# Maximum length prefix such that frequency of each character is atmost number of characters with minimum frequency

Given a string S, the task is to find the prefix of string S with the maximum possible length such that frequency of each character in the prefix is at most the number of characters in S with minimum frequency.

Examples:

Input: S = ‘aabcdaab’
Output: aabcd
Explanation:
Frequency of characters in the given string –
{a: 4, b: 2, c: 1, d: 1}
Minimum frequency in 1 and the count of minimum frequency is 2,
So frequency of each character in the prefix can be at most 2.

Input: S = ‘aaabc’
Output: aa
Explanation:
Frequency of characters in the given string –
{a: 3, b: 1, c: 1}
Minimum frequency in 1 and the count of minimum frequency is 2,
So frequency of each character in the prefix can be at most 2.

Approach:

• Initialize a hash-map to store the frequency of the characters.
• Iterate over the string and increment the frequency of the character in the hash-map.
• Find the minimum occurred character in the string and the count of such characters whose frequency is minimum.
• Initialize another hash-map to store the frequency of the characters of the possible prefix string.
• Finally, Iterate over the string from start and increment the count of the characters until the frequency of any characters is not greater than the count of the minimum frequency.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find the prefix` `// of the s such that occurrence of each` `// character is atmost the count of minimum` `// frequency in the s` `#include ` `using` `namespace` `std;`   `// Function to find the maximum` `// possible prefix of the s` `void` `MaxPrefix(string s)` `{` `    `  `    ``// Hash map to store the frequency` `    ``// of the characters in the s` `    ``map<``char``, ``int``> Dict;`   `    ``// Iterate over the s to find` `    ``// the occurrence of each Character` `    ``for``(``char` `i : s)` `    ``{` `        ``Dict[i]++;` `    ``}`   `    ``int` `minfrequency = INT_MAX;`   `    ``// Minimum frequency of the Characters` `    ``for``(``auto` `x : Dict)` `    ``{` `        ``minfrequency = min(minfrequency, x.second);` `    ``}`   `    ``int` `countminFrequency = 0;`   `    ``// Loop to find the count of minimum` `    ``// frequency in the hash-map` `    ``for``(``auto` `x: Dict)` `    ``{` `        ``if` `(x.second == minfrequency)` `            ``countminFrequency += 1;` `    ``}` `    ``map<``char``, ``int``> mapper;`   `    ``int` `indi = 0;`   `    ``// Loop to find the maximum possible` `    ``// length of the prefix in the s` `    ``for``(``char` `i: s)` `    ``{` `        ``mapper[i] += 1;`   `        ``// Condition to check if the frequency` `        ``// is greater than minimum possible freq` `        ``if` `(mapper[i] > countminFrequency)` `            ``break``;`   `        ``indi += 1;` `    ``}`   `    ``// maxprefix s and its length.` `    ``cout << (s.substr(0, indi));` `}`   `// Driver code` `int` `main()` `{`   `    ``// s is initialize.` `    ``string str = ``"aabcdaab"``;`   `    ``// str is passed in` `    ``// MaxPrefix function.` `    ``MaxPrefix(str);` `}`   `// This code is contributed by mohit kumar 29`

## Java

 `// Java implementation to find the prefix` `// of the s such that occurrence of each` `// character is atmost the count of minimum ` `// frequency in the s` `import` `java.util.*;` `import` `java.lang.*;` `import` `java.io.*;`   `class` `GFG{` `    `  `// Function to find the maximum` `// possible prefix of the s` `static` `void` `MaxPrefix(String s)` `{    ` `    `  `    ``// Hash map to store the frequency` `    ``// of the characters in the s` `    ``Map Dict = ``new` `HashMap<>();` ` `  `    ``// Iterate over the s to find` `    ``// the occurrence of each Character` `    ``for``(``char` `i : s.toCharArray())` `    ``{` `        ``Dict.put(i, Dict.getOrDefault(i, ``0``) + ``1``);` `    ``}` `     `  `    ``int` `minfrequency = Integer.MAX_VALUE;` `    `  `    ``// Minimum frequency of the Characters` `    ``for``(Integer x: Dict.values())` `    ``{` `        ``minfrequency = Math.min(minfrequency, x);    ` `    ``}` `     `  `    ``int` `countminFrequency = ``0``;` `      `  `    ``// Loop to find the count of minimum` `    ``// frequency in the hash-map` `    ``for``(Map.Entry x: Dict.entrySet())` `    ``{` `        ``if` `(x.getValue() == minfrequency)` `            ``countminFrequency += ``1``;` `    ``}` `     `  `   ``Map mapper = ``new` `HashMap<>(); ` `       `  `    ``int` `indi = ``0``;` `      `  `    ``// Loop to find the maximum possible ` `    ``// length of the prefix in the s` `    ``for``(``char` `i: s.toCharArray())` `    ``{` `        ``mapper.put(i, mapper.getOrDefault(i, ``0``) + ``1``);` `        `  `        ``// Condition to check if the frequency` `        ``// is greater than minimum possible freq` `        ``if` `(mapper.get(i) > countminFrequency)` `            ``break``;` `             `  `        ``indi += ``1``;` `    ``}` `              `  `    ``// maxprefix s and its length.` `    ``System.out.println(s.substring(``0``, indi));` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    `  `    ``// s is initialize.` `    ``String str = ``"aabcdaab"``;` `    `  `    ``// str is passed in ` `    ``// MaxPrefix function.` `    ``MaxPrefix(str); ` `}` `}`   `// This code is contributed by offbeat`

