通过将给定字符串的 [L, R] 范围内的每个字符重复其字典值次而形成的字符串长度

给定一个长度为N的字符串S和一个范围[L, R] (1 <= L, R <= N)。任务是找到通过将[L, R]范围内的每个字符重复到其字典值次而形成的字符串的长度。

例子：

Input: s = “cbbde”, l = 2, r = 5
Output: 13
Explanation: Resultant String is formed after repeating each character in range [2, 5] as shown below: 
b repeated 2 times
b repeated 2 times
d repeated 4 times
e repeated 5 times
 

Resultant string: ‘bbbbddddeeeee’
Therefore, length of resultant string so formed is 13

Input: s = “xyyz”, l = 1, r = 2
Output:  49

本机方法：可以通过形成一个临时字符串来解决该任务，在[L, R]范围内附加按字典顺序重复的字符，最后返回结果字符串的长度。
时间复杂度： O((R-L+1) * 26)
辅助空间： O((R-L+1) * 26)

高效方法：可以借助前缀数组来解决该任务，该数组将存储相应字典值的总和。
请按照以下步骤解决问题：

• 创建一个前缀数组' prefix '，用来存储当前字符的字典值的累积和
• 得到给定 [L, R] 的答案：求prefix[R]和prefix[L-1]的差异，如果 L > 0 和prefix[L] ，如果 L = 0，得到窗口的答案( R-L+1 )。
• 注意：这种方法在多个查询的情况下有效。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the length of
// recurring substring in range [l, r]
int recurringSubstring(string s, int l, int r)
{
    // Length of the string
    int N = s.size();
 
    // Variable to store the index of
    // the character in the alphabet
    int a[N];
 
    for (int i = 0; i < N; i++) {
        a[i] = (s[i] - 'a') + 1;
    }
 
    // Prefix array to store the sum
    int prefix[N];
    prefix[0] = a[0];
 
    for (int i = 1; i < N; i++) {
        prefix[i] = prefix[i - 1] + a[i];
    }
 
    l = l - 1;
    r = r - 1;
 
    // If l is greater than 0
    if (l != 0) {
        return prefix[r] - prefix[l - 1];
    }
    // If l is less or equal to 0
    else {
        return prefix[r];
    }
}
 
// Driver Code
int main()
{
    string s = "cbbde";
    int l = 2, r = 5;
 
    cout << recurringSubstring(s, l, r);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class GFG
{
   
// Function to find the length of
// recurring substring in range [l, r]
static int recurringSubstring(String s, int l, int r)
{
   
    // Length of the string
    int N = s.length();
 
    // Variable to store the index of
    // the character in the alphabet
    int []a = new int[N];
 
    for (int i = 0; i < N; i++) {
        a[i] = (s.charAt(i) - 'a') + 1;
    }
 
    // Prefix array to store the sum
    int []prefix = new int[N];
    prefix[0] = a[0];
 
    for (int i = 1; i < N; i++) {
        prefix[i] = prefix[i - 1] + a[i];
    }
 
    l = l - 1;
    r = r - 1;
 
    // If l is greater than 0
    if (l != 0) {
        return prefix[r] - prefix[l - 1];
    }
    // If l is less or equal to 0
    else {
        return prefix[r];
    }
}
 
// Driver Code
public static void main(String args[])
{
    String s = "cbbde";
    int l = 2, r = 5;
 
    System.out.println(recurringSubstring(s, l, r));
     
}
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Python program for the above approach
 
# Function to find the length of
# recurring substring in range [l, r]
def recurringSubstring(s, l, r):
 
    # Length of the string
    N = len(s)
 
    # Variable to store the index of
    # the character in the alphabet
    a = [0] * N
 
    for i in range(N):
        a[i] = (ord(s[i]) - ord('a')) + 1
 
    # Prefix array to store the sum
    prefix = [0] * N
    prefix[0] = a[0]
 
    for i in range(1, N):
        prefix[i] = prefix[i - 1] + a[i]
 
    l = l - 1
    r = r - 1
 
    # If l is greater than 0
    if (l != 0):
        return prefix[r] - prefix[l - 1]
 
    # If l is less or equal to 0
    else:
        return prefix[r]
 
# Driver Code
s = "cbbde"
l = 2
r = 5
 
print(recurringSubstring(s, l, r))
 
