# Form lexicographically smallest string with minimum replacements having equal number of 0s, 1s and 2s

Given a string str of length n (n is a multiple of 3) containing characters only from the set {0, 1, 2}. The task is to update the string such that each character has the same frequency with minimum number of operations. In a single operation, any character of the string can be replaced with any other character (also from the same set). If there are multiple strings possible then print the lexicographically smallest one.

Examples:

Input: str = “000000”
Output: 001122
Replace 3rd and 4th ‘0’ with ‘1’ and 5th and 6th ‘0’ with ‘2’ such that given condition is satisfied and it forms lexicographically smallest string

Input: str = “211200”
Output: 211200
The string already has equal number of 0s, 1s and 2s, hence there is no need to perform any operation.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: This problem can be solved using greedy approach. We need only n / 3 number of characters of each type where n is the size of the string. Iterate over the string and count the number of characters of each type. Again, iterate over the string and now check if the current character’s count is equal to n / 3 then there is no need to perform any operation on the current character.
However, if count(currentChar) != n / 3 then we may need to perform replacement operation depending on the value of the character in such a way that it maintains smallest lexicographical order as follows:

• If the current character is zero(0) and we have already processed required number of zeroes, then this character needs to be replaced with either one(1) if count[1] < (n / 3) or with two(1) if count[2] < (n / 3).
• If the current character is one(1), then we can either replace it with zero(0) if count[0] < (n / 3) or with two(2) if count[2] < (n / 3) and we have already processed required number of ones to maintain lowest lexicographical order
• If the current character is two(2), then we can either replace it with zero(0) if count[0] < (n / 3) or with two(2) if count[2] < (n / 3).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function that returns the modified lexicographically ` `// smallest string after performing minimum number ` `// of given operations ` `string formStringMinOperations(string s) ` `{ ` `    ``// Stores the initial frequencies of characters ` `    ``// 0s, 1s and 2s ` `    ``int` `count[3] = { 0 }; ` `    ``for` `(``auto``& c : s) ` `        ``count++; ` ` `  `    ``// Stores number of processed characters upto that ` `    ``// point of each type ` `    ``int` `processed[3] = { 0 }; ` ` `  `    ``// Required number of characters of each type ` `    ``int` `reqd = (``int``)s.size() / 3; ` `    ``for` `(``int` `i = 0; i < s.size(); i++) { ` ` `  `        ``// If the current type has already reqd ` `        ``// number of characters, no need to perform ` `        ``// any operation ` `        ``if` `(count[s[i] - ``'0'``] == reqd) ` `            ``continue``; ` ` `  `        ``// Process all 3 cases ` `        ``if` `(s[i] == ``'0'` `&& count[0] > reqd && ` `                     ``processed[0] >= reqd) { ` ` `  `            ``// Check for 1 first ` `            ``if` `(count[1] < reqd) { ` `                ``s[i] = ``'1'``; ` `                ``count[1]++; ` `                ``count[0]--; ` `            ``} ` ` `  `            ``// Else 2 ` `            ``else` `if` `(count[2] < reqd) { ` `                ``s[i] = ``'2'``; ` `                ``count[2]++; ` `                ``count[0]--; ` `            ``} ` `        ``} ` ` `  `        ``// Here we need to check processed[1] only ` `        ``// for 2 since 0 is less than 1 and we can ` `        ``// replace it anytime ` `        ``if` `(s[i] == ``'1'` `&& count[1] > reqd) { ` `            ``if` `(count[0] < reqd) { ` `                ``s[i] = ``'0'``; ` `                ``count[0]++; ` `                ``count[1]--; ` `            ``} ` `            ``else` `if` `(count[2] < reqd && ` `                    ``processed[1] >= reqd) { ` `                ``s[i] = ``'2'``; ` `                ``count[2]++; ` `                ``count[1]--; ` `            ``} ` `        ``} ` ` `  `        ``// Here we can replace 2 with 0 and 1 anytime ` `        ``if` `(s[i] == ``'2'` `&& count[2] > reqd) { ` `            ``if` `(count[0] < reqd) { ` `                ``s[i] = ``'0'``; ` `                ``count[0]++; ` `                ``count[2]--; ` `            ``} ` `            ``else` `if` `(count[1] < reqd) { ` `                ``s[i] = ``'1'``; ` `                ``count[1]++; ` `                ``count[2]--; ` `            ``} ` `        ``} ` ` `  `        ``// keep count of processed characters of each ` `        ``// type ` `        ``processed[s[i] - ``'0'``]++; ` `    ``} ` `    ``return` `s; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``string s = ``"011200"``; ` `    ``cout << formStringMinOperations(s); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG  ` `{ ` ` `  `    ``// Function that returns the  ` `    ``// modified lexicographically ` `    ``// smallest String after  ` `    ``// performing minimum number ` `    ``// of given