10252 - Common Permutation

題目敘述：here

解題想法：

• 統計兩個字串 a 至 z 字元個數。


• 按照 a 至 z 的順序，印出較少個數的字元。

程式碼：

          
//  鄒雨笙 UVa Online Judge 2023/11/12
//
//  Problem 10252 - Common Permutation


#include <iostream>
#include <string>
#include <algorithm>
using namespace std;


string common_permutation(const string& a, const string& b) {
    int count_a[26] = {0}, count_b[26] = {0};

    // Count occurrences of each character in both strings
    for (char ch : a) {
        count_a[ch - 'a']++;
    }
    for (char ch : b) {
        count_b[ch - 'a']++;
    }

    // Find the common characters and build the result
    string result;
    for (int i = 0; i < 26; ++i) {
        int common_count = min(count_a[i], count_b[i]);
        result.append(common_count, 'a' + i);
    }

    return result;
}

int main() {
    string line1, line2;

    while (getline(cin, line1) && getline(cin, line2)) {
        cout << common_permutation(line1, line2) << endl;
    }

    return 0;
}