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:



  • Intialize a hash-map to store the frequency of the characters.
  • Iterate over the string and increment the frequecy 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.
  • Intialize 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:

Python3

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# Python 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 occurence 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)

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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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