LeetCode #1160 — EASY

Find Words That Can Be Formed by Characters

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

The Problem

Problem Statement

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

A string is good if it can be formed by characters from chars (each character can only be used once for each word in words).

Return the sum of lengths of all good strings in words.

Example 1:

Input: words = ["cat","bt","hat","tree"], chars = "atach"
Output: 6
Explanation: The strings that can be formed are "cat" and "hat" so the answer is 3 + 3 = 6.

Example 2:

Input: words = ["hello","world","leetcode"], chars = "welldonehoneyr"
Output: 10
Explanation: The strings that can be formed are "hello" and "world" so the answer is 5 + 5 = 10.

Constraints:

1 <= words.length <= 1000
1 <= words[i].length, chars.length <= 100
words[i] and chars consist of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given an array of strings words and a string chars. A string is good if it can be formed by characters from chars (each character can only be used once for each word in words). Return the sum of lengths of all good strings in words.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

["cat","bt","hat","tree"]
"atach"

Example 2

["hello","world","leetcode"]
"welldonehoneyr"

Core Insight

What unlocks the optimal approach

Solve the problem for each string in <code>words</code> independently.
Now try to think in frequency of letters.
Count how many times each character occurs in string <code>chars</code>.
To form a string using characters from <code>chars</code>, the frequency of each character in <code>chars</code> must be greater than or equal the frequency of that character in the string to be formed.

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 #1160: Find Words That Can Be Formed by Characters
class Solution {
    public int countCharacters(String[] words, String chars) {
        int[] cnt = new int[26];
        for (int i = 0; i < chars.length(); ++i) {
            ++cnt[chars.charAt(i) - 'a'];
        }
        int ans = 0;
        for (String w : words) {
            int[] wc = new int[26];
            boolean ok = true;
            for (int i = 0; i < w.length(); ++i) {
                int j = w.charAt(i) - 'a';
                if (++wc[j] > cnt[j]) {
                    ok = false;
                    break;
                }
            }
            if (ok) {
                ans += w.length();
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1160: Find Words That Can Be Formed by Characters
func countCharacters(words []string, chars string) (ans int) {
	cnt := [26]int{}
	for _, c := range chars {
		cnt[c-'a']++
	}
	for _, w := range words {
		wc := [26]int{}
		ok := true
		for _, c := range w {
			c -= 'a'
			wc[c]++
			if wc[c] > cnt[c] {
				ok = false
				break
			}
		}
		if ok {
			ans += len(w)
		}
	}
	return
}

# Accepted solution for LeetCode #1160: Find Words That Can Be Formed by Characters
class Solution:
    def countCharacters(self, words: List[str], chars: str) -> int:
        cnt = Counter(chars)
        ans = 0
        for w in words:
            wc = Counter(w)
            if all(cnt[c] >= v for c, v in wc.items()):
                ans += len(w)
        return ans

// Accepted solution for LeetCode #1160: Find Words That Can Be Formed by Characters
struct Solution;

use std::collections::HashMap;

impl Solution {
    fn str_2_hs(s: &str) -> HashMap<char, i32> {
        let mut hs: HashMap<char, i32> = HashMap::new();
        for c in s.chars() {
            *hs.entry(c).or_default() += 1;
        }
        hs
    }

    fn count_characters(words: Vec<String>, chars: String) -> i32 {
        let chars = Self::str_2_hs(&chars);
        let mut sum = 0;
        for w in words {
            let mut chars = chars.clone();
            let mut valid = true;
            for c in w.chars() {
                let count = chars.entry(c).or_default();
                *count -= 1;
                if *count < 0 {
                    valid = false;
                    break;
                }
            }
            if valid {
                sum += w.len();
            }
        }
        sum as i32
    }
}

#[test]
fn test() {
    let words = vec_string!["cat", "bt", "hat", "tree"];
    let chars = "atach".to_string();
    assert_eq!(Solution::count_characters(words, chars), 6);
    let words = vec_string!["hello", "world", "leetcode"];
    let chars = "welldonehoneyr".to_string();
    assert_eq!(Solution::count_characters(words, chars), 10);
}

// Accepted solution for LeetCode #1160: Find Words That Can Be Formed by Characters
function countCharacters(words: string[], chars: string): number {
    const idx = (c: string) => c.charCodeAt(0) - 'a'.charCodeAt(0);
    const cnt = new Array(26).fill(0);
    for (const c of chars) {
        cnt[idx(c)]++;
    }
    let ans = 0;
    for (const w of words) {
        const wc = new Array(26).fill(0);
        let ok = true;
        for (const c of w) {
            if (++wc[idx(c)] > cnt[idx(c)]) {
                ok = false;
                break;
            }
        }
        if (ok) {
            ans += w.length;
        }
    }
    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(C)

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.