从具有更新的字符串查找范围[L，R]中的Kth个最大字符的查询

给定长度为N的字符串str和以下两种类型的Q查询：

1. (1 LRK)：从索引[L，R] (基于1的索引)的范围中找到第K^个最大字符(不明显)
2. (2 JC)：替换歼从由字符C的字符串^个字符

例子：

Input: str = “abcddef”, Q = 3, queries[][] = {{1, 2, 5, 3}, {2, 4, g}, {1, 1, 4, 3}}
Output:
c
b 
Explanation : 
Query 1: String between indices (2, 5) is “bcdd”. The third largest character is ‘c’. Therefore, c is the required output. 
Query 2: Replace S[4] by ‘g’. Therefore, S modifies to “abcgdef”. 
Query 3: String between indices (1, 4) is “abcg”. The third largest character is ‘b’. Therefore, b is the required output.

Input: str=” afcdehgk”, Q = 4, queries[][] = {{1, 2, 5, 4}, {2, 5, m}, {1, 3, 7, 2}, {1, 1, 6, 4}}
Output:
c
h
d

天真的方法：解决问题的最简单方法如下：

• 对于类型( 1 LRK )的每个查询，从索引[L，R]的范围中找到S的子字符串，然后以非递增顺序对该子字符串进行排序。在子字符串的^第K^个索引处打印字符。
• 对于每个类型(2 JC)的查询，将S中的^第J^个字符替换为C。

时间复杂度： O(Q *(N log(N)))，其中N logN是对每个子字符串进行排序的计算复杂度。
辅助空间： O(N)

下面的代码是上述方法的实现：

C++

// C++ Program to implement
// the above approach
 
#include "bits/stdc++.h"
using namespace std;
 
// Function to find the kth greatest
// character from the strijng
char find_kth_largest(string str, int k)
{
 
    // Sorting the string in
    // non-increasing Order
    sort(str.begin(), str.end(),
         greater());
    return str[k - 1];
}
 
// Function to print the K-th character
// from the substring S[l] .. S[r]
char printCharacter(string str, int l,
                    int r, int k)
{
    // 0-based indexing
    l = l - 1;
    r = r - 1;
 
    // Substring of str from the
    // indices l to r.
    string temp
        = str.substr(l, r - l + 1);
 
    // Extract kth Largest character
    char ans
        = find_kth_largest(temp, k);
 
    return ans;
}
 
// Function to replace character at
// pos of str by the character s
void updateString(string str, int pos,
                  char s)
{
    // Index of S to be updated.
    int index = pos - 1;
    char c = s;
 
    // Character to be replaced
    // at index in S
    str[index] = c;
}
 
// Driver Code
int main()
{
    // Given string
    string str = "abcddef";
 
    // Count of queries
    int Q = 3;
 
    // Queries
    cout << printCharacter(str, 1, 2, 2)
         << endl;
    updateString(str, 4, 'g');
    cout << printCharacter(str, 1, 5, 4)
         << endl;
 
    return 0;
}

Java

// Java Program to implement
// the above approach
//include "bits/stdJava.h"
import java.util.*;
class GFG{
 
// Function to find the kth greatest
// character from the strijng
static char find_kth_largest(char []str,
                             int k)
{
  // Sorting the String in
  // non-increasing Order
  Arrays.sort(str);
  reverse(str);
  return str[k - 1];
}
  static char[] reverse(char a[])
  {
    int i, n = a.length;
    char t;
    for (i = 0; i < n / 2; i++)
    {
      t = a[i];
      a[i] = a[n - i - 1];
      a[n - i - 1] = t;
    }
    return a;
}
   
// Function to print the K-th character
// from the subString S[l] .. S[r]
static char printCharacter(String str, int l,
                           int r, int k)
{
  // 0-based indexing
  l = l - 1;
  r = r - 1;
 
  // SubString of str from the
  // indices l to r.
  String temp = str.substring(l, r - l + 1);
 
  // Extract kth Largest character
  char ans =
    find_kth_largest(temp.toCharArray(), k);
 
  return ans;
}
 
// Function to replace character at
// pos of str by the character s
static void updateString(char []str,
                         int pos, char s)
{
  // Index of S to be updated.
  int index = pos - 1;
  char c = s;
 
  // Character to be replaced
  // at index in S
  str[index] = c;
}
 
// Driver Code
public static void main(String[] args)
{
  // Given String
  String str = "abcddef";
 
  // Count of queries
  int Q = 3;
 
