Length of longest prefix anagram which are common in given two strings
Last Updated :
28 Dec, 2022
Given two strings str1 and str2 of the lengths of N and M respectively, the task is to find the length of the longest anagram string that is prefix substring of both strings.
Examples:
Input: str1 = "abaabcdezzwer", str2 = "caaabbttyh"
Output: 6
Explanation:
Prefixes of length 1 of string str1 and str2 are "a", and "c".
Prefixes of length 2 of string str1 and str2 are "ab", and "ca".
Prefixes of length 3 of string str1 and str2 are "aba", and "caa".
Prefixes of length 4 of string str1 and str2 are "abaa", and "caaa".
Prefixes of length 5 of string str1 and str2 are "abaab", and "caaab".
Prefixes of length 6 of string str1 and str2 are "abaabc", and "caaabb".
Prefixes of length 7 of string str1 and str2 are "abaabcd", and "caaabbt".
Prefixes of length 8 of string str1 and str2 are "abaabcde", and "caaabbtt".
Prefixes of length 9 of string str1 and str2 are "abaabcdez", and "caaabbtty".
Prefixes of length 10 of string str1 and str2 are "abaabcdezz", and "caaabbttyh".
Prefixes of length 6 are anagram with each other only.
Input: str1 = "abcdef", str2 = "tuvwxyz"
Output: 0
Approach: The idea is to use Hashing for solving the above problem. Follow the steps below to solve the problem:
- Initialize two integer arrays freq1[] and freq2[], each of size 26, to store the count of characters in strings str1 and str2 respectively.
- Initialize a variable, say ans, to store the result.
- Iterate over the characters of the string present in indices [0, minimum(N - 1, M - 1)] and perform the following:
- Increment count of str1[i] in freq1[] array and count of str2[i] in freq2[] array by 1.
- Check if the frequency array freq1[] is the same as the frequency array freq2[], assign ans = i + 1.
- After the above steps, print the value of ans as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
#define SIZE 26
// Function to check if two arrays
// are identical or not
bool longHelper(int freq1[], int freq2[])
{
// Iterate over the range [0, SIZE]
for (int i = 0; i < SIZE; ++i) {
// If frequency any character is
// not same in both the strings
if (freq1[i] != freq2[i]) {
return false;
}
}
// Otherwise
return true;
}
// Function to find the maximum
// length of the required string
int longCommomPrefixAnagram(
string s1, string s2, int n1, int n2)
{
// Store the count of
// characters in string str1
int freq1[26] = { 0 };
// Store the count of
// characters in string str2
int freq2[26] = { 0 };
// Stores the maximum length
int ans = 0;
// Minimum length of str1 and str2
int mini_len = min(n1, n2);
for (int i = 0; i < mini_len; ++i) {
// Increment the count of
// characters of str1[i] in
// freq1[] by one
freq1[s1[i] - 'a']++;
// Increment the count of
// characters of str2[i] in
// freq2[] by one
freq2[s2[i] - 'a']++;
// Checks if prefixes are
// anagram or not
if (longHelper(freq1, freq2)) {
ans = i + 1;
}
}
// Finally print the ans
cout << ans;
}
// Driver Code
int main()
{
string str1 = "abaabcdezzwer";
string str2 = "caaabbttyh";
int N = str1.length();
int M = str2.length();
// Function Call
longCommomPrefixAnagram(str1, str2,
N, M);
return 0;
}
// Java program for the above approach
public class Main
{
static int SIZE = 26;
// Function to check if two arrays
// are identical or not
static boolean longHelper(int[] freq1, int[] freq2)
{
// Iterate over the range [0, SIZE]
for (int i = 0; i < SIZE; ++i)
{
// If frequency any character is
// not same in both the strings
if (freq1[i] != freq2[i])
{
return false;
}
}
// Otherwise
return true;
}
// Function to find the maximum
// length of the required string
static void longCommomPrefixAnagram(
String s1, String s2, int n1, int n2)
{
// Store the count of
// characters in string str1
int[] freq1 = new int[26];
// Store the count of
// characters in string str2
int[] freq2 = new int[26];
// Stores the maximum length
int ans = 0;
// Minimum length of str1 and str2
int mini_len = Math.min(n1, n2);
for (int i = 0; i < mini_len; ++i) {
// Increment the count of
// characters of str1[i] in
// freq1[] by one
freq1[s1.charAt(i) - 'a']++;
// Increment the count of
// characters of str2[i] in
// freq2[] by one
freq2[s2.charAt(i) - 'a']++;
// Checks if prefixes are
// anagram or not
if (longHelper(freq1, freq2)) {
ans = i + 1;
}
}
// Finally print the ans
System.out.print(ans);
}
// Driver code
public static void main(String[] args)
{
String str1 = "abaabcdezzwer";
String str2 = "caaabbttyh";
int N = str1.length();
int M = str2.length();
// Function Call
longCommomPrefixAnagram(str1, str2, N, M);
}
}
// This code is contributed by divyeshrrabadiya07.
