用于计算指定子字符串中出现次数最多和出现次数最少的字符的频率差异的查询

给定一个由N个小写字符组成的字符串str和一个数组Q[][] ，其中每行{l, r}形式代表一个查询。对于每个查询，任务是找出子串{str[l], … 中字符的最大频率和最小频率之差。 str[r]} 。

注意：考虑基于1的索引。

例子：

Input: N = 7, S = “abaabac”, Q[][] = {{ 2, 6 }, { 1, 7 }}
Output: 1 3
Explanation:
Query 1: ‘a’ occurs maximum number of times in the given range i.e., 3 and ‘b’ occurs minimum number of times in the given range i.e. 2. Therefore, output = 3 – 2 = 1.
Query 2: ‘a’ occurs maximum number of times in the given range i.e. 4 and ‘c’ occurs minimum number of times in the given range i.e. 1. Therefore, output = 4 – 1 = 3.

Input: N = 6, S = “aabbcc”, Q[][] = {{1, 4}, {1, 6}}
Output: 0 0
Explanation:
Query 1: ‘a’ and ‘b’ both occurs same number of times in the given range. Therefore, the output is 0.
Query 2: All ‘a’, ‘b’ and ‘c’ occurs same number of times in the given range. Therefore, the output is 0.

朴素方法：对于每个查询，找到给定范围内所有字符的频率，并取最大和最小频率之差。

时间复杂度： O((N + 26)* |Q|)
辅助空间： O(26)

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Funciton to find difference between maximum and
// minimum frequency of a character in given range
void maxDiffFreq(vector > queries,
                 string S)
{
 
    // Stores length of string
    int N = S.size();
 
    // Stores count of queries
    int Q = queries.size();
 
    // Iterate over the characters
    // of the string
    for (int i = 0; i < Q; ++i) {
 
        // Stores l-value of a query
        int l = queries[i].first - 1;
 
        // Stores r-value of a query
        int r = queries[i].second - 1;
        int freq[26] = { 0 };
 
        // Store count of every character
        // laying in range [l, r]
        for (int j = l; j <= r; j++) {
 
            // Update frequency of
            // current character
            freq[S[j] - 'a']++;
        }
 
        // Stores maximum frequency
        // of characters in given range
        int mx = 0;
 
        // Stores minimum frequency
        // of characters in given range
        int mn = 99999999;
 
        // Iterate over all possible characters
        // of the given string
        for (int j = 0; j < 26; j++) {
 
            // Update mx
            mx = max(mx, freq[j]);
 
            // If (j + 'a') is present
            if (freq[j])
                mn = min(mn, freq[j]);
        }
 
        // difference between max and min
        cout << mx - mn << endl;
    }
}
 
// Driver Code
int main()
{
 
    // Given string
    string S = "abaabac";
 
    // Given queries
    vector > queries{ { 2, 6 }, { 1, 7 } };
 
    // Function Call
    maxDiffFreq(queries, S);
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
 
  // Funciton to find difference between maximum and
  // minimum frequency of a character in given range
  static void maxDiffFreq(int [][]queries,
                          String S)
  {
 
    // Stores length of String
    int N = S.length();
 
    // Stores count of queries
    int Q = queries.length;
 
    // Iterate over the characters
    // of the String
    for (int i = 0; i < Q; ++i)    
    {
 
      // Stores l-value of a query
      int l = queries[i][0] - 1;
 
      // Stores r-value of a query
      int r = queries[i][1] - 1;
      int freq[] = new int[26];
 
      // Store count of every character
      // laying in range [l, r]
      for (int j = l; j <= r; j++) {
 
        // Update frequency of
        // current character
        freq[S.charAt(j) - 'a']++;
      }
 
      // Stores maximum frequency
      // of characters in given range
      int mx = 0;
 
      // Stores minimum frequency
      // of characters in given range
      int mn = 99999999;
 
      // Iterate over all possible characters
      // of the given String
      for (int j = 0; j < 26; j++) {
 
        // Update mx
        mx = Math.max(mx, freq[j]);
 
        // If (j + 'a') is present
        if (freq[j]>0)
          mn = Math.min(mn, freq[j]);
      }
 
      // difference between max and min
      System.out.print(mx - mn +"\n");
    }
  }
 
  // Driver Code
  public static void main(String[] args)
  {
 
    // Given String
    String S = "abaabac";
 
    // Given queries
    int [][]queries = { { 2, 6 }, { 1, 7 } };
 
    // Function Call
    maxDiffFreq(queries, S);
  }
}
// This code is contributed by 29AjayKumar

Python3

# Python 3 program for the above approach
 
# Funciton to find difference between maximum
# and minimum frequency of a character in
# given range
def maxDiffFreq(queries, S):
 