  // Queries
  System.out.print(printCharacter(str, 1,
                                  2, 2) + "\n");
  updateString(str.toCharArray(), 4, 'g');
  System.out.print(printCharacter(str, 1,
                                  5, 4) + "\n");
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python3 Program to implement
# the above approach
# Function to find the kth greatest
# character from the strrijng
def find_kth_largest(strr, k):
 
    # Sorting the in
    # non-increasing Order
    strr = sorted(strr)
    strr = strr[:: -1]
    return strr[k - 1]
 
# Function to prthe K-th character
# from the subS[l] .. S[r]
def printCharacter(strr, l, r, k):
   
    #0-based indexing
    l = l - 1
    r = r - 1
 
    # Subof strr from the
    # indices l to r.
    temp= strr[l: r - l + 1]
 
    #Extract kth Largest character
    ans = find_kth_largest(temp, k)
 
    return ans
 
# Function to replace character at
# pos of strr by the character s
def updateString(strr, pos, s):
    # Index of S to be updated.
    index = pos - 1
    c = s
 
    # Character to be replaced
    # at index in S
    strr[index] = c
 
# Driver Code
if __name__ == '__main__':
   
    # Given strring
    strr = "abcddef"
    strr=[i for i in strr]
 
    # Count of queries
    Q = 3
 
    # Queries
    print(printCharacter(strr, 1, 2, 2))
    updateString(strr, 4, 'g')
    print(printCharacter(strr, 1, 5, 4))
 
# This code is contributed by Mohit Kumar 29

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to find the kth greatest
// character from the strijng
static char find_kth_largest(char []str,
                             int k)
{
     
    // Sorting the String in
    // non-increasing Order
    Array.Sort(str);
    reverse(str);
     
    return str[k - 1];
}
 
static char[] reverse(char []a)
{
    int i, n = a.Length;
    char t;
     
    for(i = 0; i < n / 2; i++)
    {
        t = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = t;
    }
    return a;
}
 
// Function to print the K-th character
// from the subString S[l] .. S[r]
static char printchar(String str, int l,
                           int r, int k)
{
     
    // 0-based indexing
    l = l - 1;
    r = r - 1;
 
    // SubString of str from the
    // indices l to r.
    String temp = str.Substring(l, r - l + 1);
 
    // Extract kth Largest character
    char ans = find_kth_largest(
               temp.ToCharArray(), k);
 
    return ans;
}
 
// Function to replace character at
// pos of str by the character s
static void updateString(char []str,
                         int pos, char s)
{
     
    // Index of S to be updated.
    int index = pos - 1;
    char c = s;
 
    // char to be replaced
    // at index in S
    str[index] = c;
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given String
    String str = "abcddef";
 
    // Count of queries
    //int Q = 3;
 
    // Queries
    Console.Write(printchar(str, 1, 2, 2) + "\n");
    updateString(str.ToCharArray(), 4, 'g');
     
    Console.Write(printchar(str, 1, 5, 4) + "\n");
}
}
 
// This code is contributed by Amit Katiyar

输出：

a
b

高效方法：可以通过使用Fenwick树有效地预先计算所有大于或等于字符C (’a’≤C≤’z’)的字符数来优化上述方法。
请按照以下步骤解决问题：

• 创建一个Fenwick树来存储从’ a’到’z ‘的所有字符的频率
• 对于类型1的每个查询，请检查从‘z’到’a’的每个字符，是否它是第K^个最大字符。
• 为了从“Z”到“A”执行此，横动并且对于每个字符，检查是否遍历变得≥K或不是所有的字符的计数。打印为其计数变为≥k处的字符。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
 
#include "bits/stdc++.h"
using namespace std;
 
// Maximum Size of a String
const int maxn = 100005;
 
// Fenwick Tree to store the
// frequencies of 26 alphabets
int BITree[26][maxn];
 
// Size of the String.
int N;
 
// Function to update Fenwick Tree for
// Character c at index val
void update_BITree(int index, char C,
                   int val)
{
    while (index <= N) {
 
        // Add val to current node
        // Fenwick Tree
        BITree[C - 'a'][index]
            += val;
 
        // Move index to parent node
        // in update View
        index += (index & -index);
    }
}
 
// Function to get sum of frequencies
// of character c till index
int sum_BITree(int index, char C)
{
 
    // Stores the sum
    int s = 0;
    while (index) {
 
        // Add current element of
        // Fenwick tree to sum
        s += BITree[C - 'a'][index];
 
        // Move index to parent node
        // in getSum View
        index -= (index & -index);
    }
    return s;
}
 
// Function to create the Fenwick tree
void buildTree(string str)
{
    for (int i = 1; i <= N; i++) {
 
        update_BITree(i, str[i], 1);
    }
    cout << endl;
}
 
// Function to print the kth largest
// character in the range of l to r
char printCharacter(string str, int l,
                    int r, int k)
{
 
