LeetCode #1023 — MEDIUM

Camelcase Matching

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

Given an array of strings queries and a string pattern, return a boolean array answer where answer[i] is true if queries[i] matches pattern, and false otherwise.

A query word queries[i] matches pattern if you can insert lowercase English letters into the pattern so that it equals the query. You may insert a character at any position in pattern or you may choose not to insert any characters at all.

Example 1:

Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FB"
Output: [true,false,true,true,false]
Explanation: "FooBar" can be generated like this "F" + "oo" + "B" + "ar".
"FootBall" can be generated like this "F" + "oot" + "B" + "all".
"FrameBuffer" can be generated like this "F" + "rame" + "B" + "uffer".

Example 2:

Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBa"
Output: [true,false,true,false,false]
Explanation: "FooBar" can be generated like this "Fo" + "o" + "Ba" + "r".
"FootBall" can be generated like this "Fo" + "ot" + "Ba" + "ll".

Example 3:

Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBaT"
Output: [false,true,false,false,false]
Explanation: "FooBarTest" can be generated like this "Fo" + "o" + "Ba" + "r" + "T" + "est".

Constraints:

1 <= pattern.length, queries.length <= 100
1 <= queries[i].length <= 100
queries[i] and pattern consist of English letters.

Step 01

Brute Force Baseline

Problem summary: Given an array of strings queries and a string pattern, return a boolean array answer where answer[i] is true if queries[i] matches pattern, and false otherwise. A query word queries[i] matches pattern if you can insert lowercase English letters into the pattern so that it equals the query. You may insert a character at any position in pattern or you may choose not to insert any characters at all.

Baseline thinking

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

Pattern signal: Array · Two Pointers · Trie · String Matching

Example 1

["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"]
"FB"

Example 2

["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"]
"FoBa"

Example 3

["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"]
"FoBaT"

Step 02

Core Insight

What unlocks the optimal approach

Given a single pattern and word, how can we solve it?
One way to do it is using a DP (pos1, pos2) where pos1 is a pointer to the word and pos2 to the pattern and returns true if we can match the pattern with the given word.
We have two scenarios: The first one is when `word[pos1] == pattern[pos2]`, then the transition will be just DP(pos1 + 1, pos2 + 1). The second scenario is when `word[pos1]` is lowercase then we can add this character to the pattern so that the transition is just DP(pos1 + 1, pos2) The case base is `if (pos1 == n && pos2 == m) return true;` Where n and m are the sizes of the strings word and pattern respectively.

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 #1023: Camelcase Matching
class Solution {
    public List<Boolean> camelMatch(String[] queries, String pattern) {
        List<Boolean> ans = new ArrayList<>();
        for (var q : queries) {
            ans.add(check(q, pattern));
        }
        return ans;
    }

    private boolean check(String s, String t) {
        int m = s.length(), n = t.length();
        int i = 0, j = 0;
        for (; j < n; ++i, ++j) {
            while (i < m && s.charAt(i) != t.charAt(j) && Character.isLowerCase(s.charAt(i))) {
                ++i;
            }
            if (i == m || s.charAt(i) != t.charAt(j)) {
                return false;
            }
        }
        while (i < m && Character.isLowerCase(s.charAt(i))) {
            ++i;
        }
        return i == m;
    }
}

// Accepted solution for LeetCode #1023: Camelcase Matching
func camelMatch(queries []string, pattern string) (ans []bool) {
	check := func(s, t string) bool {
		m, n := len(s), len(t)
		i, j := 0, 0
		for ; j < n; i, j = i+1, j+1 {
			for i < m && s[i] != t[j] && (s[i] >= 'a' && s[i] <= 'z') {
				i++
			}
			if i == m || s[i] != t[j] {
				return false
			}
		}
		for i < m && s[i] >= 'a' && s[i] <= 'z' {
			i++
		}
		return i == m
	}
	for _, q := range queries {
		ans = append(ans, check(q, pattern))
	}
	return
}

# Accepted solution for LeetCode #1023: Camelcase Matching
class Solution:
    def camelMatch(self, queries: List[str], pattern: str) -> List[bool]:
        def check(s, t):
            m, n = len(s), len(t)
            i = j = 0
            while j < n:
                while i < m and s[i] != t[j] and s[i].islower():
                    i += 1
                if i == m or s[i] != t[j]:
                    return False
                i, j = i + 1, j + 1
            while i < m and s[i].islower():
                i += 1
            return i == m

        return [check(q, pattern) for q in queries]

// Accepted solution for LeetCode #1023: Camelcase Matching
struct Solution;

impl Solution {
    fn camel_match(queries: Vec<String>, pattern: String) -> Vec<bool> {
        queries
            .into_iter()
            .map(|query| Self::query_match(query.chars().collect(), pattern.chars().collect()))
            .collect()
    }

    fn query_match(query: Vec<char>, pattern: Vec<char>) -> bool {
        let mut j = 0;
        for i in 0..query.len() {
            if j < pattern.len() && query[i] == pattern[j] {
                j += 1;
            } else {
                if query[i].is_uppercase() {
                    return false;
                }
            }
        }
        j == pattern.len()
    }
}

#[test]
fn test() {
    let queries = vec_string![
        "FooBar",
        "FooBarTest",
        "FootBall",
        "FrameBuffer",
        "ForceFeedBack"
    ];
    let pattern = "FB".to_string();
    let res = vec![true, false, true, true, false];
    assert_eq!(Solution::camel_match(queries, pattern), res);
    let queries = vec_string![
        "FooBar",
        "FooBarTest",
        "FootBall",
        "FrameBuffer",
        "ForceFeedBack"
    ];
    let pattern = "FoBa".to_string();
    let res = vec![true, false, true, false, false];
    assert_eq!(Solution::camel_match(queries, pattern), res);
    let queries = vec_string![
        "FooBar",
        "FooBarTest",
        "FootBall",
        "FrameBuffer",
        "ForceFeedBack"
    ];
    let pattern = "FoBaT".to_string();
    let res = vec![false, true, false, false, false];
    assert_eq!(Solution::camel_match(queries, pattern), res);
}

// Accepted solution for LeetCode #1023: Camelcase Matching
function camelMatch(queries: string[], pattern: string): boolean[] {
    const check = (s: string, t: string) => {
        const m = s.length;
        const n = t.length;
        let i = 0;
        let j = 0;
        for (; j < n; ++i, ++j) {
            while (i < m && s[i] !== t[j] && s[i].codePointAt(0) >= 97) {
                ++i;
            }
            if (i === m || s[i] !== t[j]) {
                return false;
            }
        }
        while (i < m && s[i].codePointAt(0) >= 97) {
            ++i;
        }
        return i == m;
    };
    const ans: boolean[] = [];
    for (const q of queries) {
        ans.push(check(q, pattern));
    }
    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)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

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.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.