LeetCode #2185 — EASY

Counting Words With a Given Prefix

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

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The Problem

Problem Statement

You are given an array of strings words and a string pref.

Return the number of strings in words that contain pref as a prefix.

A prefix of a string s is any leading contiguous substring of s.

Example 1:

Input: words = ["pay","attention","practice","attend"], pref = "at"
Output: 2
Explanation: The 2 strings that contain "at" as a prefix are: "attention" and "attend".

Example 2:

Input: words = ["leetcode","win","loops","success"], pref = "code"
Output: 0
Explanation: There are no strings that contain "code" as a prefix.

Constraints:

1 <= words.length <= 100
1 <= words[i].length, pref.length <= 100
words[i] and pref consist of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given an array of strings words and a string pref. Return the number of strings in words that contain pref as a prefix. A prefix of a string s is any leading contiguous substring of s.

Baseline thinking

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

Pattern signal: Array · String Matching

Example 1

["pay","attention","practice","attend"]
"at"

Example 2

["leetcode","win","loops","success"]
"code"

Core Insight

What unlocks the optimal approach

Go through each word in words and increment the answer if pref is a prefix of the word.

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 #2185: Counting Words With a Given Prefix
class Solution {
    public int prefixCount(String[] words, String pref) {
        int ans = 0;
        for (String w : words) {
            if (w.startsWith(pref)) {
                ++ans;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2185: Counting Words With a Given Prefix
func prefixCount(words []string, pref string) (ans int) {
	for _, w := range words {
		if strings.HasPrefix(w, pref) {
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #2185: Counting Words With a Given Prefix
class Solution:
    def prefixCount(self, words: List[str], pref: str) -> int:
        return sum(w.startswith(pref) for w in words)

// Accepted solution for LeetCode #2185: Counting Words With a Given Prefix
impl Solution {
    pub fn prefix_count(words: Vec<String>, pref: String) -> i32 {
        words.iter().filter(|s| s.starts_with(&pref)).count() as i32
    }
}

// Accepted solution for LeetCode #2185: Counting Words With a Given Prefix
function prefixCount(words: string[], pref: string): number {
    return words.reduce((r, s) => (r += s.startsWith(pref) ? 1 : 0), 0);
}

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(m)

Approach Breakdown

BRUTE FORCE

O(n × m) time

O(1) space

At each of the n starting positions in the text, compare up to m characters with the pattern. If a mismatch occurs, shift by one and restart. Worst case (e.g., searching "aab" in "aaaa...a") checks m characters at nearly every position: O(n × m).

KMP / Z-ALGO

O(n + m) time

O(m) space

KMP and Z-algorithm preprocess the pattern in O(m) to build a failure/Z-array, then scan the text in O(n) — never backtracking. Total: O(n + m). Rabin-Karp uses rolling hashes for O(n + m) expected time. All beat the O(n × m) brute force of checking every position.

Shortcut: Preprocessing avoids backtracking → O(n + m). The failure function is the key insight.

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.