09 Sept 2026•Easy•Go / Python / TypeScript

Most Common Word

Find the most frequent case-insensitive word that is not banned.

Scan letters into lowercase words, count non-banned tokens, and update the answer whenever a word reaches a new highest frequency.

Complexity

O(n + b) time and O(n + b) space.

Solution files

Go•most-common-word/solution.go

package main

import (
	"strings"
	"unicode"
)

func mostCommonWord(paragraph string, banned []string) string {
	blocked := map[string]bool{}
	for _, word := range banned {
		blocked[word] = true
	}
	counts := map[string]int{}
	answer, best := "", 0
	words := strings.FieldsFunc(strings.ToLower(paragraph), func(character rune) bool { return !unicode.IsLetter(character) })
	for _, word := range words {
		if blocked[word] {
			continue
		}
		counts[word]++
		if counts[word] > best {
			answer, best = word, counts[word]
		}
	}
	return answer
}

Python•most-common-word/solution.py

import re
from collections import Counter


class Solution:
    def mostCommonWord(self, paragraph: str, banned: list[str]) -> str:
        blocked = set(banned)
        counts = Counter(word for word in re.findall(r"[a-z]+", paragraph.lower()) if word not in blocked)
        return counts.most_common(1)[0][0]

TypeScript•most-common-word/solution.ts

function mostCommonWord(paragraph: string, banned: string[]): string {
  const blocked = new Set(banned.map((word) => word.toLowerCase()));
  const counts = new Map<string, number>();
  let answer = "";
  let best = 0;
  for (const word of paragraph.toLowerCase().match(/[a-z]+/g) ?? []) { if (blocked.has(word)) continue; const count = (counts.get(word) ?? 0) + 1; counts.set(word, count); if (count > best) { best = count; answer = word; } }
  return answer;
}