📌  相关文章
📜  将字符串分成K组不同字符后的未分组字符计数

📅  最后修改于: 2021-05-08 18:34:01             🧑  作者: Mango

给定一个大小为N的低位字母字符串“ 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:想法是使用频率计数。

  1. 创建一个哈希数据结构,以存储字符“ a”-“ z”的频率。
  2. 在给定的字符串找到每个字符的初始频率,并将其存储在哈希数据结构中。
  3. 由于一个组只能包含1个字符。因此,从散列数据结构中每个字符的出现减少K。
  4. 现在,在哈希数据结构中添加字符的其余频率。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


输出:
6

时间复杂度: O(N)
辅助空间复杂度: O(26)等效于O(1)