    # Stores length of string
    N = len(S)
 
    # Stores count of queries
    Q = len(queries)
 
    # Iterate over the characters
    # of the string
    for i in range(Q):
 
        # Stores l-value of a query
        l = queries[i][0] - 1
 
        # Stores r-value of a query
        r = queries[i][1] - 1
        freq = [0] * 26
 
        # Store count of every character
        # laying in range [l, r]
        for j in range(l, r + 1):
 
            # Update frequency of
            # current character
            freq[ord(S[j]) - ord('a')] += 1
 
        # Stores maximum frequency
        # of characters in given range
        mx = 0
 
        # Stores minimum frequency
        # of characters in given range
        mn = 99999999
 
        # Iterate over all possible characters
        # of the given string
        for j in range(26):
 
            # Update mx
            mx = max(mx, freq[j])
 
            # If (j + 'a') is present
            if (freq[j]):
                mn = min(mn, freq[j])
 
        # Difference between max and min
        print(mx - mn)
 
# Driver Code
if __name__ == "__main__":
 
    # Given string
    S = "abaabac"
 
    # Given queries
    queries = [ [ 2, 6 ], [ 1, 7 ] ]
 
    # Function Call
    maxDiffFreq(queries, S)
 
# This code is contributed by chitranayal

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
     
    // Funciton to find difference between maximum and
    // minimum frequency of a character in given range
    static void maxDiffFreq(List> queries, string S)
    {
         
        // Stores length of string
        int N = S.Length;
       
        // Stores count of queries
        int Q = queries.Count;
       
        // Iterate over the characters
        // of the string
        for (int i = 0; i < Q; ++i)
        {
       
            // Stores l-value of a query
            int l = queries[i].Item1 - 1;
       
            // Stores r-value of a query
            int r = queries[i].Item2 - 1;
            int[] freq = new int[26];
       
            // Store count of every character
            // laying in range [l, r]
            for (int j = l; j <= r; j++)
            {
       
                // Update frequency of
                // current character
                freq[S[j] - 'a']++;
            }
       
            // Stores maximum frequency
            // of characters in given range
            int mx = 0;
       
            // Stores minimum frequency
            // of characters in given range
            int mn = 99999999;
       
            // Iterate over all possible characters
            // of the given string
            for (int j = 0; j < 26; j++)
            {
       
                // Update mx
                mx = Math.Max(mx, freq[j]);
       
                // If (j + 'a') is present
                if (freq[j] != 0)
                    mn = Math.Min(mn, freq[j]);
            }
       
            // difference between max and min
            Console.WriteLine(mx - mn);
        }
    }   
 
  // Driver code
  static void Main()
  {
       
    // Given string
    string S = "abaabac";
   
    // Given queries
    List> queries = new List>();
    queries.Add(new Tuple(2, 6));
    queries.Add(new Tuple(1, 7));
   
    // Function Call
    maxDiffFreq(queries, S);
  }
}
 
// This code is contributed by divyeshrabadiya07

输出：

1
3

高效的方法：想法是使用2D-Fenwick树来存储每个字符的频率。请按照以下步骤解决问题：

1. 创建一个二维 Fenwick 树，用于存储从‘a’到‘z’ 的每个字符的信息。
2. 然后对于每个查询，使用 Fenwick 树计算给定范围内每个字符的频率。
3. 从上面找到的频率中，得到最大和最小频率。
4. 打印最大和最小频率之间的差异作为答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to update frequency of
// a character in Fenwick tree
void update(int BIT[26][10005], int idx,
            int i, int val)
{
    while (i < 10005) {
 
        // Update frequency of (idx + 'a')
        BIT[idx][i] += val;
 
        // Update i
        i = i + (i & (-i));
    }
}
 
// Function to find the frequency of
// a character (idx + 'a') in range [1, i]
int query(int BIT[26][10005], int idx, int i)
{
 
    // Stores frequency of character, (idx + 'a')
    // in range [1, i]
    int ans = 0;
 
    while (i > 0) {
 
        // Update ans
        ans += BIT[idx][i];
 
        // Update i
        i = i - (i & (-i));
    }
    return ans;
}
 
// Function to find difference between maximum and
// minimum frequency of a character in given range
void maxDiffFreq(string s, vector > queries)
{
 
    // BIT[i][j]: Stores frequency of (i + 'a')
    // If j is a power of 2, then it stores
    // the frequency (i + 'a') of  from [1, j]
    int BIT[26][10005];
 
    // Stores length of string
    int n = s.size();
 
    // Iterate over the characters
    // of the string
    for (int i = 0; i < n; i++) {
 
        // Update the frequency of
        // s[i] in fenwick tree
        update(BIT, s[i] - 'a', i + 1, 1);
    }
 