## Python3

 `# Python3 implementation to find the` `# prefix of the string such that ` `# occurrence of each character is` `# atmost the count of minimum ` `# frequency in the string`   `# Function to find the maximum` `# possible prefix of the string` `def` `MaxPrefix(string):` `    `  `    ``# Hash map to store the frequency` `    ``# of the characters in the string` `    ``Dict` `=` `{}` `    ``maxprefix ``=` `0` `    `  `    ``# Iterate over the string to find` `    ``# the occurrence of each Character` `    ``for` `i ``in` `string:` `        ``Dict``[i] ``=` `Dict``.get(i, ``0``) ``+` `1` `    `  `    ``# Minimum frequency of the Characters` `    ``minfrequency ``=` `min``(``Dict``.values())` `    ``countminFrequency ``=` `0` `    `  `    ``# Loop to find the count of minimum` `    ``# frequency in the hash-map` `    ``for` `x ``in` `Dict``:` `        ``if` `(``Dict``[x] ``=``=` `minfrequency):` `            ``countminFrequency ``+``=` `1` `    `  `    ``mapper ``=` `{}` `    ``indi ``=` `0` `    `  `    ``# Loop to find the maximum possible ` `    ``# length of the prefix in the string    ` `    ``for` `i ``in` `string:` `        ``mapper[i] ``=` `mapper.get(i, ``0``) ``+` `1` `        `  `        ``# Condition to check if the frequency` `        ``# is greater than minimum possible freq` `        ``if` `(mapper[i] > countminFrequency):` `            ``break` `        ``indi ``+``=` `1` `            `  `    ``# maxprefix string and its length.` `    ``print``(string[:indi])`   `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    `  `    ``# String is initialize.` `    ``str` `=` `'aabcdaab'` `    ``# str is passed in MaxPrefix function.` `    ``MaxPrefix(``str``)`

## C#

 `// C# implementation to find the` `// prefix of the s such that ` `// occurrence of each character is` `// atmost the count of minimum ` `// frequency in the s` `using` `System;` `using` `System.Collections;` `using` `System.Collections.Generic;` `using` `System.Linq;`   `class` `GFG{` ` `  `// Function to find the maximum` `// possible prefix of the s` `static` `void` `MaxPrefix(``string` `s)` `{    ` `    `  `    ``// Hash map to store the frequency` `    ``// of the characters in the s` `    ``Dictionary<``char``,` `               ``int``> Dict = ``new` `Dictionary<``char``,` `                                          ``int``>();`   `    ``// Iterate over the s to find` `    ``// the occurrence of each Character` `    ``foreach``(``char` `i ``in` `s)` `    ``{` `        ``if` `(Dict.ContainsKey(i))` `        ``{` `            ``Dict[i]++;` `        ``}` `        ``else` `        ``{` `            ``Dict[i] = 1;` `        ``}` `    ``}` `    `  `    ``int` `minfrequency = Int32.MaxValue;` `     `  `    ``// Minimum frequency of the Characters` `    ``foreach``(``int` `x ``in` `Dict.Values.ToList())` `    ``{` `        ``minfrequency = Math.Min(minfrequency, x);    ` `    ``}` `    `  `    ``int` `countminFrequency = 0;` `     `  `    ``// Loop to find the count of minimum` `    ``// frequency in the hash-map` `    ``foreach``(``char` `x ``in` `Dict.Keys.ToList())` `    ``{` `        ``if` `(Dict[x] == minfrequency)` `            ``countminFrequency += 1;` `    ``}` `    `  `    ``Dictionary<``char``,` `               ``int``> mapper = ``new` `Dictionary<``char``,` `                                            ``int``>(); ` `    ``int` `indi = 0;` `     `  `    ``// Loop to find the maximum possible ` `    ``// length of the prefix in the s` `    ``foreach``(``char` `i ``in` `s)` `    ``{` `        ``if` `(mapper.ContainsKey(i))` `        ``{` `            ``mapper[i]++;` `        ``}` `        ``else` `        ``{` `            ``mapper[i] = 1;` `        ``}` `         `  `        ``// Condition to check if the frequency` `        ``// is greater than minimum possible freq` `        ``if` `(mapper[i] > countminFrequency)` `            ``break``;` `            `  `        ``indi += 1;` `    ``}` `             `  `    ``// maxprefix s and its length.` `    ``Console.Write(s.Substring(0, indi));` `}`   `// Driver Code` `public` `static` `void` `Main(``string``[] args)` `{` `    `  `    ``// s is initialize.` `    ``string` `str = ``"aabcdaab"``;` `    `  `    ``// str is passed in ` `    ``// MaxPrefix function.` `    ``MaxPrefix(str);` `}` `}`   `// This code is contributed by rutvik_56`

## Javascript

 ``

Output:

`aabcd`

Performance Analysis:

• Time Complexity: In the above-given approach, there is one loop to find the frequency of each character in the string which takes O(N) time in the worst case. Therefore, the time complexity for this approach will be O(N).
• Space Complexity: In the above-given approach, there is extra space used to store the frequency of characters. Therefore, the space complexity for the above approach will be O(N)

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