LeetCode #1255 — HARD

Maximum Score Words Formed by Letters

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

Given a list of words, list of single letters (might be repeating) and score of every character.

Return the maximum score of any valid set of words formed by using the given letters (words[i] cannot be used two or more times).

It is not necessary to use all characters in letters and each letter can only be used once. Score of letters 'a', 'b', 'c', ... ,'z' is given by score[0], score[1], ... , score[25] respectively.

Example 1:

Input: words = ["dog","cat","dad","good"], letters = ["a","a","c","d","d","d","g","o","o"], score = [1,0,9,5,0,0,3,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0]
Output: 23
Explanation:
Score  a=1, c=9, d=5, g=3, o=2
Given letters, we can form the words "dad" (5+1+5) and "good" (3+2+2+5) with a score of 23.
Words "dad" and "dog" only get a score of 21.

Example 2:

Input: words = ["xxxz","ax","bx","cx"], letters = ["z","a","b","c","x","x","x"], score = [4,4,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,5,0,10]
Output: 27
Explanation:
Score  a=4, b=4, c=4, x=5, z=10
Given letters, we can form the words "ax" (4+5), "bx" (4+5) and "cx" (4+5) with a score of 27.
Word "xxxz" only get a score of 25.

Example 3:

Input: words = ["leetcode"], letters = ["l","e","t","c","o","d"], score = [0,0,1,1,1,0,0,0,0,0,0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,0]
Output: 0
Explanation:
Letter "e" can only be used once.

Constraints:

1 <= words.length <= 14
1 <= words[i].length <= 15
1 <= letters.length <= 100
letters[i].length == 1
score.length == 26
0 <= score[i] <= 10
words[i], letters[i] contains only lower case English letters.

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: Given a list of words, list of single letters (might be repeating) and score of every character. Return the maximum score of any valid set of words formed by using the given letters (words[i] cannot be used two or more times). It is not necessary to use all characters in letters and each letter can only be used once. Score of letters 'a', 'b', 'c', ... ,'z' is given by score[0], score[1], ... , score[25] respectively.

Baseline thinking

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

Pattern signal: Array · Hash Map · Dynamic Programming · Backtracking · Bit Manipulation

Example 1

["dog","cat","dad","good"]
["a","a","c","d","d","d","g","o","o"]
[1,0,9,5,0,0,3,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0]

Example 2

["xxxz","ax","bx","cx"]
["z","a","b","c","x","x","x"]
[4,4,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,5,0,10]

Example 3

["leetcode"]
["l","e","t","c","o","d"]
[0,0,1,1,1,0,0,0,0,0,0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,0]

Related Problems

Maximum Good People Based on Statements (maximum-good-people-based-on-statements)

Step 02

Core Insight

What unlocks the optimal approach

Note that words.length is small. This means you can iterate over every subset of words (2^N).

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

Largest constraint values

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 #1255: Maximum Score Words Formed by Letters
class Solution {
    public int maxScoreWords(String[] words, char[] letters, int[] score) {
        int[] cnt = new int[26];
        for (int i = 0; i < letters.length; ++i) {
            cnt[letters[i] - 'a']++;
        }
        int n = words.length;
        int ans = 0;
        for (int i = 0; i < 1 << n; ++i) {
            int[] cur = new int[26];
            for (int j = 0; j < n; ++j) {
                if (((i >> j) & 1) == 1) {
                    for (int k = 0; k < words[j].length(); ++k) {
                        cur[words[j].charAt(k) - 'a']++;
                    }
                }
            }
            boolean ok = true;
            int t = 0;
            for (int j = 0; j < 26; ++j) {
                if (cur[j] > cnt[j]) {
                    ok = false;
                    break;
                }
                t += cur[j] * score[j];
            }
            if (ok && ans < t) {
                ans = t;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1255: Maximum Score Words Formed by Letters
func maxScoreWords(words []string, letters []byte, score []int) (ans int) {
	cnt := [26]int{}
	for _, c := range letters {
		cnt[c-'a']++
	}
	n := len(words)
	for i := 0; i < 1<<n; i++ {
		cur := [26]int{}
		for j := 0; j < n; j++ {
			if i>>j&1 == 1 {
				for _, c := range words[j] {
					cur[c-'a']++
				}
			}
		}
		ok := true
		t := 0
		for i, v := range cur {
			if v > cnt[i] {
				ok = false
				break
			}
			t += v * score[i]
		}
		if ok && ans < t {
			ans = t
		}
	}
	return
}