    // Stores count of queries
    int Q = queries.size();
 
    // Iterate over all the queries
    for (int i = 0; i < Q; ++i) {
 
        // Stores maximum frequency of
        // a character in range [l, r]
        int mx = 0;
 
        // Stores minimum frequency of
        // a character in range [l, r]
 
        int mn = INT_MAX;
        int l = queries[i].first;
        int r = queries[i].second;
 
        // Iterate over all possible characters
        for (int j = 0; j < 26; j++) {
 
            // Stores frequency of (j + 'a')
            // in range [1, r]
            int p = query(BIT, j, r);
 
            // Stores frequency of (j + 'a')
            // in range [1, l - 1]
            int q = query(BIT, j, l - 1);
 
            // Update mx
            mx = max(mx, p - q);
 
            // If a character (i + 'a') present
            // in range [l, r]
            if (p > 0) {
 
                // Update mn
                mn = min(mn, p - q);
            }
        }
 
        // Print the difference between
        // max and min freq
        cout << mx - mn << endl;
    }
}
 
// Driver Code
int main()
{
 
    // Given string
    string S = "abaabac";
 
    // Given queries
    vector > queries
        = { { 2, 6 }, { 1, 7 } };
 
    // Function Call
    maxDiffFreq(S, queries);
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG
{
 
// Function to update frequency of
// a character in Fenwick tree
static void update(int BIT[][], int idx,
            int i, int val)
{
    while (i < 10005)
    {
 
        // Update frequency of (idx + 'a')
        BIT[idx][i] += val;
 
        // Update i
        i = i + (i & (-i));
    }
}
 
// Function to find the frequency of
// a character (idx + 'a') in range [1, i]
static int query(int BIT[][], int idx, int i)
{
 
    // Stores frequency of character, (idx + 'a')
    // in range [1, i]
    int ans = 0;
 
    while (i > 0) {
 
        // Update ans
        ans += BIT[idx][i];
 
        // Update i
        i = i - (i & (-i));
    }
    return ans;
}
 
// Function to find difference between maximum and
// minimum frequency of a character in given range
static void maxDiffFreq(String s, int [][]queries)
{
 
    // BIT[i][j]: Stores frequency of (i + 'a')
    // If j is a power of 2, then it stores
    // the frequency (i + 'a') of  from [1, j]
    int[][] BIT = new int[26][10005];
 
    // Stores length of String
    int n = s.length();
 
    // Iterate over the characters
    // of the String
    for (int i = 0; i < n; i++) {
 
        // Update the frequency of
        // s[i] in fenwick tree
        update(BIT, s.charAt(i) - 'a', i + 1, 1);
    }
 
    // Stores count of queries
    int Q = queries.length;
 
    // Iterate over all the queries
    for (int i = 0; i < Q; ++i) {
 
        // Stores maximum frequency of
        // a character in range [l, r]
        int mx = 0;
 
        // Stores minimum frequency of
        // a character in range [l, r]
 
        int mn = Integer.MAX_VALUE;
        int l = queries[i][0];
        int r = queries[i][1];
 
        // Iterate over all possible characters
        for (int j = 0; j < 26; j++) {
 
            // Stores frequency of (j + 'a')
            // in range [1, r]
            int p = query(BIT, j, r);
 
            // Stores frequency of (j + 'a')
            // in range [1, l - 1]
            int q = query(BIT, j, l - 1);
 
            // Update mx
            mx = Math.max(mx, p - q);
 
            // If a character (i + 'a') present
            // in range [l, r]
            if (p > 0) {
 
                // Update mn
                mn = Math.min(mn, p - q);
            }
        }
 
        // Print the difference between
        // max and min freq
        System.out.print(mx - mn +"\n");
    }
}
 
// Driver Code
public static void main(String[] args)
{
 
    // Given String
    String S = "abaabac";
 
    // Given queries
    int [][]queries
        = { { 2, 6 }, { 1, 7 } };
 
    // Function Call
    maxDiffFreq(S, queries);
}
}
 
// This code is contributed by shikhasingrajput

蟒蛇3

# Python3 program for the above approach
import sys
 
# Function to update frequency of
# a character in Fenwick tree
def update(BIT, idx, i, val) :
    while (i < 10005) :
     
      # Update frequency of (idx + 'a')
      BIT[idx][i] += val
     
      # Update i
      i = i + (i & (-i))
       
# Function to find the frequency of
# a character (idx + 'a') in range [1, i]
def query(BIT, idx, i) :
 
    # Stores frequency of character, (idx + 'a')
    # in range [1, i]
    ans = 0
    while (i > 0) :
     
      # Update ans
      ans += BIT[idx][i]
     
      # Update i
      i = i - (i & (-i))
 
    return ans
     
# Function to find difference between maximum and
# minimum frequency of a character in given range
def maxDiffFreq(s, queries) :
 