operations ` `    ``static` `String formStringMinOperations(``char``[] s)  ` `    ``{ ` `         `  `        ``// Stores the initial frequencies  ` `        ``// of characters 0s, 1s and 2s ` `        ``int` `count[] = ``new` `int``[``3``]; ` `        ``for` `(``char` `c : s)  ` `        ``{ ` `            ``count[(``int``)c - ``48``] += ``1``; ` `        ``} ` ` `  `        ``// Stores number of processed characters  ` `        ``// upto that point of each type ` `        ``int` `processed[] = ``new` `int``[``3``]; ` ` `  `        ``// Required number of characters of each type ` `        ``int` `reqd = (``int``) s.length / ``3``; ` `        ``for` `(``int` `i = ``0``; i < s.length; i++)  ` `        ``{ ` ` `  `            ``// If the current type has already  ` `            ``// reqd number of characters, no  ` `            ``// need to perform any operation ` `            ``if` `(count[s[i] - ``'0'``] == reqd)  ` `            ``{ ` `                ``continue``; ` `            ``} ` ` `  `            ``// Process all 3 cases ` `            ``if` `(s[i] == ``'0'` `&& count[``0``] > reqd ` `                    ``&& processed[``0``] >= reqd)  ` `            ``{ ` ` `  `                ``// Check for 1 first ` `                ``if` `(count[``1``] < reqd)  ` `                ``{ ` `                    ``s[i] = ``'1'``; ` `                    ``count[``1``]++; ` `                    ``count[``0``]--; ` `                ``}  ` `                 `  `                ``// Else 2 ` `                ``else` `if` `(count[``2``] < reqd)  ` `                ``{ ` `                    ``s[i] = ``'2'``; ` `                    ``count[``2``]++; ` `                    ``count[``0``]--; ` `                ``} ` `            ``} ` ` `  `            ``// Here we need to check processed[1] only ` `            ``// for 2 since 0 is less than 1 and we can ` `            ``// replace it anytime ` `            ``if` `(s[i] == ``'1'` `&& count[``1``] > reqd)  ` `            ``{ ` `                ``if` `(count[``0``] < reqd)  ` `                ``{ ` `                    ``s[i] = ``'0'``; ` `                    ``count[``0``]++; ` `                    ``count[``1``]--; ` `                ``}  ` `                ``else` `if` `(count[``2``] < reqd ` `                        ``&& processed[``1``] >= reqd)  ` `                ``{ ` `                    ``s[i] = ``'2'``; ` `                    ``count[``2``]++; ` `                    ``count[``1``]--; ` `                ``} ` `            ``} ` ` `  `            ``// Here we can replace 2 with 0 and 1 anytime ` `            ``if` `(s[i] == ``'2'` `&& count[``2``] > reqd)  ` `            ``{ ` `                ``if` `(count[``0``] < reqd) ` `                ``{ ` `                    ``s[i] = ``'0'``; ` `                    ``count[``0``]++; ` `                    ``count[``2``]--; ` `                ``}  ` `                ``else` `if` `(count[``1``] < reqd)  ` `                ``{ ` `                    ``s[i] = ``'1'``; ` `                    ``count[``1``]++; ` `                    ``count[``2``]--; ` `                ``} ` `            ``} ` ` `  `            ``// keep count of processed  ` `            ``// characters of each type ` `            ``processed[s[i] - ``'0'``]++; ` `        ``} ` `        ``return` `String.valueOf(s); ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``String s = ``"011200"``; ` `        ``System.out.println(formStringMinOperations(s.toCharArray())); ` `    ``} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python3 implementation of the approach  ` `import` `math ` ` `  `# Function that returns the modified  ` `# lexicographically smallest string after  ` `# performing minimum number of given operations  ` `def` `formStringMinOperations(ss): ` `     `  `    ``# Stores the initial frequencies of  ` `    ``# characters 0s, 1s and 2s  ` `    ``count ``=` `[``0``] ``*` `3``; ` `    ``s ``=` `list``(ss); ` `    ``for` `i ``in` `range``(``len``(s)):  ` `        ``count[``ord``(s[i]) ``-` `ord``(``'0'``)] ``+``=` `1``;  ` ` `  `    ``# Stores number of processed characters  ` `    ``# upto that point of each type  ` `    ``processed ``=` `[``0``] ``*` `3``;  ` ` `  `    ``# Required number of characters of each type  ` `    ``reqd ``=` `math.floor(``len``(s) ``/` `3``);  ` `    ``for` `i ``in` `range``(``len``(s)):  ` ` `  `        ``# If the current type has already reqd  ` `        ``# number of characters, no need to  ` `        ``# perform any operation  ` `        ``if` `(count[``ord``(s[i]) ``-` `ord``(``'0'``)] ``=``=` `reqd):  ` `            ``continue``;  ` ` `  `        ``# Process all 3 cases  ` `        ``if` `(s[i] ``=``=` `'0'` `and` `count[``0``] > reqd ``and` `                            ``processed[``0``] >``=` `reqd): ` ` `  `            ``# Check for 1 first  ` `            ``if` `(count[``1``] < reqd): ` `                ``s[i] ``=` `'1'``;  ` `                ``count[``1``] ``+``=` `1``;  ` `                ``count[``0``] ``-``=` `1``;  ` ` `  `            ``# Else 2  ` `            ``elif` `(count[``2``] < reqd):  ` `                ``s[i] ``=` `'2'``;  ` `                ``count[``2``] ``+``=` `1``;  ` `                ``count[``0``] ``-``=` `1``;  ` ` `  `        ``# Here we need to check processed[1] only  ` `        ``# for 2 since 0 is less than 1 and we can  ` `        ``# replace it anytime  ` `        ``if` `(s[i] ``=``=` `'1'` `and` `count[``1``] > reqd): ` `            ``if` `(count[``0``] < reqd): ` `                ``s[i] ``=` `'0'``;  ` `                ``count[``0``] ``+``=` `1``;  ` `                ``count[``1``] ``-``=` `1``;  ` `            ``elif` `(count[``2``] < reqd ``and`  `                  ``processed[``1``] >``=` `reqd): ` `                ``s[i] ``=` `'2'``;  ` `                ``count[``2``] ``+``=` `1``;  ` `                ``count[``1``] ``-``=` `1``;  ` ` `  `        ``# Here we can replace 2 with 0 and 1 anytime  ` `        ``if` `(s[i] ``=``=` `'2'` `and` `count[``2``] > reqd): ` `            ``if` `(count[``0``] < reqd): ` `                ``s[i] ``=` `'0'``;  ` `                ``count[``0``] ``+``=` `1``;  ` `                ``count[``2``] ``-``=` `1``;  ` `            ``elif` `(count[``1``] < reqd): ` `                ``s[i] ``=` `'1'``;  ` `                ``count[``1``] ``+``=` `1``;  ` `                ``count[``2``] ``-``=` `1``;  ` ` `  `        ``# keep count of processed characters  ` `        ``# of each type  ` `        ``processed[``ord``(s[i]) ``-` `ord``(``'0'``)] ``+``=` `1``;  ` `    ``return` `''.join(s);  ` ` `  `# Driver Code  ` `s ``=` `"011200"``;  ` `print``(formStringMinOperations(s));  ` ` `  `# This code is contributed by mits `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// Function that returns the  ` `    ``// modified lexicographically ` `    ``// smallest String after  ` `    ``// performing minimum number ` `    ``// of given operations ` `    ``static` `String formStringMinOperations(``char``[] s)  ` `    ``{ ` `         `  `        ``// Stores the initial frequencies  ` `        ``// of characters 0s, 1s and 2s ` `        ``int` `[]count = ``new` `int``[3]; ` `        ``foreach` `(``char` `c ``in` `s)  ` `        ``{ ` `            ``count[(``int``)c - 48] += 1; ` `        ``} ` ` `  `        ``// Stores number of processed characters  ` `        ``// upto that point of each type ` `        ``int` `[]processed = ``new` `int``[3]; ` ` `  `        ``// Required number of characters of each type ` `        ``int` `reqd = (``int``) s.Length / 3; ` `        ``for` `(``int` `i = 0; i < s.Length; i++)  ` `        ``{ ` ` `  `            ``// If the current type has already  ` `            ``// reqd number of characters, no  ` `            ``// need to perform any operation ` `            ``if` `(count[s[i] - ``'0'``] == reqd)  ` `            ``{ ` `                ``continue``; ` `            ``} ` ` `  `            ``// Process all 3 cases ` `            ``if` `(s[i] == ``'0'` `&& count[0] > reqd ` `                    ``&& processed[0] >= reqd)  ` `            ``{ ` ` `  `                ``// Check for 1 first ` `                ``if` `(count[1] < reqd)  ` `                ``{ ` `                    ``s[i] = ``'1'``; ` `                    ``count[1]++; ` `                    ``count[0]--; ` `                ``}  ` `                 `  `                ``// Else 2 ` `                ``else` `if` `(count[2] < reqd)  ` `                ``{ ` `                    ``s[i] = ``'2'``; ` `                    ``count[2]++; ` `                    ``count[0]--; ` `                ``} ` `            ``} ` ` `  `            ``// Here we need to check processed[1] only ` `            ``// for 2 since 0 is less than 1 and we can ` `            ``// replace it anytime ` `            ``if` `(s[i] == ``'1'` `&& count[1] > reqd)  ` `            ``{ ` `                ``if` `(count[0] < reqd)  ` `                ``{ ` `                    ``s[i] = ``'0'``; ` `                    ``count[0]++; ` `                    ``count[1]--; ` `                ``}  ` `                ``else` `if` `(count[2] < reqd ` `                        ``&& processed[1] >= reqd)  ` `                ``{ ` `                    ``s[i] = ``'2'``; ` `                    ``count[2]++; ` `                    ``count[1]--; ` `                ``} ` `            ``} ` ` `  `            ``// Here we can replace 2 with 0 and 1 anytime ` `            ``if` `(s[i] == ``'2'` `&& count[2] > reqd)  ` `            ``{ ` `                ``if` `(count[0] < reqd) ` `                ``{ ` `                    ``s[i] = ``'0'``; ` `                    ``count[0]++; ` `                    ``count[2]--; ` `                ``}  ` `                ``else` `if` `(count[1] < reqd)  ` `                ``{ ` `                    ``s[i] = ``'1'``; ` `                    ``count[1]++; ` `                    ``count[2]--; ` `                ``} ` `            ``} ` ` `  `            ``// keep count of processed  ` `            ``// characters of each type ` `            ``processed[s[i] - ``'0'``]++; ` `        ``} ` `        ``return` `String.Join(``""``,s); ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `Main(String[] args)  ` `    ``{ ` `        ``String s = ``"011200"``; ` `        ``Console.WriteLine(formStringMinOperations(s.ToCharArray())); ` `    ``} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## PHP

