LeetCode //C - 383. Ransom Note

383. Ransom Note

Given two strings ransomNote and magazine, return true if ransomNote can be constructed by using the letters from magazine and false otherwise.

Each letter in magazine can only be used once in ransomNote.

Example 1:

Input: ransomNote = "a", magazine = "b"
Output: false

Example 2:

Input: ransomNote = "aa", magazine = "ab"
Output: false

Example 3:

Input: ransomNote = "aa", magazine = "aab"
Output: true

Constraints:
  • 1 <= ransomNote.length, magazine.length <= 1 0 5 10^5 105
  • ransomNote and magazine consist of lowercase English letters.

From: LeetCode

Link: 383. Ransom Note


Solution:

Ideas:

In this function, magazineLetters is an array of 26 integers (for each letter in the English alphabet), initialized to 0. We first count the frequency of each letter in magazine. Then, for each letter in ransomNote, we check if it exists in magazine (by checking magazineLetters). If it exists, we decrement its count; otherwise, we return false. If all letters in ransomNote are successfully found in magazine, we return true.

Code:
c 复制代码
bool canConstruct(char* ransomNote, char* magazine) {
    int magazineLetters[26] = {0}; // Array to store the frequency of each letter in magazine

    // Count the frequency of each letter in magazine
    for (int i = 0; magazine[i] != '\0'; i++) {
        magazineLetters[magazine[i] - 'a']++;
    }

    // Check if each letter in ransomNote can be constructed from magazine
    for (int i = 0; ransomNote[i] != '\0'; i++) {
        if (magazineLetters[ransomNote[i] - 'a'] > 0) {
            magazineLetters[ransomNote[i] - 'a']--; // Use the letter and decrement its count
        } else {
            return false; // If a letter in ransomNote is not found in magazine, return false
        }
    }

    return true; // All letters in ransomNote are found in magazine
}
相关推荐
YuTaoShao6 分钟前
【LeetCode 热题 100】73. 矩阵置零——(解法二)空间复杂度 O(1)
java·算法·leetcode·矩阵
Heartoxx8 分钟前
c语言-指针(数组)练习2
c语言·数据结构·算法
大熊背22 分钟前
图像处理专业书籍以及网络资源总结
人工智能·算法·microsoft
满分观察网友z26 分钟前
别怕树!一层一层剥开它的心:用BFS/DFS优雅计算层平均值(637. 二叉树的层平均值)
算法
杰克尼1 小时前
1. 两数之和 (leetcode)
数据结构·算法·leetcode
YuTaoShao2 小时前
【LeetCode 热题 100】56. 合并区间——排序+遍历
java·算法·leetcode·职场和发展
二进制person6 小时前
Java SE--方法的使用
java·开发语言·算法
OneQ6667 小时前
C++讲解---创建日期类
开发语言·c++·算法
JoJo_Way7 小时前
LeetCode三数之和-js题解
javascript·算法·leetcode
码农不惑7 小时前
2025.06.27-14.44 C语言开发:Onvif(二)
c语言·开发语言