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

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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 the first string and finds the maximum count. Now traverse through the second string and check if the maximum occurring character occurs the same number of times or not.

Algorithm:

  • Create an array count of size 256 to keep the count of individual characters and initialize the array as 0.
  • Iterate through the first string s1 and for each character, increment the count of that character in the count array.
  • Find the maximum occurring character and its count in the count array using a loop:
                   a. Initialize variables mx_cnt and mx_chr to 0 and NULL character, respectively.
                   b. Iterate through the count array and for each character, if its count is greater than mx_cnt, update                                      mx_cnt and mx_chr to the count and the character, respectively.
  • Iterate through the second string s2 and for each character, if it matches mx_chr, decrement mx_cnt.
  • If mx_cnt is 0, return 1 to indicate that the maximum occurring character in the first string is present in the second string the same number of times as it occurs in the first string.
  • If mx_cnt is not 0, return 0 to indicate that the condition is not met.


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
import java.io.*;
public 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 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

                    

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

                    

Javascript

<script>
 
// Javascript program to check the problem
     
    let ASCIISIZE = 256;
function match(s1,s2)
{
    // Create array to keep the
// count of individual characters
// and initialize the array as 0
let count = new Array(ASCIISIZE);
for(let i=0;i<ASCIISIZE;i++)
{
    count[i]=0;
}
// Construct character count
// array from the input string.
let s3 = s1.split("");
for (let i = 0; i < s3.length; i++)
    count[s3[i]]++;
   
// Count occurrences of
// maximum occurring character
let mx_cnt = 0;
let mx_chr = 0;
for (let 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
let s4 = s2.split("");
for (let 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
let s1 = "geekforgeeks",
           s2 = "geekisgeeky";
let p = match(s1, s2);
if (p == 1)
    document.write("Yes ");
else
    document.write("No");
     
     
// This code is contributed by avanitrachhadiya2155
 
</script>

                    

Output: 
Yes

 

Time Complexity: O(m + n), where m is the size of string s1 and n is the size of string s2. 

Auxiliary Space: O(256), no extra space required, so it is a constant.



Last Updated : 15 Mar, 2023
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