str1, str2, k):
    # If both strings are
    # not of equal length
    # then return false

    n = len(str1);
    if (len(str2) != n):
        return False;

    hash_str1 = [0]*(MAX_CHAR);

    # Store the occurrence of
    # all characters in a hash_array
    for i in range(n):
        hash_str1[ord(str1[i]) - ord('a')]+=1;

    # Store the occurrence of all
    # characters in a hash_array
    count = 0;
    for i in range(n):
        if (hash_str1[ord(str2[i]) - ord('a')] > 0):
            hash_str1[ord(str2[i]) - ord('a')]-=1;
        else:
            count+=1;

        if (count > k):
            return False;

    # Return true if count is
    # less than or equal to k
    return True;

# Driver code
str1 = "fodr";
str2 = "gork";
k = 2;
if (areKAnagrams(str1, str2, k) == True):
    print("Yes");
else:
    print("No");

# This code is contributed by mits

C#

// Optimized C# program to check if two strings
// are k anagram or not.
using System;

 class GFG {

    static  int MAX_CHAR = 26;

    // Function to check if str1 and str2 are k-anagram
    // or not
    static bool areKAnagrams(String str1, String str2,
                                                int k)
    {
        // If both strings are not of equal
        // [i] then return false
        int n = str1.Length;
        if (str2.Length != n)
            return false;

        int[] hash_str1 = new int[MAX_CHAR];

        // Store the occurrence of all characters
        // in a hash_array
        for (int i = 0; i < n ; i++)
            hash_str1[str1[i]-'a']++;

        // Store the occurrence of all characters
        // in a hash_array
        int count = 0;
        for (int i = 0; i < n ; i++)
        {
            if (hash_str1[str2[i]-'a'] > 0)
                hash_str1[str2[i]-'a']--;
            else
                count++;

            if (count > k)
                return false;
        }

        // Return true if count is less than or
        // equal to k
        return true;
    }

    // Driver code
     static void Main()
    {
        String str1 = "fodr";
        String str2 = "gork";
        int k = 2;

        if (areKAnagrams(str1, str2, k) == true)
            Console.Write("Yes");
        else
            Console.Write("No");
    }
}
// This code is contributed by Anuj_67

Javascript

<script>

    // Optimized Javascript program
    // to check if two strings
    // are k anagram or not.

    let MAX_CHAR = 26;

    // Function to check if str1 and
    // str2 are k-anagram
    // or not
    function areKAnagrams(str1, str2, k)
    {
        // If both strings are not of equal
        // [i] then return false
        let n = str1.length;
        if (str2.length != n)
            return false;

        let hash_str1 = Array(MAX_CHAR);
        hash_str1.fill(0);

        // Store the occurrence of all characters
        // in a hash_array
        for (let i = 0; i < n ; i++)
            hash_str1[str1[i].charCodeAt()-
            'a'.charCodeAt()]++;

        // Store the occurrence of all characters
        // in a hash_array
        let count = 0;
        for (let i = 0; i < n ; i++)
        {
            if (hash_str1[str2[i].charCodeAt()-
            'a'.charCodeAt()] > 0)
                hash_str1[str2[i].charCodeAt()-
                'a'.charCodeAt()]--;
            else
                count++;

            if (count > k)
                return false;
        }

        // Return true if count is less than or
        // equal to k
        return true;
    }

    let str1 = "fodr";
    let str2 = "gork";
    let k = 2;

    if (areKAnagrams(str1, str2, k) == true)
      document.write("Yes");
    else
      document.write("No");

</script>

PHP

<?php
// Optimized PHP program
// to check if two strings
// are k anagram or not.
$MAX_CHAR = 26;

// Function to check if str1
// and str2 are k-anagram or not
function areKAnagrams($str1,
                      $str2, $k)
{
    global $MAX_CHAR;
    // If both strings are
    // not of equal length
    // then return false

    $n = strlen($str1);
    if (strlen($str2) != $n)
        return false;

    $hash_str1 = array(0);

    // Store the occurrence of
    // all characters in a hash_array
    for ($i = 0; $i < $n ; $i++)
        $hash_str1[$str1[$i] - 'a']++;

    // Store the occurrence of all
    // characters in a hash_array
    $count = 0;
    for ($i = 0; $i < $n ; $i++)
    {
        if ($hash_str1[$str2[$i] - 'a'] > 0)
            $hash_str1[$str2[$i] - 'a']--;
        else
            $count++;

        if ($count > $k)
            return false;
    }

    // Return true if count is
    // less than or equal to k
    return true;
}

// Driver code
$str1 = "fodr";
$str2 = "gork";
$k = 2;
if (areKAnagrams($str1, $str2, $k) == true)
    echo "Yes";
else
    echo "No";

// This code is contributed by ajit
?>

Output

Yes

Time complexity: O(n)
Auxiliary Space: O(1)

 This article is contributed by Sahil Chhabra (akku). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.