    // Stores the count of
    // charcters
    int count = 0;
 
    // Stores the required
    // character
    char ans;
 
    for (char C = 'z'; C >= 'a'; C--) {
 
        // Calculate frequency of
        // C in the given range
        int times = sum_BITree(r, C)
                    - sum_BITree(l - 1, C);
 
        // Increase count
        count += times;
 
        // If count exceeds K
        if (count >= k) {
 
            // Required character
            // found
            ans = C;
            break;
        }
    }
 
    return ans;
}
 
// Function to update character
// at pos by character s
void updateTree(string str, int pos,
                char s)
{
 
    // 0 based index system
    int index = pos;
    update_BITree(index, str[index], -1);
 
    str[index] = s;
    update_BITree(index, s, 1);
}
 
// Driver Code
int main()
{
    string str = "abcddef";
    N = str.size();
 
    // Makes the string 1-based indexed
    str = '#' + str;
 
    // Number of queries
    int Q = 3;
 
    // Construct the Fenwick Tree
    buildTree(str);
 
    cout << printCharacter(str, 1, 2, 2)
         << endl;
    updateTree(str, 4, 'g');
    cout << printCharacter(str, 1, 5, 4)
         << endl;
 
    return 0;
}

Java

// Java Program to implement
// the above approach
 
//include "bits/stdJava.h"
import java.util.*;
class GFG{
 
// Maximum Size of a String
static int maxn = 100005;
 
// Fenwick Tree to store the
// frequencies of 26 alphabets
static int [][]BITree = new int[26][maxn];
 
// Size of the String.
static int N;
 
// Function to update Fenwick Tree for
// Character c at index val
static void update_BITree(int index,
                          char C, int val)
{
  while (index <= N)
  {
    // Add val to current node
    // Fenwick Tree
    BITree[C - 'a'][index] += val;
 
    // Move index to parent node
    // in update View
    index += (index & -index);
  }
}
 
// Function to get sum of frequencies
// of character c till index
static int sum_BITree(int index, char C)
{
 
  // Stores the sum
  int s = 0;
  while (index > 0)
  {
    // Add current element of
    // Fenwick tree to sum
    s += BITree[C - 'a'][index];
 
    // Move index to parent node
    // in getSum View
    index -= (index & -index);
  }
  return s;
}
 
// Function to create the Fenwick tree
static void buildTree(String str)
{
  for (int i = 1; i <= N; i++)
  {
    update_BITree(i, str.charAt(i), 1);
  }
  System.out.println();
}
 
// Function to print the kth largest
// character in the range of l to r
static char printCharacter(String str, int l,
                           int r, int k)
{
  // Stores the count of
  // charcters
  int count = 0;
 
  // Stores the required
  // character
  char ans = 0;
 
  for (char C = 'z'; C >= 'a'; C--)
  {
    // Calculate frequency of
    // C in the given range
    int times = sum_BITree(r, C) -
      sum_BITree(l - 1, C);
 
    // Increase count
    count += times;
 
    // If count exceeds K
    if (count >= k)
    {
      // Required character
      // found
      ans = C;
      break;
    }
  }
  return ans;
}
 
// Function to update character
// at pos by character s
static void updateTree(String str,
                       int pos, char s)
{
    // 0 based index system
    int index = pos;
    update_BITree(index,
                  str.charAt(index), -1);
    str = str.substring(0, index) + s +
          str.substring(index + 1);
    update_BITree(index, s, 1);
}
 
// Driver Code
public static void main(String[] args)
{
  String str = "abcddef";
  N = str.length();
 
  // Makes the String 1-based indexed
  str = '/' + str;
 
  // Number of queries
  int Q = 3;
 
  // Conthe Fenwick Tree
  buildTree(str);
 
  System.out.print(printCharacter(str, 1,
                                  2, 2) + "\n");
  updateTree(str, 4, 'g');
  System.out.print(printCharacter(str, 1,
                                  5, 4) + "\n");
 
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python3 Program to implement
# the above approach
 
# Maximum Size of a String
maxn = 100005
 
# Fenwick Tree to store the
# frequencies of 26 alphabets
BITree = [[0 for x in range(maxn)]
             for y in range(26)]
 
# Size of the String.
N = 0
 
# Function to update Fenwick Tree for
# Character c at index val
def update_BITree(index, C, val):
 
    while (index <= N):
 
