PROBLEM LINKS

Practice
Contest

DIFFICULTY

CAKEWALK-SIMPLE

PREREQUISITES

Simple Math

PROBLEM

A pair of strings R and S is called granama if each letter 'a' - 'z' is used the same number of times in both R and S. However, Chef mistakenly considers them as granama if there are no letters which are used in R but not used in S (or vice-versa).

Determine whether Chef correctly classified two strings R and S as granama or not granama.

QUICK EXPLANATION

The answer is NO if both R and S have the same set of letters and there is at least one letter which is used different number of times in R and S. Otherwise, the answer is YES.

EXPLANATION

The most intuitive way to solve this problem is to simulate both methods of classifying whether a pair of strings is granama: the correct method and Chef's wrong method. If both methods return the same answer, output "YES", else output "NO".

read(R, S)
if correct_method(R, S) = wrong_method(R, S):
    println("YES")
else:
    println("NO")

Correct Method

There are several ways to implement the correct method.

• Use two tables: freqR[26] and freqS[26]. Populate the tables with the frequencies of letters 'a' - 'z' in R and S. The answer is "YES" if and only if the frequency of each letter is equal in both R and S.
  

  function correct_method(R, S):
      for i = 0; i < length(R); i++:
          freqR[R[i] - 'a']++
      for i = 0; i < length(S); i++:
          freqS[S[i] - 'a']++
      res = "YES"
      for i = 0; i < 26; i++:
          if freqR[i] ≠ freqS[i]:
              res = "NO"
      return res
  

• Sort both R and S based on their letters. If they become equal, the answer is "YES", otherwise "NO".

Chef's Wrong Method

Chef will consider a pair of strings as granama if they have the same set of letters. Therefore, we can use two boolean tables: usedR[26] and usedS[26]. For each letter in R and S, mark it as used in the corresponding table. The answer is "YES" if and only if the flag of each letter is equal in both R and S.

function wrong_method(R, S):
    for i = 0; i < length(R); i++:
        usedR[R[i] - 'a'] = true
    for i = 0; i < length(S); i++:
        freqS[S[i] - 'a'] = true
    res = "YES"
    for i = 0; i < 26; i++:
        if usedR[i] ≠ usedS[i]:
            res = "NO"
    return res

Concise Solution

The above method suffices to get Accepted on this problem. However, there is a concise solution mentioned in the Quick Explanation section based on a clever observation. The proof is left as an exercise for the readers.

SETTER'S SOLUTION

Will be provided soon.

TESTER'S SOLUTION

Can be found here.