 ` ``\$reqd` `&&  ` `                             ``\$processed``[0] >= ``\$reqd``) ` `        ``{  ` ` `  `            ``// Check for 1 first  ` `            ``if` `(``\$count``[1] < ``\$reqd``) ` `            ``{  ` `                ``\$s``[``\$i``] = ``'1'``;  ` `                ``\$count``[1]++;  ` `                ``\$count``[0]--;  ` `            ``}  ` ` `  `            ``// Else 2  ` `            ``else` `if` `(``\$count``[2] < ``\$reqd``)  ` `            ``{  ` `                ``\$s``[``\$i``] = ``'2'``;  ` `                ``\$count``[2]++;  ` `                ``\$count``[0]--;  ` `            ``}  ` `        ``}  ` ` `  `        ``// Here we need to check processed[1] only  ` `        ``// for 2 since 0 is less than 1 and we can  ` `        ``// replace it anytime  ` `        ``if` `(``\$s``[``\$i``] == ``'1'` `&& ``\$count``[1] > ``\$reqd``)  ` `        ``{  ` `            ``if` `(``\$count``[0] < ``\$reqd``)  ` `            ``{  ` `                ``\$s``[``\$i``] = ``'0'``;  ` `                ``\$count``[0]++;  ` `                ``\$count``[1]--;  ` `            ``}  ` `            ``else` `if` `(``count``[2] < ``\$reqd` `&&  ` `                                ``\$processed``[1] >= ``\$reqd``)  ` `            ``{  ` `                ``\$s``[``\$i``] = ``'2'``;  ` `                ``\$count``[2]++;  ` `                ``\$count``[1]--;  ` `            ``}  ` `        ``}  ` ` `  `        ``// Here we can replace 2 with 0 and 1 anytime  ` `        ``if` `(``\$s``[``\$i``] == ``'2'` `&& ``\$count``[2] > ``\$reqd``) ` `        ``{  ` `            ``if` `(``\$count``[0] < ``\$reqd``) ` `            ``{  ` `                ``\$s``[``\$i``] = ``'0'``;  ` `                ``\$count``[0]++;  ` `                ``\$count``[2]--;  ` `            ``}  ` `            ``else` `if` `(``\$count``[1] < ``\$reqd``)  ` `            ``{  ` `                ``\$s``[``\$i``] = ``'1'``;  ` `                ``\$count``[1]++;  ` `                ``\$count``[2]--;  ` `            ``}  ` `        ``}  ` ` `  `        ``// keep count of processed characters  ` `        ``// of each type  ` `        ``\$processed``[``\$s``[``\$i``] - ``'0'``]++;  ` `    ``}  ` `    ``return` `\$s``;  ` `}  ` ` `  `// Driver Code  ` `\$s` `= ``"011200"``;  ` `echo` `formStringMinOperations(``\$s``);  ` ` `  `// This code is contributed by Ryuga ` `?> `

Output:

```011202
```

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

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 Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.