Given two strings, print all the common characters in lexicographical order. If there are no common letters, print -1. All letters are lower case.
Examples:
Input : string1 : geeks string2 : forgeeks Output : eegks Explanation: The letters that are common between the two strings are e(2 times), k(1 time) and s(1 time). Hence the lexicographical output is "eegks" Input : string1 : hhhhhello string2 : gfghhmh Output : hhh
The idea is to use character count arrays.
- Count occurrences of all characters from ‘a’ to ‘z’ in the first and second strings. Store these counts in two arrays a1[] and a2[].
- Traverse a1[] and a2[] (Note size of both is 26). For every index i, print character ‘a’ + i number of times equal min(a1[i], a2[i]).
Below is the implementation of the above steps.
C++
// C++ program to print common characters // of two Strings in alphabetical order #include<bits/stdc++.h> using namespace std;
int main()
{ string s1 = "geeksforgeeks" ;
string s2 = "practiceforgeeks" ;
// to store the count of
// letters in the first string
int a1[26] = {0};
// to store the count of
// letters in the second string
int a2[26] = {0};
int i , j;
char ch;
char ch1 = 'a' ;
int k = ( int )ch1, m;
// for each letter present, increment the count
for (i = 0 ; i < s1.length() ; i++)
{
a1[( int )s1[i] - k]++;
}
for (i = 0 ; i < s2.length() ; i++)
{
a2[( int )s2[i] - k]++;
}
for (i = 0 ; i < 26 ; i++)
{
// the if condition guarantees that
// the element is common, that is,
// a1[i] and a2[i] are both non zero
// means that the letter has occurred
// at least once in both the strings
if (a1[i] != 0 and a2[i] != 0)
{
// print the letter for a number
// of times that is the minimum
// of its count in s1 and s2
for (j = 0 ; j < min(a1[i] , a2[i]) ; j++)
{
m = k + i;
ch = ( char )(k + i);
cout << ch;
}
}
}
return 0;
} |
Java
// Java program to print common characters // of two Strings in alphabetical order import java.io.*;
import java.util.*;
// Function to find similar characters public class Simstrings
{ static final int MAX_CHAR = 26 ;
static void printCommon(String s1, String s2)
{
// two arrays of length 26 to store occurrence
// of a letters alphabetically for each string
int [] a1 = new int [MAX_CHAR];
int [] a2 = new int [MAX_CHAR];
int length1 = s1.length();
int length2 = s2.length();
for ( int i = 0 ; i < length1 ; i++)
a1[s1.charAt(i) - 'a' ] += 1 ;
for ( int i = 0 ; i < length2 ; i++)
a2[s2.charAt(i) - 'a' ] += 1 ;
// If a common index is non-zero, it means
// that the letter corresponding to that
// index is common to both strings
for ( int i = 0 ; i < MAX_CHAR ; i++)
{
if (a1[i] != 0 && a2[i] != 0 )
{
// Find the minimum of the occurrence
// of the character in both strings and print
// the letter that many number of times
for ( int j = 0 ; j < Math.min(a1[i], a2[i]) ; j++)
System.out.print((( char )(i + 'a' )));
}
}
}
// Driver code
public static void main(String[] args) throws IOException
{
String s1 = "geeksforgeeks" , s2 = "practiceforgeeks" ;
printCommon(s1, s2);
}
} |
Python3
# Python3 program to print common characters # of two Strings in alphabetical order # Initializing size of array MAX_CHAR = 26
# Function to find similar characters def printCommon( s1, s2):
# two arrays of length 26 to store occurrence
# of a letters alphabetically for each string
a1 = [ 0 for i in range (MAX_CHAR)]
a2 = [ 0 for i in range (MAX_CHAR)]
length1 = len (s1)
length2 = len (s2)
for i in range ( 0 ,length1):
a1[ ord (s1[i]) - ord ( 'a' )] + = 1
for i in range ( 0 ,length2):
a2[ ord (s2[i]) - ord ( 'a' )] + = 1
# If a common index is non-zero, it means
# that the letter corresponding to that
# index is common to both strings
for i in range ( 0 ,MAX_CHAR):
if (a1[i] ! = 0 and a2[i] ! = 0 ):
# Find the minimum of the occurrence
# of the character in both strings and print
# the letter that many number of times
for j in range ( 0 , min (a1[i],a2[i])):
ch = chr ( ord ( 'a' ) + i)
print (ch, end = '')
# Driver code if __name__ = = "__main__" :
s1 = "geeksforgeeks"
s2 = "practiceforgeeks"
printCommon(s1, s2);
# This Code is contributed by Abhishek Sharma |
C#
// C# program to print common characters // of two Strings in alphabetical order using System;
// Function to find similar characters public class Simstrings
{ static int MAX_CHAR = 26;
static void printCommon( string s1, string s2)
{
// two arrays of length 26 to store occurrence
// of a letters alphabetically for each string
int [] a1 = new int [MAX_CHAR];
int [] a2 = new int [MAX_CHAR];
int length1 = s1.Length;
int length2 = s2.Length;
for ( int i = 0 ; i < length1 ; i++)
a1[s1[i] - 'a' ] += 1;
for ( int i = 0 ; i < length2 ; i++)
a2[s2[i] - 'a' ] += 1;
// If a common index is non-zero, it means
// that the letter corresponding to that
// index is common to both strings
for ( int i = 0 ; i < MAX_CHAR ; i++)
{
if (a1[i] != 0 && a2[i] != 0)
{
// Find the minimum of the occurrence
// of the character in both strings and print
// the letter that many number of times
for ( int j = 0 ; j < Math.Min(a1[i], a2[i]) ; j++)
Console.Write((( char )(i + 'a' )));
}
}
}
// Driver code
public static void Main()
{
string s1 = "geeksforgeeks" , s2 = "practiceforgeeks" ;
printCommon(s1, s2);
}
} |
Javascript
<script> // Javascript program to print common characters // of two Strings in alphabetical order let MAX_CHAR = 26;
// Function to find similar characters
function printCommon(s1,s2)
{
// two arrays of length 26 to store occurrence
// of a letters alphabetically for each string
let a1 = new Array(MAX_CHAR);
let a2 = new Array(MAX_CHAR);
for (let i=0;i<MAX_CHAR;i++)
{
a1[i]=0;
a2[i]=0;
}
let length1 = s1.length;
let length2 = s2.length;
for (let i = 0 ; i < length1 ; i++)
a1[s1[i].charCodeAt(0) - 'a' .charCodeAt(0)] += 1;
for (let i = 0 ; i < length2 ; i++)
a2[s2[i].charCodeAt(0) - 'a' .charCodeAt(0)] += 1;
// If a common index is non-zero, it means
// that the letter corresponding to that
// index is common to both strings
for (let i = 0 ; i < MAX_CHAR ; i++)
{
if (a1[i] != 0 && a2[i] != 0)
{
// Find the minimum of the occurrence
// of the character in both strings and print
// the letter that many number of times
for (let j = 0 ; j < Math.min(a1[i], a2[i]) ; j++)
document.write((String.fromCharCode(i + 'a' .charCodeAt(0))));
}
}
}
// Driver code
let s1 = "geeksforgeeks" , s2 = "practiceforgeeks" ;
printCommon(s1, s2);
// This code is contributed by avanitrachhadiya2155
</script> |
Output
eeefgkors
Time Complexity: If we consider n = length(larger string), then this algorithm runs in O(n) complexity.
Auxiliary Space: O(1).
Recommended Articles