Frequency maps and two pointers

Two patterns that solve half the easy problems

Interview problems repeat themselves: behind fresh wording, a small set of patterns does most of the work, and recognizing one turns a blank stare into a plan. This lesson drills the two highest-frequency patterns for strings and arrays — each replaces a naive O(n²) nested loop with an O(n) or O(n log n) idea, which is usually the exact improvement the interviewer is waiting for you to name.

Frequency map (lesson 7-2's counting idiom): count occurrences first, then answer questions in a second pass. First non-repeating character, anagram checks, and most "count the..." problems fall to it.

Anagram check: two words are anagrams if sorting their characters gives the same result.

char[] a = s1.toCharArray();
char[] b = s2.toCharArray();
Arrays.sort(a);
Arrays.sort(b);
boolean anagram = Arrays.equals(a, b);

toCharArray() explodes a String into a char array, and java.util.Arrays provides sort, equals, and toString for arrays. These three static helpers appear in nearly every interview solution you will write.

Two pointers

Many array and string tasks compare or move elements from both ends: reverse in place, check a palindrome without building a copy, find a pair in sorted data. The two-pointer pattern keeps one index at each end and walks them toward each other:

int left = 0;
int right = arr.length - 1;
while (left < right) {
  // work with arr[left] and arr[right]: swap, compare...
  left++;
  right--;
}

Each element is visited once, so the whole pass is O(n) with no extra array — the two things this pattern is prized for. The loop condition left < right is the standard stop: the pointers have met or crossed, meaning every pair has been handled.