LeetCode #2273 — EASY

Find Resultant Array After Removing Anagrams

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given a 0-indexed string array words, where words[i] consists of lowercase English letters.

In one operation, select any index i such that 0 < i < words.length and words[i - 1] and words[i] are anagrams, and delete words[i] from words. Keep performing this operation as long as you can select an index that satisfies the conditions.

Return words after performing all operations. It can be shown that selecting the indices for each operation in any arbitrary order will lead to the same result.

An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase using all the original letters exactly once. For example, "dacb" is an anagram of "abdc".

Example 1:

Input: words = ["abba","baba","bbaa","cd","cd"]
Output: ["abba","cd"]
Explanation:
One of the ways we can obtain the resultant array is by using the following operations:
- Since words[2] = "bbaa" and words[1] = "baba" are anagrams, we choose index 2 and delete words[2].
  Now words = ["abba","baba","cd","cd"].
- Since words[1] = "baba" and words[0] = "abba" are anagrams, we choose index 1 and delete words[1].
  Now words = ["abba","cd","cd"].
- Since words[2] = "cd" and words[1] = "cd" are anagrams, we choose index 2 and delete words[2].
  Now words = ["abba","cd"].
We can no longer perform any operations, so ["abba","cd"] is the final answer.

Example 2:

Input: words = ["a","b","c","d","e"]
Output: ["a","b","c","d","e"]
Explanation:
No two adjacent strings in words are anagrams of each other, so no operations are performed.

Constraints:

1 <= words.length <= 100
1 <= words[i].length <= 10
words[i] consists of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed string array words, where words[i] consists of lowercase English letters. In one operation, select any index i such that 0 < i < words.length and words[i - 1] and words[i] are anagrams, and delete words[i] from words. Keep performing this operation as long as you can select an index that satisfies the conditions. Return words after performing all operations. It can be shown that selecting the indices for each operation in any arbitrary order will lead to the same result. An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase using all the original letters exactly once. For example, "dacb" is an anagram of "abdc".

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

["abba","baba","bbaa","cd","cd"]

Example 2

["a","b","c","d","e"]

Core Insight

What unlocks the optimal approach

Instead of removing each repeating anagram, try to find all the strings in words which will not be present in the final answer.
For every index i, find the largest index j < i such that words[j] will be present in the final answer.
Check if words[i] and words[j] are anagrams. If they are, then it can be confirmed that words[i] will not be present in 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

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
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

Upper-end input sizes

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 #2273: Find Resultant Array After Removing Anagrams
class Solution {
    public List<String> removeAnagrams(String[] words) {
        List<String> ans = new ArrayList<>();
        ans.add(words[0]);
        for (int i = 1; i < words.length; ++i) {
            if (check(words[i - 1], words[i])) {
                ans.add(words[i]);
            }
        }
        return ans;
    }

    private boolean check(String s, String t) {
        if (s.length() != t.length()) {
            return true;
        }
        int[] cnt = new int[26];
        for (int i = 0; i < s.length(); ++i) {
            ++cnt[s.charAt(i) - 'a'];
        }
        for (int i = 0; i < t.length(); ++i) {
            if (--cnt[t.charAt(i) - 'a'] < 0) {
                return true;
            }
        }
        return false;
    }
}

// Accepted solution for LeetCode #2273: Find Resultant Array After Removing Anagrams
func removeAnagrams(words []string) []string {
	ans := []string{words[0]}
	check := func(s, t string) bool {
		if len(s) != len(t) {
			return true
		}
		cnt := [26]int{}
		for _, c := range s {
			cnt[c-'a']++
		}
		for _, c := range t {
			cnt[c-'a']--
			if cnt[c-'a'] < 0 {
				return true
			}
		}
		return false
	}
	for i, t := range words[1:] {
		if check(words[i], t) {
			ans = append(ans, t)
		}
	}
	return ans
}

# Accepted solution for LeetCode #2273: Find Resultant Array After Removing Anagrams
class Solution:
    def removeAnagrams(self, words: List[str]) -> List[str]:
        def check(s: str, t: str) -> bool:
            if len(s) != len(t):
                return True
            cnt = Counter(s)
            for c in t:
                cnt[c] -= 1
                if cnt[c] < 0:
                    return True
            return False

        return [words[0]] + [t for s, t in pairwise(words) if check(s, t)]

// Accepted solution for LeetCode #2273: Find Resultant Array After Removing Anagrams
impl Solution {
    pub fn remove_anagrams(words: Vec<String>) -> Vec<String> {
        fn check(s: &str, t: &str) -> bool {
            if s.len() != t.len() {
                return true;
            }
            let mut cnt = [0; 26];
            for c in s.bytes() {
                cnt[(c - b'a') as usize] += 1;
            }
            for c in t.bytes() {
                let idx = (c - b'a') as usize;
                cnt[idx] -= 1;
                if cnt[idx] < 0 {
                    return true;
                }
            }
            false
        }

        let mut ans = vec![words[0].clone()];
        for i in 1..words.len() {
            if check(&words[i - 1], &words[i]) {
                ans.push(words[i].clone());
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #2273: Find Resultant Array After Removing Anagrams
function removeAnagrams(words: string[]): string[] {
    const ans: string[] = [words[0]];
    const check = (s: string, t: string): boolean => {
        if (s.length !== t.length) {
            return true;
        }
        const cnt: number[] = Array(26).fill(0);
        for (const c of s) {
            ++cnt[c.charCodeAt(0) - 97];
        }
        for (const c of t) {
            if (--cnt[c.charCodeAt(0) - 97] < 0) {
                return true;
            }
        }
        return false;
    };
    for (let i = 1; i < words.length; ++i) {
        if (check(words[i - 1], words[i])) {
            ans.push(words[i]);
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

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

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(L)

Space

O(|\Sigma|)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

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.