以给定字符串的字符频率降序生成数字

给定长度为N的字符串Str (由小写字母组成)，任务是按给定字符串中字符频率的降序生成数字。如果两个字符具有相同的频率，则首先显示具有较小ASCII值的字符。分配给字符{a，b，…。，y，z}的数字分别为{1,2，…。，25，26}。
注意：对于分配了大于9的值的字符，将其取模10。
例子：

Input: N = 6, Str = “aaabbd” 
Output: 124 
Explanation: 
Given characters and their respective frequencies are: 

• a = 3
• b = 2
• d = 1

Since the number needs to be generated in increasing order of their frequencies, the final generated number is 124.
Input: N = 6, Str = “akkzzz” 
Output: 611 
Explanation: 
Given characters and their respective frequencies are: 

• a = 1
• k = 2
• z = 3

For z, value to assigned = 26 
Hence, the corresponding digit assigned = 26 % 10 = 6 
For k, value to assigned = 11 
Hence, the corresponding digit assigned = 11 % 10 = 1 
Since the number needs to be generated in increasing order of their frequencies, the final generated number is 611. 

方法：
请按照以下步骤解决问题：

• 初始化Map并存储每个字符的频率。
• 遍历地图并将所有{字符，频率}对插入向量对中。
• 对向量进行排序，以使频率较高的对首先出现，而频率相同的对中，ASCII值较小的对首先出现。
• 遍历此向量并找到与每个字符相对应的数字。
• 打印生成的最终号码。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Custom comparator for sorting
bool comp(pair& p1,
          pair& p2)
{
 
    // If frequency is same
    if (p1.second == p2.second)
 
        // Character with lower ASCII
        // value appears first
        return p1.first < p2.first;
 
    // Otherwise character with higher
    // frequency appears first
    return p1.second > p2.second;
}
 
// Function to sort map accordingly
string sort(map& m)
{
 
    // Declaring vector of pairs
    vector > a;
 
    // Output string to store the result
    string out;
 
    // Traversing map and pushing
    // pairs to vector
    for (auto x : m) {
 
        a.push_back({ x.first, x.second });
    }
 
    // Using custom comparator
    sort(a.begin(), a.end(), comp);
 
    // Traversing the Vector
    for (auto x : a) {
 
        // Get the possible digit
        // from assigned value
        int k = x.first - 'a' + 1;
 
        // Ensures k does not exceed 9
        k = k % 10;
 
        // Apppend each digit
        out = out + to_string(k);
    }
 
    // Returning final result
    return out;
}
 
// Function to generate and return
// the required number
string formString(string s)
{
    // Stores the frequencies
    map mp;
 
    for (int i = 0; i < s.length(); i++)
        mp[s[i]]++;
 
    // Sort map in required order
    string res = sort(mp);
 
    // Return the final result
    return res;
}
 
// Driver Code
int main()
{
    int N = 4;
 
    string Str = "akkzzz";
 
    cout << formString(Str);
 
    return 0;
}

Python3

# Python3 Program to implement
# the above approach
 
# Function to sort map
# accordingly
def sort(m):
 
    # Declaring vector
    # of pairs
    a = {}
 
    # Output string to
    # store the result
    out = ""
 
    # Traversing map and
    # pushing pairs to vector
    for x in m:
        a[x] = []
 
        a[x].append(m[x])
 
    # Character with lower ASCII
    # value appears first
    a = dict(sorted(a.items(),
                  key = lambda x : x[0]))
 
    # Character with higher
    # frequency appears first
    a = dict(sorted(a.items(),
                    reverse = True,
                    key = lambda x : x[1]))
 
    # Traversing the Vector
    for x in a:
 
        # Get the possible digit
        # from assigned value
        k = ord(x[0]) - ord('a') + 1
 
        # Ensures k does
        # not exceed 9
        k = k % 10
 
        # Apppend each digit
        out = out + str(k)
 
    # Returning final result
    return out
 
# Function to generate and return
# the required number
def formString(s):
 
    # Stores the frequencies
    mp = {}
    for i in range(len(s)):
        if s[i] in mp:
            mp[s[i]] += 1
        else:
            mp[s[i]] = 1
 
    # Sort map in
    # required order
    res = sort(mp)
 
    # Return the
    # final result
    return res
 
# Driver Code
N = 4
Str = "akkzzz"
print(formString(Str))
 
# This code is contributed by avanitrachhadiya2155

输出：

611

时间复杂度： O(NlogN)
辅助空间： O(N)