Check if max occurring character of one string appears same no. of times in other

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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; 
} 

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 

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 occurence 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 

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 

Output:

Yes

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

Recommended Posts: