InterviewBit: Self Permutation Solution

Approach

Permutation exists if the letter counts in both string are equal

Implementation

1function permuteStrings(A, B) {
2  const [countA, countB] = [new Uint8Array(26), new Uint8Array(26)]
3
4  A.split("").forEach(char => countA[char.charCodeAt(0) - "a".charCodeAt(0)]++)
5  B.split("").forEach(char => countB[char.charCodeAt(0) - "a".charCodeAt(0)]++)
6
7  for (let i = 0; i < 26; i++) {
8    if (countA[i] !== countB[i]) return 0
9  }
10
11  return 1
12}

References

Original problem

Similar problems

Amazing Subarrays

Palindrome String

Minimum Characters required to make a String Palindromic