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📜  将字符串分成 K 组不同字符后未分组字符的计数

📅  最后修改于: 2021-10-27 09:10:20             🧑  作者: Mango

给定一个大小为N的较低字母字符串“S”和一个整数K ;任务是找到仍将取消组合,划分给定的字符串成明显不同的字符K组后的字符个数。
例子:

方法:这个想法是使用频率计数。

  1. 创建一个 Hashing 数据结构,用于存储字符’a’-‘z’ 的频率。
  2. 找到给定字符串中每个字符的初始频率并将其存储在散列数据结构中。
  3. 由于一个组只能包含 1 次出现的字符。因此,从散列数据结构中每个字符的出现处递减 K。
  4. 现在添加散列数据结构中字符的剩余频率。这将是保持未分组的所需字符数。

下面是上述方法的实现:

C++
// C++ code to implement the above approach
 
#include 
using namespace std;
 
void findUngroupedElement(string s,
                          int k)
{
 
    int n = s.length();
 
    // create array where
    // index represents alphabets
    int b[26];
 
    for (int i = 0; i < 26; i++)
        b[i] = 0;
 
    // fill count of every
    // alphabet to corresponding
    // array index
    for (int i = 0; i < n; i++) {
        char p = s.at(i);
        b[p - 'a'] += 1;
    }
 
    int sum = 0;
 
    // count for every element
    // how much is exceeding
    // from no. of groups then
    // sum them
    for (int i = 0; i < 26; i++) {
        if (b[i] > k)
            sum += b[i] - k;
    }
 
    // print answer
    cout << sum << endl;
}
 
// Driver code
int main()
{
    string s = "stayinghomesaveslife";
    int k = 1;
 
    findUngroupedElement(s, k);
 
    return 0;
}


Java
// Java code to implement the above approach
import java.util.*;
 
class GFG{
 
static void findUngroupedElement(String s,
                                 int k)
{
    int n = s.length();
 
    // Create array where
    // index represents alphabets
    int []b = new int[26];
 
    for(int i = 0; i < 26; i++)
        b[i] = 0;
 
    // Fill count of every
    // alphabet to corresponding
    // array index
    for(int i = 0; i < n; i++)
    {
        char p = s.charAt(i);
        b[p - 'a'] += 1;
    }
 
    int sum = 0;
 
    // Count for every element
    // how much is exceeding
    // from no. of groups then
    // sum them
    for(int i = 0; i < 26; i++)
    {
        if (b[i] > k)
            sum += b[i] - k;
    }
 
    // Print answer
    System.out.println(sum);
}
 
// Driver code
public static void main(String srgs[])
{
    String s = "stayinghomesaveslife";
    int k = 1;
 
    findUngroupedElement(s, k);
}
}
 
// This code is contributed by ANKITKUMAR34


Python3
# Python3 code to implement the above approach
def findUngroupedElement(s, k):
 
    n = len(s);
 
    # Create array where
    # index represents alphabets
    b = [0] * 26
 
    # Fill count of every
    # alphabet to corresponding
    # array index
    for i in range(n):
        p = s[i]
        b[ord(p) - ord('a')] += 1
 
    sum = 0;
 
    # Count for every element
    # how much is exceeding
    # from no. of groups then
    # sum them
    for i in range(26):
        if (b[i] > k):
            sum += b[i] - k
 
    # Print answer
    print(sum)
 
# Driver code
s = "stayinghomesaveslife"
k = 1
 
findUngroupedElement(s, k)
 
# This code is contributed by ANKITKUMAR34


C#
// C# code to implement the above approach
using System;
 
class GFG{
 
static void findUngroupedElement(String s,
                                 int k)
{
    int n = s.Length;
 
    // Create array where
    // index represents alphabets
    int []b = new int[26];
 
    for(int i = 0; i < 26; i++)
        b[i] = 0;
 
    // Fill count of every
    // alphabet to corresponding
    // array index
    for(int i = 0; i < n; i++)
    {
        char p = s[i];
        b[p - 'a'] += 1;
    }
 
    int sum = 0;
 
    // Count for every element
    // how much is exceeding
    // from no. of groups then
    // sum them
    for(int i = 0; i < 26; i++)
    {
        if (b[i] > k)
            sum += b[i] - k;
    }
 
    // Print answer
    Console.WriteLine(sum);
}
 
// Driver code
public static void Main(String []srgs)
{
    String s = "stayinghomesaveslife";
    int k = 1;
 
    findUngroupedElement(s, k);
}
}
 
// This code is contributed by Rajput-Ji


Javascript


输出:

6

时间复杂度: O(N)
辅助空间复杂度: O(26) 相当于 O(1)

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