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.
1) Count occurrences of all characters from ‘a’ to ‘z’ in first and second strings. Store these counts in two arrays a1[] and a2[].
2) 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 implementation of 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); } } |
Output:
eeefgkors
Time Complexity: If we consider n = length(larger string), then this algorithm runs in O(n) complexity.
This article is contributed by Deepak Srivatsav. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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