How do you solve Weighted Word Mapping on LeetCode?

Weighted Word Mapping (LeetCode #3838) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Mapping weights to characters efficiently reduces complexity.

What is the time complexity of Weighted Word Mapping?

The optimal solution for Weighted Word Mapping runs in O(n) time and uses O(n) space. Each word is processed in linear time, leading to O(n) overall complexity.

What is the brute force solution for Weighted Word Mapping?

The brute force approach for Weighted Word Mapping is Brute Force, with O(n²) time complexity and O(1) space complexity. Calculate the weight of each word by summing the weights of its characters. Then, map the result modulo 26 to a character in reverse alphabetical order. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Weighted Word Mapping?

Weighted Word Mapping uses the following concepts: Array, String, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Weighted Word Mapping?

Clarify the problem requirements before coding. Think about edge cases like empty strings or single character words. Explain your thought process while coding.

What are common mistakes when solving Weighted Word Mapping?

Forgetting to apply modulo 26. Incorrectly mapping the result to characters.

#3838

Weighted Word Mapping

Easy

ArrayStringSimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Sum character weights directly using the weights array and compute the result in a single pass for each word, achieving linear time complexity.

⚙️

Algorithm

3 steps

1Step 1: Initialize an empty result string.
2Step 2: For each word, compute the total weight using the weights array.
3Step 3: Map the total weight modulo 26 to the corresponding character and append to the result.

solution.py6 lines

1def weighted_word_mapping(words, weights):
2    result = ''
3    for word in words:
4        weight_sum = sum(weights[ord(c) - ord('a')] for c in word)
5        result += chr(122 - (weight_sum % 26))
6    return result

ℹ

Complexity note: Each word is processed in linear time, leading to O(n) overall complexity.

1Mapping weights to characters efficiently reduces complexity.
2Understanding character encoding helps in direct calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.