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Sherlock considers a string to be valid if all characters of the string appear the same number of times. It is also valid if he can remove just character at index in the string, and the remaining characters will occur the same number of times. Given a string , determine if it is valid. If so, return YES, otherwise return NO.

Example

This is a valid string because frequencies are .

This is a valid string because we can remove one and have of each character in the remaining string.

This string is not valid as we can only remove occurrence of . That leaves character frequencies of .

Function Description

Complete the isValid function in the editor below.

isValid has the following parameter(s):

  • string s: a string

Returns

  • string: either YES or NO

Input Format

A single string .

Constraints

  • Each character

Sample Input 0

aabbcd

Sample Output 0

NO

Explanation 0

Given , we would need to remove two characters, both c and d aabb or a and b abcd, to make it valid. We are limited to removing only one character, so is invalid.

Sample Input 1

aabbccddeefghi

Sample Output 1

NO

Explanation 1

Frequency counts for the letters are as follows:

{'a': 2, 'b': 2, 'c': 2, 'd': 2, 'e': 2, 'f': 1, 'g': 1, 'h': 1, 'i': 1}

There are two ways to make the valid string:

  • Remove characters with a frequency of : .
  • Remove characters of frequency : .

Neither of these is an option.

Sample Input 2

abcdefghhgfedecba

Sample Output 2

YES

Explanation 2

All characters occur twice except for which occurs times. We can delete one instance of to have a valid string.

Limbajul de programare folosit: python3

Cod:

#!/bin/python3

from collections import Counter

def isValid(s):
    freq = Counter(s)
    cnt = Counter(freq.values())
    if len(cnt) == 1:
        return 'YES'
    if len(cnt) > 2:
        return 'NO'
    (f1, c1), (f2, c2) = sorted(cnt.items())
    if f1 == 1 and c1 == 1:
        return 'YES'
    if f2 == f1 + 1 and c2 == 1:
        return 'YES'
    return 'NO'

if __name__ == '__main__':
    s = input().strip()
    print(isValid(s))

Scor obtinut: 1.0

Submission ID: 464591309

Link challenge: https://www.hackerrank.com/challenges/sherlock-and-valid-string/problem

Sherlock and the Valid String