LeetCode #2085 — EASY

Count Common Words With One Occurrence

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

The Problem

Problem Statement

Given two string arrays words1 and words2, return the number of strings that appear exactly once in each of the two arrays.

Example 1:

Input: words1 = ["leetcode","is","amazing","as","is"], words2 = ["amazing","leetcode","is"]
Output: 2
Explanation:
- "leetcode" appears exactly once in each of the two arrays. We count this string.
- "amazing" appears exactly once in each of the two arrays. We count this string.
- "is" appears in each of the two arrays, but there are 2 occurrences of it in words1. We do not count this string.
- "as" appears once in words1, but does not appear in words2. We do not count this string.
Thus, there are 2 strings that appear exactly once in each of the two arrays.

Example 2:

Input: words1 = ["b","bb","bbb"], words2 = ["a","aa","aaa"]
Output: 0
Explanation: There are no strings that appear in each of the two arrays.

Example 3:

Input: words1 = ["a","ab"], words2 = ["a","a","a","ab"]
Output: 1
Explanation: The only string that appears exactly once in each of the two arrays is "ab".

Constraints:

1 <= words1.length, words2.length <= 1000
1 <= words1[i].length, words2[j].length <= 30
words1[i] and words2[j] consists only of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: Given two string arrays words1 and words2, return the number of strings that appear exactly once in each of the two arrays.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

["leetcode","is","amazing","as","is"]
["amazing","leetcode","is"]

Example 2

["b","bb","bbb"]
["a","aa","aaa"]

Example 3

["a","ab"]
["a","a","a","ab"]

Core Insight

What unlocks the optimal approach

Could you try every word?
Could you use a hash map to achieve a good complexity?

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 #2085: Count Common Words With One Occurrence
class Solution {
    public int countWords(String[] words1, String[] words2) {
        Map<String, Integer> cnt1 = new HashMap<>();
        Map<String, Integer> cnt2 = new HashMap<>();
        for (var w : words1) {
            cnt1.merge(w, 1, Integer::sum);
        }
        for (var w : words2) {
            cnt2.merge(w, 1, Integer::sum);
        }
        int ans = 0;
        for (var e : cnt1.entrySet()) {
            if (e.getValue() == 1 && cnt2.getOrDefault(e.getKey(), 0) == 1) {
                ++ans;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2085: Count Common Words With One Occurrence
func countWords(words1 []string, words2 []string) (ans int) {
	cnt1 := map[string]int{}
	cnt2 := map[string]int{}
	for _, w := range words1 {
		cnt1[w]++
	}
	for _, w := range words2 {
		cnt2[w]++
	}
	for w, v := range cnt1 {
		if v == 1 && cnt2[w] == 1 {
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #2085: Count Common Words With One Occurrence
class Solution:
    def countWords(self, words1: List[str], words2: List[str]) -> int:
        cnt1 = Counter(words1)
        cnt2 = Counter(words2)
        return sum(v == 1 and cnt2[w] == 1 for w, v in cnt1.items())

// Accepted solution for LeetCode #2085: Count Common Words With One Occurrence
struct Solution;

use std::collections::HashMap;

impl Solution {
    fn count_words(words1: Vec<String>, words2: Vec<String>) -> i32 {
        let mut hm1: HashMap<String, usize> = HashMap::new();
        let mut hm2: HashMap<String, usize> = HashMap::new();
        for w in words1 {
            *hm1.entry(w).or_default() += 1;
        }
        for w in words2 {
            *hm2.entry(w).or_default() += 1;
        }
        let mut res = 0;
        for (k, v) in hm1 {
            if v == 1 && *hm2.get(&k).unwrap_or(&0) == 1 {
                res += 1;
            }
        }
        res
    }
}

#[test]
fn test() {
    let words1 = vec_string!["leetcode", "is", "amazing", "as", "is"];
    let words2 = vec_string!["amazing", "leetcode", "is"];
    let res = 2;
    assert_eq!(Solution::count_words(words1, words2), res);
    let words1 = vec_string!["b", "bb", "bbb"];
    let words2 = vec_string!["a", "aa", "aaa"];
    let res = 0;
    assert_eq!(Solution::count_words(words1, words2), res);
    let words1 = vec_string!["a", "ab"];
    let words2 = vec_string!["a", "a", "a", "ab"];
    let res = 1;
    assert_eq!(Solution::count_words(words1, words2), res);
}

// Accepted solution for LeetCode #2085: Count Common Words With One Occurrence
function countWords(words1: string[], words2: string[]): number {
    const cnt1 = new Map<string, number>();
    const cnt2 = new Map<string, number>();
    for (const w of words1) {
        cnt1.set(w, (cnt1.get(w) ?? 0) + 1);
    }
    for (const w of words2) {
        cnt2.set(w, (cnt2.get(w) ?? 0) + 1);
    }
    let ans = 0;
    for (const [w, v] of cnt1) {
        if (v === 1 && cnt2.get(w) === 1) {
            ++ans;
        }
    }
    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(n + m)

Space

O(n + m)

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.