Check if a string has all characters with same frequency with one variation allowed

Given a string of lowercase alphabets, find if it can be converted to a Valid String by removing 1 or 0 characters. A “valid” string is a string str such that for all distinct characters in str each such character occurs the same number of times in it.

Examples :

Input : string str = "abbca"
Output : Yes
We can make it valid by removing "c"

Input : string str = "aabbcd"
Output : No
We need to remove at least two characters
to make it valid.

Input : string str = "abbccd"
Output : No

We are allowed to traverse string only once.



The idea is to use a frequency array that stores frequencies of all characters. Once we have frequencies of all characters in an array, we check if count of total different and non zero values are not more than 2. Also, one of the counts of two allowed different frequencies must be less than or equal to 2. Below is the implementation of idea.

C++

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// C++ program to check if a string can be made
// valid by removing at most 1 character.
#include<bits/stdc++.h>
using namespace std;
  
// Assuming only lower case characters
const int CHARS = 26;
  
/* To check a string S can be converted to a “valid”
   string by removing less than or equal to one
   character. */
bool isValidString(string str)
{
    int freq[CHARS] = {0};
  
    // freq[] : stores the  frequency of each
    // character of a string
    for (int i=0; i<str.length(); i++)
        freq[str[i]-'a']++;
  
    // Find first character with non-zero frequency
    int i, freq1 = 0, count_freq1 = 0;
    for (i=0; i<CHARS; i++)
    {
        if (freq[i] != 0)
        {
            freq1  = freq[i];
            count_freq1 = 1;
            break;
        }
    }
  
    // Find a character with frequency different
    // from freq1.
    int j, freq2 = 0, count_freq2 = 0;
    for (j=i+1; j<CHARS; j++)
    {
        if (freq[j] != 0)
        {
            if (freq[j] == freq1)
               count_freq1++;
            else
            {
                count_freq2 = 1;
                freq2 = freq[j];
                break;
            }
        }
    }
  
    // If we find a third non-zero frequency
    // or count of both frequencies become more
    // than 1, then return false
    for (int k=j+1; k<CHARS; k++)
    {
        if (freq[k] != 0)
        {
            if (freq[k] == freq1)
               count_freq1++;
            if (freq[k] == freq2)
               count_freq2++;
            else  // If we find a third non-zero freq
               return false;
        }
  
        // If counts of both frequencies is more than 1
        if (count_freq1 > 1 && count_freq2 > 1)
           return false;
    }
  
    // Return true if we reach here
    return true;
}
  
// Driver code
int main()
{
    char str[] = "abcbc";
  
    if (isValidString(str))
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
    return 0;
}

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Java

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// Java program to check if a string can be made
// valid by removing at most 1 character.
public class GFG {
  
// Assuming only lower case characters
    static int CHARS = 26;
  
    /* To check a string S can be converted to a “valid”
   string by removing less than or equal to one
   character. */
    static boolean isValidString(String str) {
        int freq[] = new int[CHARS];
  
        // freq[] : stores the  frequency of each
        // character of a string
        for (int i = 0; i < str.length(); i++) {
            freq[str.charAt(i) - 'a']++;
        }
  
        // Find first character with non-zero frequency
        int i, freq1 = 0, count_freq1 = 0;
        for (i = 0; i < CHARS; i++) {
            if (freq[i] != 0) {
                freq1 = freq[i];
                count_freq1 = 1;
                break;
            }
        }
  
        // Find a character with frequency different
        // from freq1.
        int j, freq2 = 0, count_freq2 = 0;
        for (j = i + 1; j < CHARS; j++) {
            if (freq[j] != 0) {
                if (freq[j] == freq1) {
                    count_freq1++;
                } else {
                    count_freq2 = 1;
                    freq2 = freq[j];
                    break;
                }
            }
        }
  
        // If we find a third non-zero frequency
        // or count of both frequencies become more
        // than 1, then return false
        for (int k = j + 1; k < CHARS; k++) {
            if (freq[k] != 0) {
                if (freq[k] == freq1) {
                    count_freq1++;
                }
                if (freq[k] == freq2) {
                    count_freq2++;
                } else // If we find a third non-zero freq
                {
                    return false;
                }
            }
  
            // If counts of both frequencies is more than 1
            if (count_freq1 > 1 && count_freq2 > 1) {
                return false;
            }
        }
  
        // Return true if we reach here
        return true;
    }
  
// Driver code
    public static void main(String[] args) {
        String str = "abcbc";
  
        if (isValidString(str)) {
            System.out.println("YES");
        } else {
            System.out.println("NO");
        }
    }
}
// This code is contributed by PrinciRaj1992

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Python3

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# Python 3 program to check if 
# a string can be made
# valid by removing at most 1 character.
  