        # Add val to current node
        # Fenwick Tree
        BITree[ord(C) -
               ord('a')][index]+= val
 
        # Move index to parent node
        # in update View
        index += (index & -index)
 
# Function to get sum of
# frequencies of character
# c till index
def sum_BITree(index, C):
 
    # Stores the sum
    s = 0
    while (index):
 
        # Add current element of
        # Fenwick tree to sum
        s += BITree[ord(C) -
                    ord('a')][index]
 
        # Move index to parent node
        # in getSum View
        index -= (index & -index)
    return s
 
# Function to create
# the Fenwick tree
def buildTree(st):
   
    for i in range(1,
                   N + 1):
        update_BITree(i,
                      st[i], 1)
     
    print()
 
# Function to print the
# kth largest character
# in the range of l to r
def printCharacter(st, l,
                   r, k):
   
    # Stores the count of
    # charcters
    count = 0
 
    for C in range(ord('z'),
                   ord('a') -
                   1, -1):
 
        # Calculate frequency of
        # C in the given range
        times = (sum_BITree(r, chr(C)) -
                 sum_BITree(l - 1, chr(C)))
 
        # Increase count
        count += times
 
        # If count exceeds K
        if (count >= k):
 
            # Required character
            # found
            ans = chr( C)
            break
    
    return ans
 
# Function to update character
# at pos by character s
def updateTree(st, pos, s):
   
    # 0 based index system
    index = pos;
    update_BITree(index,
                  st[index], -1)
 
    st.replace(st[index], s, 1)
    update_BITree(index, s, 1)
 
# Driver Code
if __name__ == "__main__":
   
    st = "abcddef"
    N = len(st)
 
    # Makes the string
    # 1-based indexed
    st = '#' + st
 
    # Number of queries
    Q = 3
 
    # Construct the Fenwick Tree
    buildTree(st)
 
    print (printCharacter(st, 1,
                          2, 2))
    updateTree(st, 4, 'g')
    print (printCharacter(st, 1,
                          5, 4))
 
# This code is contributed by Chitranayal

C＃

// C# Program to implement
// the above approach
using System;
class GFG{
 
// Maximum Size of a String
static int maxn = 100005;
 
// Fenwick Tree to store the
// frequencies of 26 alphabets
static int [,]BITree = new int[26, maxn];
 
// Size of the String.
static int N;
 
// Function to update Fenwick Tree for
// char c at index val
static void update_BITree(int index,
                          char C, int val)
{
  while (index <= N)
  {
    // Add val to current node
    // Fenwick Tree
    BITree[C - 'a', index] += val;
 
    // Move index to parent node
    // in update View
    index += (index & -index);
  }
}
 
// Function to get sum of frequencies
// of character c till index
static int sum_BITree(int index, char C)
{
  // Stores the sum
  int s = 0;
  while (index > 0)
  {
    // Add current element of
    // Fenwick tree to sum
    s += BITree[C - 'a', index];
 
    // Move index to parent node
    // in getSum View
    index -= (index & -index);
  }
  return s;
}
 
// Function to create the Fenwick tree
static void buildTree(String str)
{
  for (int i = 1; i <= N; i++)
  {
    update_BITree(i, str[i], 1);
  }
  Console.WriteLine();
}
 
// Function to print the kth largest
// character in the range of l to r
static char printchar(String str, int l,
                      int r, int k)
{
  // Stores the count of
  // charcters
  int count = 0;
 
  // Stores the required
  // character
  char ans = (char)0;
 
  for (char C = 'z'; C >= 'a'; C--)
  {
    // Calculate frequency of
    // C in the given range
    int times = sum_BITree(r, C) -
                sum_BITree(l - 1, C);
 
    // Increase count
    count += times;
 
    // If count exceeds K
    if (count >= k)
    {
      // Required character
      // found
      ans = C;
      break;
    }
  }
  return ans;
}
 
// Function to update character
// at pos by character s
static void updateTree(String str,
                       int pos, char s)
{
  // 0 based index system
  int index = pos;
  update_BITree(index,
                str[index], -1);
  str = str.Substring(0, index) + s +
    str.Substring(index + 1);
  update_BITree(index, s, 1);
}
 
// Driver Code
public static void Main(String[] args)
{
  String str = "abcddef";
  N = str.Length;
 
  // Makes the String 1-based indexed
  str = '/' + str;
 
  // Number of queries
  int Q = 3;
 
  // Conthe Fenwick Tree
  buildTree(str);
 
  Console.Write(printchar(str, 1, 2, 2) + "\n");
  updateTree(str, 4, 'g');
  Console.Write(printchar(str, 1, 5, 4) + "\n");
}
}
 
// This code is contributed by Rajput-Ji

输出：

a
b

时间复杂度： O(QlogN + NlogN)
辅助空间： O(26 * maxn)，其中maxn表示字符串的最大可能长度。