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Check if a string has all characters with same frequency with one variation allowed
  • Difficulty Level : Medium
  • Last Updated : 01 Apr, 2021

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 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++




// 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;
}

Java




// 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

Python3




# 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

C#




// 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

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. 
 



C++




// C++ program to check if a string can be made
// valid by removing at most 1 character using hashmap.
#include <bits/stdc++.h>
using namespace std;
 
// To check a string S can be converted to a variation
// string 
bool checkForVariation(string str)
{
    if(str.empty() || str.length() != 0)
    {
        return true;
    }
    map<char, int> mapp;
   
    // Run loop form 0 to length of string
    for(int i = 0; i < str.length(); i++)
    {
        mapp[str[i]]++;
    }
   
    // declaration of variables
    bool first = true, second = true;
    int val1 = 0, val2 = 0;
    int countOfVal1 = 0, countOfVal2 = 0;
     
    map<char, int>::iterator itr;
    for (itr = mapp.begin(); itr != mapp.end(); ++itr)
    {
        int i = itr->first;
         
        // 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
int main() {
    if(checkForVariation("abcbcvf"))
        cout << "true" << endl;
    else
        cout << "false" << endl;
     
    return 0;
}
 
// This code is contributed by avanitrachhadiya2155

Java




// 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"));
    }
}

Python3




# Python program to check if a string can be made
# valid by removing at most 1 character using hashmap.
 
# To check a string S can be converted to a variation
# string
def checkForVariation(strr):
    if(len(strr) == 0):
        return True
     
    mapp = {}
     
    # Run loop form 0 to length of string   
    for i in range(len(strr)):
        if strr[i] in mapp:
            mapp[strr[i]] += 1
        else:
            mapp[strr[i]] = 1
     
    # declaration of variables
    first = True
    second = True
    val1 = 0
    val2 = 0
    countOfVal1 = 0
    countOfVal2 = 0
     
    for itr in mapp:
        i = itr
         
        # if first is true than countOfVal1 increase
        if(first):
            val1 = i
            first = False
            countOfVal1 += 1
            continue
         
        if(i == val1):
            countOfVal1 += 1
            continue
         
        # if second is true than countOfVal2 increase
        if(second):
            val2 = i
            countOfVal2 += 1
            second = False
            continue
         
        if(i == val2):
            countOfVal2 += 1
            continue
    if(countOfVal1 > 1 and countOfVal2 > 1):
        return False
     
    else:
        return True
 
# Driver code
print(checkForVariation("abcbc"))
 
# This code is contributed by rag2127

C#




// C# program to check if a string can be made
// valid by removing at most 1 character using hashmap.
using System;
using System.Collections.Generic;
 
public class AllCharsWithSameFrequencyWithOneVarAllowed
{
     
    // To check a string S can be converted to a variation
    // string
    public static bool checkForVariation(String str)
    {
        if(str == null || str.Length != 0)
        {
            return true;
        }
         
        Dictionary<char, int> map = new Dictionary<char, int>();
         
        // Run loop form 0 to length of string
        for(int i = 0; i < str.Length; i++)
        {
            if(map.ContainsKey(str[i]))
                map[str[i]] = map[str[i]]+1;
            else
                map.Add(str[i], 1);
        }
 
        // declaration of variables
        bool first = true, second = true;
        int val1 = 0, val2 = 0;
        int countOfVal1 = 0, countOfVal2 = 0;
         
        foreach(KeyValuePair<char, int> itr in map)
        {
            int i = itr.Key;
             
            // 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)
    {
             
        Console.WriteLine(checkForVariation("abcbc"));
    }
}
 
// This code is contributed by 29AjayKumar

Output: 

true

Another method: Using Built-in Python Functions

  • Calculate the frequencies of all characters using Counter() function.
  • Convert these frequencies to the list.
  • Calculate again the frequencies of this list using Counter.
  • If the length of Counter is 1 then return true.
  • If the length of Counter is 2 if min  value is 1 then return true.
  • Else return False.

Below is the implementation:

Python3




# Python program
from collections import Counter
 
# To check a string S can be
# converted to a variation
# string
def checkForVariation(strr):
   
    freq = Counter(strr)
     
    # Converting these values to list
    valuelist = list(freq.values())
     
    # Counting frequencies again
    ValueCounter = Counter(valuelist)
    if(len(ValueCounter) == 1):
        return True
    elif(len(ValueCounter) == 2 and
         min(ValueCounter.values()) == 1):
        return True
       
    # If no conditions satisfied return false
    return False
 
 
# Driver code
string = "abcbc"
 
# passing string to checkForVariation Function
print(checkForVariation(string))
 
# This code is contributed by vikkycirus

Output:

true

Time Complexity: O(n)

Space Complexity: O(n)

This article is contributed by Nishant_singh(pintu). 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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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