Given two strings, we need to take the character which has the maximum occurrence in the first string and then we have to check if that particular character is present in the second string the same number of times as it is present in the first string.
Examples:
Input : s1 = "sssgeek", s2 = "geeksss"
Output : Yes
Max occurring character in s1 is
's'. It occurs same number of times
in s2.
Input : geekyarticle
gfggfggfg
Output : No
Store counts of characters in first string and find the maximum count. Now traverse through the second string and check if the maximum occurring character occurs same number of times or not.
Below program to illustrate the above problem
C++
// C++ program to check the problem #include <bits/stdc++.h> using namespace std; #define ll long long #define ASCIISIZE 256 int match(string s1, string s2) { // Create array to keep the count of individual // characters and initialize the array as 0 int count[ASCIISIZE] = { 0 }; // Construct character count array from the input // string. for (int i = 0; i < s1.length(); i++) count[s1[i]]++; // Count occurrences of maximum occurring character int mx_cnt = 0, mx_chr; for (int i = 0; i < ASCIISIZE; i++) { if (count[i] > mx_cnt) { mx_cnt = count[i]; mx_chr = i; } } // look if that character is present, the same // number of times it is present in second string for (int i = 0; i < s2.length(); i++) if (mx_chr == s2[i]) mx_cnt--; // check if sum is greater or equal to number // return 1 if (mx_cnt == 0) return 1; } // Driver program to test the above function int main() { string s1 = "geekforgeeks", s2 = "geekisgeeky"; if (match(s1, s2)) cout << "Yes "; else cout << "No"; return 0; } |
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Java
// Java program to check the problem class GFG { static int ASCIISIZE = 256; static int match(String s1, String s2) { // Create array to keep the // count of individual characters // and initialize the array as 0 int count[] = new int[ASCIISIZE]; // Construct character count // array from the input string. char []s3 = s1.toCharArray(); for (int i = 0; i < s3.length; i++) count[s3[i]]++; // Count occurrences of // maximum occurring character int mx_cnt = 0; int mx_chr = 0; for (int i = 0; i < ASCIISIZE; i++) { if (count[i] > mx_cnt) { mx_cnt = count[i]; mx_chr = i; } } // look if that character is // present, the same number // of times it is present in // second string char []s4 = s2.toCharArray(); for (int i = 0; i < s4.length; i++) if (mx_chr == s4[i]) mx_cnt--; // check if sum is greater or // equal to number return 1 if (mx_cnt == 0) return 1; else return 0; } // Driver Code public static void main(String[] args) { String s1 = "geekforgeeks", s2 = "geekisgeeky"; int p = match(s1, s2); if (p == 1) System.out.println("Yes "); else System.out.println("No"); } } // This code is contributed // by ChitraNayal |
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Python3
# Python3 program to # check the problem # define function for Check # if max occurring character # of one string appears same # no. of times in other def match(s1, s2) : # declare empty list count_list = [] # iterate through each # character of the string for char in s1 : # find occurrence of # the character count = s1.count(char) # append tuple value # to the list count_list.append((count,char)) # return tuple of max count max_occ = max(count_list) # store max count in mx_cnt mx_cnt = max_occ[0] # store max count # character in mx_chr mx_chr = max_occ[1] # look if max count character # is present in s1, the same # number of times it is present # in second string s2 or not # if present return True # otherwise False. if mx_cnt == s2.count(mx_chr) : return True else : return False # Driver Code if __name__ == "__main__" : s1 = "geeksforgeeks" s2 = "geekisgeeky" if match(s1,s2) : print("Yes") else : print("No") # This code is contributed # by Ankit Rai |
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C#
// C# program to check the problem using System; class GFG { static int ASCIISIZE = 256; static int match(String s1, String s2) { // Create array to keep the // count of individual characters // and initialize the array as 0 int []count = new int[ASCIISIZE]; // Construct character count // array from the input string. for (int i = 0; i < s1.Length; i++) count[s1[i]]++; // Count occurrences of // maximum occurring character int mx_cnt = 0; int mx_chr = 0; for (int i = 0; i < ASCIISIZE; i++) { if (count[i] > mx_cnt) { mx_cnt = count[i]; mx_chr = i; } } // look if that character is // present, the same number // of times it is present // in second string for (int i = 0; i < s2.Length; i++) if (mx_chr == s2[i]) mx_cnt--; // check if sum is greater // or equal to number return 1 if (mx_cnt == 0) return 1; else return 0; } // Driver Code public static void Main() { String s1 = "geekforgeeks", s2 = "geekisgeeky"; int p = match(s1, s2); if (p == 1) Console.Write("Yes "); else Console.Write("No"); } } // This code is contributed // by ChitraNayal |
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Output:
Yes
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