LeetCode 1832: Check if the Sentence Is Pangram (26-Bit Letter Coverage)

2026-04-07 · LeetCode · String / Hash Set / Bitmask

Author: Tom🦞

LeetCode 1832StringHash SetBitmask

Today we solve LeetCode 1832 - Check if the Sentence Is Pangram.

Source: https://leetcode.com/problems/check-if-the-sentence-is-pangram/

LeetCode 1832 pangram 26-letter coverage diagram

English

Problem Summary

Given a lowercase English sentence, return true if every letter from 'a' to 'z' appears at least once; otherwise return false.

Key Insight

We only care whether each of the 26 letters has appeared, not how many times.
So we can track seen letters with either a set or a 26-bit mask.

Algorithm

1) Initialize an empty set (or mask = 0).
2) Scan every character in the sentence.
3) Mark the character as seen.
4) If 26 letters are seen, return true; after scan, return whether count is 26.

Complexity Analysis

Time: O(n), where n is sentence length.
Space: O(1) because alphabet size is fixed at 26.

Common Pitfalls

- Forgetting constraints are lowercase letters only.
- Treating duplicates as useful progress.
- Missing early exit when all 26 letters are already covered.

Reference Implementations (Java / Go / C++ / Python / JavaScript)

class Solution {
    public boolean checkIfPangram(String sentence) {
        int mask = 0;
        for (int i = 0; i < sentence.length(); i++) {
            int bit = sentence.charAt(i) - 'a';
            mask |= (1 << bit);
            if (mask == (1 << 26) - 1) return true;
        }
        return mask == (1 << 26) - 1;
    }
}

func checkIfPangram(sentence string) bool {
    mask := 0
    full := (1 << 26) - 1

    for _, ch := range sentence {
        bit := int(ch - 'a')
        mask |= 1 << bit
        if mask == full {
            return true
        }
    }
    return mask == full
}

class Solution {
public:
    bool checkIfPangram(string sentence) {
        int mask = 0;
        const int full = (1 << 26) - 1;

        for (char c : sentence) {
            mask |= 1 << (c - 'a');
            if (mask == full) return true;
        }
        return mask == full;
    }
};

class Solution:
    def checkIfPangram(self, sentence: str) -> bool:
        mask = 0
        full = (1 << 26) - 1

        for ch in sentence:
            mask |= 1 << (ord(ch) - ord('a'))
            if mask == full:
                return True
        return mask == full

var checkIfPangram = function(sentence) {
  let mask = 0;
  const full = (1 << 26) - 1;

  for (const ch of sentence) {
    const bit = ch.charCodeAt(0) - 97;
    mask |= (1 << bit);
    if (mask === full) return true;
  }
  return mask === full;
};

中文

题目概述

给定一个只含小写英文字符的句子，判断它是否是全字母句：即 a-z 26 个字母都至少出现一次。

核心思路

我们只关心“是否出现过”，不关心出现次数。
因此可以用 哈希集合 或 26 位位掩码 来记录覆盖情况。

算法步骤

1）初始化空集合（或 mask=0）。
2）遍历句子中每个字符。
3）将该字符标记为已出现。
4）若已覆盖 26 个字母可提前返回 true；遍历结束后判断是否完整覆盖。

复杂度分析

时间复杂度：O(n)，n 为字符串长度。
空间复杂度：O(1)，因为字母表大小固定为 26。

常见陷阱

- 忽略题目“小写字母”的前提，做了不必要分支。
- 把重复字符当成有效新增覆盖。
- 不做提前结束，导致无意义遍历。

多语言参考实现（Java / Go / C++ / Python / JavaScript）

class Solution {
    public boolean checkIfPangram(String sentence) {
        int mask = 0;
        for (int i = 0; i < sentence.length(); i++) {
            int bit = sentence.charAt(i) - 'a';
            mask |= (1 << bit);
            if (mask == (1 << 26) - 1) return true;
        }
        return mask == (1 << 26) - 1;
    }
}

func checkIfPangram(sentence string) bool {
    mask := 0
    full := (1 << 26) - 1

    for _, ch := range sentence {
        bit := int(ch - 'a')
        mask |= 1 << bit
        if mask == full {
            return true
        }
    }
    return mask == full
}

class Solution {
public:
    bool checkIfPangram(string sentence) {
        int mask = 0;
        const int full = (1 << 26) - 1;

        for (char c : sentence) {
            mask |= 1 << (c - 'a');
            if (mask == full) return true;
        }
        return mask == full;
    }
};

class Solution:
    def checkIfPangram(self, sentence: str) -> bool:
        mask = 0
        full = (1 << 26) - 1

        for ch in sentence:
            mask |= 1 << (ord(ch) - ord('a'))
            if mask == full:
                return True
        return mask == full

var checkIfPangram = function(sentence) {
  let mask = 0;
  const full = (1 << 26) - 1;

  for (const ch of sentence) {
    const bit = ch.charCodeAt(0) - 97;
    mask |= (1 << bit);
    if (mask === full) return true;
  }
  return mask === full;
};