LeetCode #1684 — EASY

Count the Number of Consistent Strings

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

The Problem

Problem Statement

You are given a string allowed consisting of distinct characters and an array of strings words. A string is consistent if all characters in the string appear in the string allowed.

Return the number of consistent strings in the array words.

Example 1:

Input: allowed = "ab", words = ["ad","bd","aaab","baa","badab"]
Output: 2
Explanation: Strings "aaab" and "baa" are consistent since they only contain characters 'a' and 'b'.

Example 2:

Input: allowed = "abc", words = ["a","b","c","ab","ac","bc","abc"]
Output: 7
Explanation: All strings are consistent.

Example 3:

Input: allowed = "cad", words = ["cc","acd","b","ba","bac","bad","ac","d"]
Output: 4
Explanation: Strings "cc", "acd", "ac", and "d" are consistent.

Constraints:

1 <= words.length <= 10⁴
1 <= allowed.length <=26
1 <= words[i].length <= 10
The characters in allowed are distinct.
words[i] and allowed contain only lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a string allowed consisting of distinct characters and an array of strings words. A string is consistent if all characters in the string appear in the string allowed. Return the number of consistent strings in the array words.

Baseline thinking

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

Pattern signal: Array · Hash Map · Bit Manipulation

Example 1

"ab"
["ad","bd","aaab","baa","badab"]

Example 2

"abc"
["a","b","c","ab","ac","bc","abc"]

Example 3

"cad"
["cc","acd","b","ba","bac","bad","ac","d"]

Core Insight

What unlocks the optimal approach

A string is incorrect if it contains a character that is not allowed
Constraints are small enough for brute force

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 #1684: Count the Number of Consistent Strings
class Solution {
    public int countConsistentStrings(String allowed, String[] words) {
        boolean[] s = new boolean[26];
        for (char c : allowed.toCharArray()) {
            s[c - 'a'] = true;
        }
        int ans = 0;
        for (String w : words) {
            if (check(w, s)) {
                ++ans;
            }
        }
        return ans;
    }

    private boolean check(String w, boolean[] s) {
        for (int i = 0; i < w.length(); ++i) {
            if (!s[w.charAt(i) - 'a']) {
                return false;
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #1684: Count the Number of Consistent Strings
func countConsistentStrings(allowed string, words []string) (ans int) {
	s := [26]bool{}
	for _, c := range allowed {
		s[c-'a'] = true
	}
	check := func(w string) bool {
		for _, c := range w {
			if !s[c-'a'] {
				return false
			}
		}
		return true
	}
	for _, w := range words {
		if check(w) {
			ans++
		}
	}
	return ans
}

# Accepted solution for LeetCode #1684: Count the Number of Consistent Strings
class Solution:
    def countConsistentStrings(self, allowed: str, words: List[str]) -> int:
        s = set(allowed)
        return sum(all(c in s for c in w) for w in words)

// Accepted solution for LeetCode #1684: Count the Number of Consistent Strings
impl Solution {
    pub fn count_consistent_strings(allowed: String, words: Vec<String>) -> i32 {
        let n = words.len();
        let mut make = [false; 26];
        for c in allowed.as_bytes() {
            make[(c - b'a') as usize] = true;
        }
        let mut ans = n as i32;
        for word in words.iter() {
            for c in word.as_bytes().iter() {
                if !make[(c - b'a') as usize] {
                    ans -= 1;
                    break;
                }
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #1684: Count the Number of Consistent Strings
function countConsistentStrings(allowed: string, words: string[]): number {
    const set = new Set([...allowed]);
    const n = words.length;
    let ans = n;
    for (const word of words) {
        for (const c of word) {
            if (!set.has(c)) {
                ans--;
                break;
            }
        }
    }
    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(m)

Space

O(C)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → 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.

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.