LeetCode #2514 — HARD

Count Anagrams

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

You are given a string s containing one or more words. Every consecutive pair of words is separated by a single space ' '.

A string t is an anagram of string s if the ith word of t is a permutation of the ith word of s.

  • For example, "acb dfe" is an anagram of "abc def", but "def cab" and "adc bef" are not.

Return the number of distinct anagrams of s. Since the answer may be very large, return it modulo 109 + 7.

Example 1:

Input: s = "too hot"
Output: 18
Explanation: Some of the anagrams of the given string are "too hot", "oot hot", "oto toh", "too toh", and "too oht".

Example 2:

Input: s = "aa"
Output: 1
Explanation: There is only one anagram possible for the given string.

Constraints:

  • 1 <= s.length <= 105
  • s consists of lowercase English letters and spaces ' '.
  • There is single space between consecutive words.

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given a string s containing one or more words. Every consecutive pair of words is separated by a single space ' '. A string t is an anagram of string s if the ith word of t is a permutation of the ith word of s. For example, "acb dfe" is an anagram of "abc def", but "def cab" and "adc bef" are not. Return the number of distinct anagrams of s. Since the answer may be very large, return it modulo 109 + 7.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Math

Example 1

"too hot"

Example 2

"aa"

Related Problems

  • Group Anagrams (group-anagrams)
  • Count Ways to Build Rooms in an Ant Colony (count-ways-to-build-rooms-in-an-ant-colony)
Step 02

Core Insight

What unlocks the optimal approach

  • For each word, can you count the number of permutations possible if all characters are distinct?
  • How to reduce overcounting when letters are repeated?
  • The product of the counts of distinct permutations of all words will give the final answer.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2514: Count Anagrams
import java.math.BigInteger;

class Solution {
    private static final int MOD = (int) 1e9 + 7;

    public int countAnagrams(String s) {
        int n = s.length();
        long[] f = new long[n + 1];
        f[0] = 1;
        for (int i = 1; i <= n; ++i) {
            f[i] = f[i - 1] * i % MOD;
        }
        long p = 1;
        for (String w : s.split(" ")) {
            int[] cnt = new int[26];
            for (int i = 0; i < w.length(); ++i) {
                ++cnt[w.charAt(i) - 'a'];
            }
            p = p * f[w.length()] % MOD;
            for (int v : cnt) {
                p = p * BigInteger.valueOf(f[v]).modInverse(BigInteger.valueOf(MOD)).intValue()
                    % MOD;
            }
        }
        return (int) p;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(n)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP
O(n) time
O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

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