LeetCode #3838 — EASY

Weighted Word Mapping

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

The Problem

Problem Statement

You are given an array of strings words, where each string represents a word containing lowercase English letters.

You are also given an integer array weights of length 26, where weights[i] represents the weight of the i^th lowercase English letter.

The weight of a word is defined as the sum of the weights of its characters.

For each word, take its weight modulo 26 and map the result to a lowercase English letter using reverse alphabetical order (0 -> 'z', 1 -> 'y', ..., 25 -> 'a').

Return a string formed by concatenating the mapped characters for all words in order.

Example 1:

Input: words = ["abcd","def","xyz"], weights = [5,3,12,14,1,2,3,2,10,6,6,9,7,8,7,10,8,9,6,9,9,8,3,7,7,2]

Output: "rij"

Explanation:

The weight of "abcd" is 5 + 3 + 12 + 14 = 34. The result modulo 26 is 34 % 26 = 8, which maps to 'r'.
The weight of "def" is 14 + 1 + 2 = 17. The result modulo 26 is 17 % 26 = 17, which maps to 'i'.
The weight of "xyz" is 7 + 7 + 2 = 16. The result modulo 26 is 16 % 26 = 16, which maps to 'j'.

Thus, the string formed by concatenating the mapped characters is "rij".

Example 2:

Input: words = ["a","b","c"], weights = [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]

Output: "yyy"

Explanation:

Each word has weight 1. The result modulo 26 is 1 % 26 = 1, which maps to 'y'.

Thus, the string formed by concatenating the mapped characters is "yyy".

Example 3:

Input: words = ["abcd"], weights = [7,5,3,4,3,5,4,9,4,2,2,7,10,2,5,10,6,1,2,2,4,1,3,4,4,5]

Output: "g"

Explanation:

The weight of "abcd" is 7 + 5 + 3 + 4 = 19. The result modulo 26 is 19 % 26 = 19, which maps to 'g'.

Thus, the string formed by concatenating the mapped characters is "g".

Constraints:

1 <= words.length <= 100
1 <= words[i].length <= 10
weights.length == 26
1 <= weights[i] <= 100
words[i] consists of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given an array of strings words, where each string represents a word containing lowercase English letters. You are also given an integer array weights of length 26, where weights[i] represents the weight of the ith lowercase English letter. The weight of a word is defined as the sum of the weights of its characters. For each word, take its weight modulo 26 and map the result to a lowercase English letter using reverse alphabetical order (0 -> 'z', 1 -> 'y', ..., 25 -> 'a'). Return a string formed by concatenating the mapped characters for all words in order.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

["abcd","def","xyz"]
[5,3,12,14,1,2,3,2,10,6,6,9,7,8,7,10,8,9,6,9,9,8,3,7,7,2]

Example 2

["a","b","c"]
[1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]

Example 3

["abcd"]
[7,5,3,4,3,5,4,9,4,2,2,7,10,2,5,10,6,1,2,2,4,1,3,4,4,5]

Step 02

Core Insight

What unlocks the optimal approach

For each word, sum character weights using <code>weights[c - 'a']</code>
Take the sum modulo <code>26</code>
Map the value to a character using reverse order: <code>char = 'z' - value</code>
Append all mapped characters in order to form the result string

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 #3838: Weighted Word Mapping
class Solution {
    public String mapWordWeights(String[] words, int[] weights) {
        var ans = new StringBuilder();
        for (var w : words) {
            int s = 0;
            for (char c : w.toCharArray()) {
                s = (s + weights[c - 'a']) % 26;
            }
            ans.append((char) ('a' + (25 - s)));
        }
        return ans.toString();
    }
}

// Accepted solution for LeetCode #3838: Weighted Word Mapping
func mapWordWeights(words []string, weights []int) string {
	ans := make([]byte, 0, len(words))
	for _, w := range words {
		s := 0
		for i := 0; i < len(w); i++ {
			s = (s + weights[int(w[i]-'a')]) % 26
		}
		ans = append(ans, byte('a'+(25-s)))
	}
	return string(ans)
}

# Accepted solution for LeetCode #3838: Weighted Word Mapping
class Solution:
    def mapWordWeights(self, words: List[str], weights: List[int]) -> str:
        ans = []
        for w in words:
            s = sum(weights[ord(c) - ord('a')] for c in w)
            ans.append(ascii_lowercase[25 - s % 26])
        return ''.join(ans)

// Accepted solution for LeetCode #3838: Weighted Word Mapping
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3838: Weighted Word Mapping
// class Solution {
//     public String mapWordWeights(String[] words, int[] weights) {
//         var ans = new StringBuilder();
//         for (var w : words) {
//             int s = 0;
//             for (char c : w.toCharArray()) {
//                 s = (s + weights[c - 'a']) % 26;
//             }
//             ans.append((char) ('a' + (25 - s)));
//         }
//         return ans.toString();
//     }
// }

// Accepted solution for LeetCode #3838: Weighted Word Mapping
function mapWordWeights(words: string[], weights: number[]): string {
    const ans: string[] = [];
    for (const w of words) {
        let s = 0;
        for (const c of w) {
            s = (s + weights[c.charCodeAt(0) - 97]) % 26;
        }
        ans.push(String.fromCharCode(97 + (25 - s)));
    }
    return ans.join('');
}

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 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.