    # BIT[i][j]: Stores frequency of (i + 'a')
    # If j is a power of 2, then it stores
    # the frequency (i + 'a') of  from [1][j]
    BIT = [[0 for i in range(10005)] for j in range(26)]
     
    # Stores length of String
    n = len(s)
     
    # Iterate over the characters
    # of the String
    for i in range(n) :
     
      # Update the frequency of
      # s[i] in fenwick tree
      update(BIT, ord(s[i]) - ord('a'), i + 1, 1)
     
    # Stores count of queries
    Q = len(queries)
     
    # Iterate over all the queries
    for i in range(Q) :
     
      # Stores maximum frequency of
      # a character in range [l, r]
      mx = 0
     
      # Stores minimum frequency of
      # a character in range [l, r]
      mn = sys.maxsize
      l = queries[i][0]
      r = queries[i][1]
     
      # Iterate over all possible characters
      for j in range(26) :
     
        # Stores frequency of (j + 'a')
        # in range [1, r]
        p = query(BIT, j, r)
     
        # Stores frequency of (j + 'a')
        # in range [1, l - 1]
        q = query(BIT, j, l - 1)
     
        # Update mx
        mx = max(mx, p - q)
     
        # If a character (i + 'a') present
        # in range [l, r]
        if (p > 0) :
     
          # Update mn
          mn = min(mn, p - q)
     
      # Print the difference between
      # max and min freq
      print(mx - mn)
       
# Given String
S = "abaabac"
 
# Given queries
queries = [ [ 2, 6 ], [ 1, 7 ] ]
 
# Function Call
maxDiffFreq(S, queries)
 
# This code is contributed by divyesh072019.

C＃

// C# program for the above approach
using System;
public class GFG
{
 
  // Function to update frequency of
  // a character in Fenwick tree
  static void update(int [,]BIT, int idx,
                     int i, int val)
  {
    while (i < 10005)
    {
 
      // Update frequency of (idx + 'a')
      BIT[idx,i] += val;
 
      // Update i
      i = i + (i & (-i));
    }
  }
 
  // Function to find the frequency of
  // a character (idx + 'a') in range [1, i]
  static int query(int [,]BIT, int idx, int i)
  {
 
    // Stores frequency of character, (idx + 'a')
    // in range [1, i]
    int ans = 0;
    while (i > 0)
    {
 
      // Update ans
      ans += BIT[idx,i];
 
      // Update i
      i = i - (i & (-i));
    }
    return ans;
  }
 
  // Function to find difference between maximum and
  // minimum frequency of a character in given range
  static void maxDiffFreq(String s, int [,]queries)
  {
 
    // BIT[i,j]: Stores frequency of (i + 'a')
    // If j is a power of 2, then it stores
    // the frequency (i + 'a') of  from [1, j]
    int[,] BIT = new int[26, 10005];
 
    // Stores length of String
    int n = s.Length;
 
    // Iterate over the characters
    // of the String
    for (int i = 0; i < n; i++)
    {
 
      // Update the frequency of
      // s[i] in fenwick tree
      update(BIT, s[i] - 'a', i + 1, 1);
    }
 
    // Stores count of queries
    int Q = queries.GetLength(0);
 
    // Iterate over all the queries
    for (int i = 0; i < Q; ++i)
    {
 
      // Stores maximum frequency of
      // a character in range [l, r]
      int mx = 0;
 
      // Stores minimum frequency of
      // a character in range [l, r]
      int mn = int.MaxValue;
      int l = queries[i, 0];
      int r = queries[i, 1];
 
      // Iterate over all possible characters
      for (int j = 0; j < 26; j++)
      {
 
        // Stores frequency of (j + 'a')
        // in range [1, r]
        int p = query(BIT, j, r);
 
        // Stores frequency of (j + 'a')
        // in range [1, l - 1]
        int q = query(BIT, j, l - 1);
 
        // Update mx
        mx = Math.Max(mx, p - q);
 
        // If a character (i + 'a') present
        // in range [l, r]
        if (p > 0)
        {
 
          // Update mn
          mn = Math.Min(mn, p - q);
        }
      }
 
      // Print the difference between
      // max and min freq
      Console.Write(mx - mn +"\n");
    }
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
 
    // Given String
    String S = "abaabac";
 
    // Given queries
    int [,]queries
      = { { 2, 6 }, { 1, 7 } };
 
    // Function Call
    maxDiffFreq(S, queries);
  }
}
 
// This code is contributed by shikhasingrajput

输出：

1
3

时间复杂度： O(|Q| * log(N) * 26)
辅助空间： O(N * 26)

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