# Python3 program for the above approach
SIZE = 26
# Function to check if two arrays
# are identical or not
def longHelper(freq1, freq2):
# Iterate over the range [0, SIZE]
for i in range(26):
# If frequency any character is
# not same in both the strings
if (freq1[i] != freq2[i]):
return False
# Otherwise
return True
# Function to find the maximum
# length of the required string
def longCommomPrefixAnagram(s1, s2, n1, n2):
# Store the count of
# characters in str1
freq1 = [0]*26
# Store the count of
# characters in str2
freq2 = [0]*26
# Stores the maximum length
ans = 0
# Minimum length of str1 and str2
mini_len = min(n1, n2)
for i in range(mini_len):
# Increment the count of
# characters of str1[i] in
# freq1[] by one
freq1[ord(s1[i]) - ord('a')] += 1
# Increment the count of
# characters of stord(r2[i]) in
# freq2[] by one
freq2[ord(s2[i]) - ord('a')] += 1
# Checks if prefixes are
# anagram or not
if (longHelper(freq1, freq2)):
ans = i + 1
# Finally print the ans
print (ans)
# Driver Code
if __name__ == '__main__':
str1 = "abaabcdezzwer"
str2 = "caaabbttyh"
N = len(str1)
M = len(str2)
# Function Call
longCommomPrefixAnagram(str1, str2, N, M)
# This code is contributed by mohit kumar 29.
// C# program for the above approach
using System;
class GFG
{
static int SIZE = 26;
// Function to check if two arrays
// are identical or not
static bool longHelper(int[] freq1, int[] freq2)
{
// Iterate over the range [0, SIZE]
for (int i = 0; i < SIZE; ++i)
{
// If frequency any character is
// not same in both the strings
if (freq1[i] != freq2[i])
{
return false;
}
}
// Otherwise
return true;
}
// Function to find the maximum
// length of the required string
static void longCommomPrefixAnagram(
string s1, string s2, int n1, int n2)
{
// Store the count of
// characters in string str1
int[] freq1 = new int[26];
// Store the count of
// characters in string str2
int[] freq2 = new int[26];
// Stores the maximum length
int ans = 0;
// Minimum length of str1 and str2
int mini_len = Math.Min(n1, n2);
for (int i = 0; i < mini_len; ++i) {
// Increment the count of
// characters of str1[i] in
// freq1[] by one
freq1[s1[i] - 'a']++;
// Increment the count of
// characters of str2[i] in
// freq2[] by one
freq2[s2[i] - 'a']++;
// Checks if prefixes are
// anagram or not
if (longHelper(freq1, freq2)) {
ans = i + 1;
}
}
// Finally print the ans
Console.Write(ans);
}
// Driver code
static void Main() {
string str1 = "abaabcdezzwer";
string str2 = "caaabbttyh";
int N = str1.Length;
int M = str2.Length;
// Function Call
longCommomPrefixAnagram(str1, str2, N, M);
}
}
// This code is contributed by divyesh072019.
<script>
// Javascript program for the above approach
let SIZE = 26;
// Function to check if two arrays
// are identical or not
function longHelper(freq1, freq2)
{
// Iterate over the range [0, SIZE]
for(let i = 0; i < SIZE; ++i)
{
// If frequency any character is
// not same in both the strings
if (freq1[i] != freq2[i])
{
return false;
}
}
// Otherwise
return true;
}
// Function to find the maximum
// length of the required string
function longCommomPrefixAnagram(s1, s2, n1, n2)
{
// Store the count of
// characters in string str1
let freq1 = new Array(26);
freq1.fill(0);
// Store the count of
// characters in string str2
let freq2 = new Array(26);
freq2.fill(0);
// Stores the maximum length
let ans = 0;
// Minimum length of str1 and str2
let mini_len = Math.min(n1, n2);
for(let i = 0; i < mini_len; ++i)
{
// Increment the count of
// characters of str1[i] in
// freq1[] by one
freq1[s1[i].charCodeAt() -
'a'.charCodeAt()]++;
// Increment the count of
// characters of str2[i] in
// freq2[] by one
freq2[s2[i].charCodeAt() -
'a'.charCodeAt()]++;
// Checks if prefixes are
// anagram or not
if (longHelper(freq1, freq2))
{
ans = i + 1;
}
}
// Finally print the ans
document.write(ans);
}
// Driver code
let str1 = "abaabcdezzwer";
let str2 = "caaabbttyh";
let N = str1.length;
let M = str2.length;
// Function Call
longCommomPrefixAnagram(str1, str2, N, M);
// This code is contributed by rameshtravel07
</script>
Time Complexity: O(N*26)
Auxiliary Space: O(1)
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