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

• Difficulty Level : Medium
• Last Updated : 23 May, 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.

## C++

 `// C++ program to check if a string can be made``// valid by removing at most 1 character.``#include ``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;``}` `// This code is contributed by Aditya Kumar (adityakumar129)`

## C

 `// C program to check if a string can be made``// valid by removing at most 1 character.``#include ``#include ``#include ` `// 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(``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 code``int` `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

 `// 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)`

## 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`

## Javascript

 ``

Output:

`YES`

Time Complexity: O(N), where N is the length of the given string.

Auxiliary Space: O(26), 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++

 `// C++ program to check if a string can be made``// valid by removing at most 1 character using hashmap.``#include ``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 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 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`

## Javascript

 ``

Output:

`true`

Time Complexity: O(NlogN), where N is the length of the given string.

Auxiliary Space: O(N)

#### 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``    ``else` `if``(``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)

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