# Lexicographically smallest substring with maximum occurrences containing a’s and b’s only

Given a string ( containing characters from ‘0’ to ‘9’) and two digits and . The task is to find the substring in the given string with maximum occurrences and containing a’s and b’s only. If there are two or more such substrings with same frequencies then print the lexicographically smallest. If there does not exists any such substring then print -1.

Examples

```Input : str = "47", a = 4, b = 7
Output : 4

Input : str = "23", a = 4, b = 7
Output : -1```

The idea is to observe that we need to find the substring with maximum number of occurrences. So, if we consider substrings that contains both a’s and b’s then the number of occurrences will be less than if we consider the substrings with single digits ‘a’ and ‘b’ individually.

So, the idea is to calculate the frequency of digits of ‘a’ and ‘b’ in the string and the one with maximum frequency will be the answer.

Note: If both digits have same frequency then the digit which is lexicographically smaller among ‘a’ and ‘b’ will be the answer.

Below is the implementation of the above approach:

## C++

 `// CPP program to Find the lexicographically``// smallest substring in a given string with``// maximum frequency and contains a's and b's only` `#include ``using` `namespace` `std;` `// Function to Find the lexicographically``// smallest substring in a given string with``// maximum frequency and contains a's and b's only.``int` `maxFreq(string s, ``int` `a, ``int` `b)``{``    ``// To store frequency of digits``    ``int` `fre[10] = { 0 };` `    ``// size of string``    ``int` `n = s.size();` `    ``// Take lexicographically larger digit in b``    ``if` `(a > b)``        ``swap(a, b);` `    ``// get frequency of each character``    ``for` `(``int` `i = 0; i < n; i++)``        ``fre[s[i] - ``'0'``]++;` `    ``// If no such string exits``    ``if` `(fre[a] == 0 and fre[b] == 0)``        ``return` `-1;` `    ``// Maximum frequency``    ``else` `if` `(fre[a] >= fre[b])``        ``return` `a;``    ``else``        ``return` `b;``}` `// Driver program``int` `main()``{``    ``int` `a = 4, b = 7;` `    ``string s = ``"47744"``;` `    ``cout << maxFreq(s, a, b);` `    ``return` `0;``}`

## Java

 `// Java program to Find the lexicographically``// smallest substring in a given string with``// maximum frequency and contains a's and b's only` `import` `java.io.*;` `class` `GFG {`  `// Function to Find the lexicographically``// smallest substring in a given string with``// maximum frequency and contains a's and b's only.``static` `int` `maxFreq(String s, ``int` `a, ``int` `b)``{``    ``// To store frequency of digits``    ``int` `fre[] =  ``new` `int``[``10``];` `    ``// size of string``    ``int` `n = s.length();` `    ``// Take lexicographically larger digit in b``    ``if` `(a > b)``    ``{``        ``int` `temp = a;``        ``a =b;``        ``b = temp;``    ` `    ``}` `    ``// get frequency of each character``    ``for` `(``int` `i = ``0``; i < n; i++)``        ``fre[s.charAt(i) - ``'0'``]++;` `    ``// If no such string exits``    ``if` `(fre[a] == ``0` `&& fre[b] == ``0``)``        ``return` `-``1``;` `    ``// Maximum frequency``    ``else` `if` `(fre[a] >= fre[b])``        ``return` `a;``    ``else``        ``return` `b;``}` `// Driver program` `    ``public` `static` `void` `main (String[] args) {``    ` `    ``int` `a = ``4``, b = ``7``;` `    ``String s = ``"47744"``;` `    ``System.out.print(maxFreq(s, a, b));``    ``}``}``// This code is contributed by inder_verma`

## Python3

 `# Python 3 program to Find the lexicographically``# smallest substring in a given string with``# maximum frequency and contains a's and b's only` `# Function to Find the lexicographically``# smallest substring in a given string with``# maximum frequency and contains a's and b's only.``def` `maxFreq(s, a, b):``    ``# To store frequency of digits``    ``fre ``=` `[``0` `for` `i ``in` `range``(``10``)]` `    ``# size of string``    ``n ``=` `len``(s)` `    ``# Take lexicographically larger digit in b``    ``if` `(a > b):``        ``swap(a, b)` `    ``# get frequency of each character``    ``for` `i ``in` `range``(``0``,n,``1``):``        ``a ``=` `ord``(s[i]) ``-` `ord``(``'0'``)``        ``fre[a] ``+``=` `1` `    ``# If no such string exits``    ``if` `(fre[a] ``=``=` `0` `and` `fre[b] ``=``=` `0``):``        ``return` `-``1` `    ``# Maximum frequency``    ``elif` `(fre[a] >``=` `fre[b]):``        ``return` `a``    ``else``:``        ``return` `b` `# Driver program``if` `__name__ ``=``=` `'__main__'``:``    ``a ``=` `4``    ``b ``=` `7` `    ``s ``=` `"47744"` `    ``print``(maxFreq(s, a, b))` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `// C# program to Find the lexicographically``// smallest substring in a given string with``// maximum frequency and contains a's and b's only``using` `System;` `class` `GFG {`  `// Function to Find the lexicographically``// smallest substring in a given string with``// maximum frequency and contains a's and b's only.``static` `int` `maxFreq(``string` `s, ``int` `a, ``int` `b)``{``    ``// To store frequency of digits``    ``int` `[]fre = ``new` `int``[10];` `    ``// size of string``    ``int` `n = s.Length;` `    ``// Take lexicographically larger digit in b``    ``if` `(a > b)``    ``{``        ``int` `temp = a;``        ``a =b;``        ``b = temp;``    ` `    ``}` `    ``// get frequency of each character``    ``for` `(``int` `i = 0; i < n; i++)``        ``fre[s[i] - ``'0'``]++;` `    ``// If no such string exits``    ``if` `(fre[a] == 0 && fre[b] == 0)``        ``return` `-1;` `    ``// Maximum frequency``    ``else` `if` `(fre[a] >= fre[b])``        ``return` `a;``    ``else``        ``return` `b;``}` `// Driver program` `    ``public` `static` `void` `Main () {``    ` `    ``int` `a = 4, b = 7;` `    ``string` `s = ``"47744"``;` `     ``Console.WriteLine(maxFreq(s, a, b));``    ``}``}``// This code is contributed by inder_verma`

## PHP

 ` ``\$b``)``    ``{``        ``\$xx` `= ``\$a``;``        ``\$a` `= ``\$b``;``        ``\$b` `= ``\$xx``;}` `     ``// get frequency of each character``    ``for` `(``\$i` `= 0; ``\$i` `< ``\$n``; ``\$i``++)``    ``{``        ``\$a` `= ord(``\$s``[``\$i``]) - ord(``'0'``);``        ``\$fre``[``\$a``] += 1;``    ``}` `    ``// If no such string exits``    ``if` `(``\$fre``[``\$a``] == 0 ``and` `\$fre``[``\$b``] == 0)``        ``return` `-1;` `    ``// Maximum frequency``    ``else` `if` `(``\$fre``[``\$a``] >= ``\$fre``[``\$b``])``        ``return` `\$a``;``    ``else``        ``return` `\$b``;``}` `// Driver Code``\$a` `= 4;``\$b` `= 7;` `\$s` `= ``"47744"``;` `print``(maxFreq(``\$s``, ``\$a``, ``\$b``));` `// This code is contributed by mits``?>`

## Javascript

 ``

Output
`4`

Complexity Analysis:

• Time Complexity: O(n), where n is the length of the string s.
• Auxiliary Space: O(10)

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