# Assuming only lower case characters
CHARS = 26
  
# To check a string S can be converted to a “valid”
# string by removing less than or equal to one
# character.
      
def isValidString(str):
  
    freq = [0]*CHARS
  
    # freq[] : stores the frequency of each
    # character of a string
    for i in range(len(str)):
        freq[ord(str[i])-ord('a')] += 1
  
    # Find first character with non-zero frequency
    freq1 = 0
    count_freq1 = 0
    for i in range(CHARS):
      
        if (freq[i] != 0):
          
            freq1 = freq[i]
            count_freq1 = 1
            break
  
    # Find a character with frequency different
    # from freq1.
    freq2 = 0
    count_freq2 = 0
    for j in range(i+1,CHARS):
      
        if (freq[j] != 0):
      
            if (freq[j] == freq1):
                count_freq1 += 1
            else:
              
                count_freq2 = 1
                freq2 = freq[j]
                break
  
    # If we find a third non-zero frequency
    # or count of both frequencies become more
    # than 1, then return false
    for k in range(j+1,CHARS):
      
        if (freq[k] != 0):
          
            if (freq[k] == freq1):
                count_freq1 += 1
            if (freq[k] == freq2):
                count_freq2 += 1
  
            # If we find a third non-zero freq
            else:
                return False
  
        # If counts of both frequencies is more than 1
        if (count_freq1 > 1 and count_freq2 > 1):
            return False
  
    # Return true if we reach here
    return True
  
# Driver code
if __name__ == "__main__":
    str= "abcbc"
  
    if (isValidString(str)):
        print("YES")
    else:
        print("NO")
          
# this code is contributed by 
# ChitraNayal

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C#

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// C# program to check if a string can be made 
// valid by removing at most 1 character. 
using System;
public class GFG { 
  
// Assuming only lower case characters 
    static int CHARS = 26; 
  
    /* To check a string S can be converted to a “valid” 
string by removing less than or equal to one 
character. */
    static bool isValidString(String str) { 
        int []freq = new int[CHARS]; 
        int i=0;
        // freq[] : stores the frequency of each 
        // character of a string 
        for ( i= 0; i < str.Length; i++) { 
            freq[str[i] - 'a']++; 
        
  
        // Find first character with non-zero frequency 
        int freq1 = 0, count_freq1 = 0; 
        for (i = 0; i < CHARS; i++) { 
            if (freq[i] != 0) { 
                freq1 = freq[i]; 
                count_freq1 = 1; 
                break
            
        
  
        // Find a character with frequency different 
        // from freq1. 
        int j, freq2 = 0, count_freq2 = 0; 
        for (j = i + 1; j < CHARS; j++) { 
            if (freq[j] != 0) { 
                if (freq[j] == freq1) { 
                    count_freq1++; 
                } else
                    count_freq2 = 1; 
                    freq2 = freq[j]; 
                    break
                
            
        
  
        // If we find a third non-zero frequency 
        // or count of both frequencies become more 
        // than 1, then return false 
        for (int k = j + 1; k < CHARS; k++) { 
            if (freq[k] != 0) { 
                if (freq[k] == freq1) { 
                    count_freq1++; 
                
                if (freq[k] == freq2) { 
                    count_freq2++; 
                } else // If we find a third non-zero freq 
                
                    return false
                
            
  
            // If counts of both frequencies is more than 1 
            if (count_freq1 > 1 && count_freq2 > 1) { 
                return false
            
        
  
        // Return true if we reach here 
        return true
    
  
// Driver code 
    public static void Main() { 
        String str = "abcbc"
  
        if (isValidString(str)) { 
            Console.WriteLine("YES"); 
        } else
            Console.WriteLine("NO"); 
        
    
  
// This code is contributed by 29AjayKumar

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Output:

YES

We traverse string only once. Also the three loops after the first loop run CHARS times in total.
Another method: (Using HashMap)
Below is the implementation.

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// Java program to check if a string can be made
// valid by removing at most 1 character using hashmap.
import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
  
public class AllCharsWithSameFrequencyWithOneVarAllowed {
      
    // To check a string S can be converted to a variation
    // string 
    public static boolean checkForVariation(String str) {
        if(str == null || str.isEmpty()) {
            return true;
        }
          
        Map<Character, Integer> map = new HashMap<>();
          
        // Run loop form 0 to length of string
        for(int i = 0; i < str.length(); i++) {
            map.put(str.charAt(i), map.getOrDefault(str.charAt(i), 0) + 1);
        }
        Iterator<Integer> itr = map.values().iterator();
          
        // declaration of variables
        boolean first = true, second = true;
        int val1 = 0, val2 = 0;
        int countOfVal1 = 0, countOfVal2 = 0;
          
        while(itr.hasNext()) {
            int i = itr.next();
              
            // if first is true than countOfVal1 increase
            if(first) {
                val1 = i;
                first = false;
                countOfVal1++;
                continue;
            }
              
            if(i == val1) {
                countOfVal1++;
                continue;
            }
              
            // if second is true than countOfVal2 increase
            if(second) {
                val2 = i;
                countOfVal2++;
                second = false;
                continue;
            }
              
            if(i == val2) {
                countOfVal2++;
                continue;
            }
              
            return false;
        }
          
        if(countOfVal1 > 1 && countOfVal2 > 1) {
            return false;
        }else {
            return true;
        }
          
    }
      
    // Driver code
    public static void main(String[] args)
    {
              
        System.out.println(checkForVariation("abcbc"));
    }
}

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Output:

true

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