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

05 Jul, 2022
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. 
Implementation:
C++
C
Java
Python3
C#
Javascript
// 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 charactersconst 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 codeint main(){    char str[] = "abcbc";     if (isValidString(str))        cout << "YES" << endl;    else        cout << "NO" << endl;    return 0;} // This code is contributed by Aditya Kumar (adityakumar129)
// C program to check if a string can be made// valid by removing at most 1 character.#include <stdbool.h>#include <stdio.h>#include <string.h> // Assuming only lower case charactersconst 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(char str[]){    int freq[CHARS];    for (int i = 0; i < CHARS; i++)        freq[i] = 0;     // freq[] : stores the  frequency of each character of a    // string    for (int i = 0; i < strlen(str); 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 codeint main(){    char str[] = "abcbc";     if (isValidString(str))        printf("YES\n");    else        printf("NO\n");    return 0;} // This code is contributed by Aditya Kumar (adityakumar129)
// 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 Aditya Kumar (adityakumar129)
# Python 3 program to check if# a string can be made# valid by removing at most 1 character. # Assuming only lower case charactersCHARS = 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 codeif __name__ == "__main__":    str= "abcbc"     if (isValidString(str)):        print("YES")    else:        print("NO")         # this code is contributed by# ChitraNayal
// 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 onecharacter. */    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
<script> // JavaScript program to check if a string can be made// valid by removing at most 1 character. // Assuming only lower case characterslet CHARS = 26; /* To check a string S can be converted to a “valid”   string by removing less than or equal to one   character. */function isValidString(str){    let freq = new Array(CHARS);    for(let i=0;i<CHARS;i++)    {        freq[i]=0;    }          // freq[] : stores the  frequency of each        // character of a string        for (let i = 0; i < str.length; i++) {            freq[str[i].charCodeAt(0) - 'a'.charCodeAt(0)]++;        }          // Find first character with non-zero frequency        let 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.        let 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 (let 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 codelet str = "abcbc";          if (isValidString(str)) {            document.write("YES");        } else {            document.write("NO");        } // This code is contributed by ab2127 </script>
YES
Time Complexity: O(N), where N is the length of the given string.Auxiliary Space: O(1), no other extra space is required, so it is a constant.
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++
Java
Python3
C#
Javascript
// 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 codeint main() {    if(checkForVariation("abcbcvf"))        cout << "true" << endl;    else        cout << "false" << endl;         return 0;} // This code is contributed by avanitrachhadiya2155
// 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"));    }}
# 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# stringdef 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 codeprint(checkForVariation("abcbc")) # This code is contributed by rag2127
// 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
<script> // JavaScript 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    // stringfunction checkForVariation(str){    if(str == null || str.length==0) {            return true;        }                  let map = new Map();                  // Run loop form 0 to length of string        for(let i = 0; i < str.length; i++) {            if(!map.has(str[i]))                map.set(str[i],0);            map.set(str[i], map.get(str[i]) + 1);        }                           // declaration of variables        let first = true, second = true;        let val1 = 0, val2 = 0;        let countOfVal1 = 0, countOfVal2 = 0;                  for(let [key, value] of map.entries()) {            let i = value;                          // 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 codedocument.write(checkForVariation("abcbc"));  // This code is contributed by patel2127 </script>
true
Time Complexity: O(NlogN), where N is the length of the given string.Auxiliary Space: O(N)
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 programfrom collections import Counter # To check a string S can be# converted to a variation# stringdef 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 codestring = "abcbc" # passing string to checkForVariation Functionprint(checkForVariation(string)) # This code is contributed by vikkycirus
True
Time Complexity: O(n)Space Complexity: O(n)
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