# Accepted solution for LeetCode #1255: Maximum Score Words Formed by Letters
class Solution:
    def maxScoreWords(
        self, words: List[str], letters: List[str], score: List[int]
    ) -> int:
        cnt = Counter(letters)
        n = len(words)
        ans = 0
        for i in range(1 << n):
            cur = Counter(''.join([words[j] for j in range(n) if i >> j & 1]))
            if all(v <= cnt[c] for c, v in cur.items()):
                t = sum(v * score[ord(c) - ord('a')] for c, v in cur.items())
                ans = max(ans, t)
        return ans

// Accepted solution for LeetCode #1255: Maximum Score Words Formed by Letters
struct Solution;

impl Solution {
    fn max_score_words(words: Vec<String>, letters: Vec<char>, score: Vec<i32>) -> i32 {
        let mut count: Vec<i32> = vec![0; 26];
        for c in letters {
            count[(c as u8 - b'a') as usize] += 1;
        }
        let n = words.len();
        let scores: Vec<i32> = words
            .iter()
            .map(|s| s.bytes().map(|b| score[(b - b'a') as usize]).sum())
            .collect();
        let mut res = 0;
        Self::dfs(0, 0, &mut count, &mut res, &words, &scores, n);
        res
    }

    fn dfs(
        start: usize,
        sum: i32,
        count: &mut Vec<i32>,
        max: &mut i32,
        words: &[String],
        scores: &[i32],
        n: usize,
    ) {
        if start == n {
            *max = (*max).max(sum);
        } else {
            Self::dfs(start + 1, sum, count, max, words, scores, n);
            Self::update(count, &words[start], -1);
            if Self::check(count, &words[start]) {
                Self::dfs(start + 1, sum + scores[start], count, max, words, scores, n);
            }
            Self::update(count, &words[start], 1);
        }
    }

    fn update(count: &mut Vec<i32>, s: &str, val: i32) {
        for c in s.chars() {
            count[(c as u8 - b'a') as usize] += val;
        }
    }

    fn check(count: &mut Vec<i32>, s: &str) -> bool {
        for c in s.chars() {
            if count[(c as u8 - b'a') as usize] < 0 {
                return false;
            }
        }
        true
    }
}

#[test]
fn test() {
    let words = vec_string!["dog", "cat", "dad", "good"];
    let letters = vec!['a', 'a', 'c', 'd', 'd', 'd', 'g', 'o', 'o'];
    let score = vec![
        1, 0, 9, 5, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
    ];
    let res = 23;
    assert_eq!(Solution::max_score_words(words, letters, score), res);
    let words = vec_string!["xxxz", "ax", "bx", "cx"];
    let letters = vec!['z', 'a', 'b', 'c', 'x', 'x', 'x'];
    let score = vec![
        4, 4, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 10,
    ];
    let res = 27;
    assert_eq!(Solution::max_score_words(words, letters, score), res);
    let words = vec_string!["leetcode"];
    let letters = vec!['l', 'e', 't', 'c', 'o', 'd'];
    let score = vec![
        0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0,
    ];
    let res = 0;
    assert_eq!(Solution::max_score_words(words, letters, score), res);
}

// Accepted solution for LeetCode #1255: Maximum Score Words Formed by Letters
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1255: Maximum Score Words Formed by Letters
// class Solution {
//     public int maxScoreWords(String[] words, char[] letters, int[] score) {
//         int[] cnt = new int[26];
//         for (int i = 0; i < letters.length; ++i) {
//             cnt[letters[i] - 'a']++;
//         }
//         int n = words.length;
//         int ans = 0;
//         for (int i = 0; i < 1 << n; ++i) {
//             int[] cur = new int[26];
//             for (int j = 0; j < n; ++j) {
//                 if (((i >> j) & 1) == 1) {
//                     for (int k = 0; k < words[j].length(); ++k) {
//                         cur[words[j].charAt(k) - 'a']++;
//                     }
//                 }
//             }
//             boolean ok = true;
//             int t = 0;
//             for (int j = 0; j < 26; ++j) {
//                 if (cur[j] > cnt[j]) {
//                     ok = false;
//                     break;
//                 }
//                 t += cur[j] * score[j];
//             }
//             if (ok && ans < t) {
//                 ans = t;
//             }
//         }
//         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

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.