# This code is contributed by gfgking

C#

// C# program for the above approach
using System;
class GFG
{
   
// Function to find the length of
// recurring substring in range [l, r]
static int recurringSubstring(string s, int l, int r)
{
   
    // Length of the string
    int N = s.Length;
 
    // Variable to store the index of
    // the character in the alphabet
    int []a = new int[N];
 
    for (int i = 0; i < N; i++) {
        a[i] = (s[i] - 'a') + 1;
    }
 
    // Prefix array to store the sum
    int []prefix = new int[N];
    prefix[0] = a[0];
 
    for (int i = 1; i < N; i++) {
        prefix[i] = prefix[i - 1] + a[i];
    }
 
    l = l - 1;
    r = r - 1;
 
    // If l is greater than 0
    if (l != 0) {
        return prefix[r] - prefix[l - 1];
    }
   
    // If l is less or equal to 0
    else {
        return prefix[r];
    }
}
 
// Driver Code
public static void Main()
{
    string s = "cbbde";
    int l = 2, r = 5;
 
    Console.Write(recurringSubstring(s, l, r));
     
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

// Javascript program for the above approach
 
// Function to find the length of
// recurring substring in range [l, r]
function recurringSubstring(s, l, r)
{
 
    // Length of the string
    let N = s.length;
 
    // Variable to store the index of
    // the character in the alphabet
    let a = [];
 
    for (let i = 0; i < N; i++) {
        a[i] = (s.charCodeAt(s[i]) - s.charCodeAt('a')) + 1;
    }
 
    // Prefix array to store the sum
    let prefix = [];
    prefix[0] = a[0];
 
    for (let i = 1; i < N; i++) {
        prefix[i] = prefix[i - 1] + a[i];
    }
 
    l = l - 1;
    r = r - 1;
 
    // If l is greater than 0
    if (l != 0) {
        return prefix[r] - prefix[l - 1];
    }
     
    // If l is less or equal to 0
    else {
        return prefix[r];
    }
}
 
// Driver Code
let s = "cbbde";
let l = 2, r = 5;
 
document.write(recurringSubstring(s, l, r));
 
// This code is contributed by Samim Hossain Mondal

输出

13

时间复杂度： O(N)，其中 N 是字符串的长度
辅助空间： O(N)

空间优化方法：上述方法可以进一步优化，方法是去掉前缀数组，简单地迭代给定范围，并将相应的字典值添加到答案中。
请按照以下步骤解决问题：

• 遍历 [L, R] 范围，并将相应的字典值添加到答案中。
  

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the length of
// recurring substring in range [l, r]
int recurringSubstring(string s, int l, int r)
{
    // Length of the string
    int N = s.size();
 
    // Length of resultant string
    int ans = 0;
    for (int i = l - 1; i <= r - 1; i++) {
 
        // Add lexicographic value of s[i]
        ans += (s[i] - 'a' + 1);
    }
    return ans;
}
 
// Driver Code
int main()
{
    string s = "cbbde";
    int l = 2, r = 5;
 
    cout << recurringSubstring(s, l, r);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class GFG
{
// Function to find the length of
// recurring substring in range [l, r]
static int recurringSubstring(String s, int l, int r)
{
    // Length of the string
    int N = s.length();
 
    // Length of resultant string
    int ans = 0;
    for (int i = l - 1; i <= r - 1; i++) {
 
        // Add lexicographic value of s[i]
        ans += (s.charAt(i) - 'a' + 1);
    }
    return ans;
}
 
// Driver Code
public static void main(String args[])
{
    String s = "cbbde";
    int l = 2, r = 5;
 
    System.out.println(recurringSubstring(s, l, r));
}
}
// This code is contributed by Samim Hossain Mondal.

Python3

# Python code for the above approach
 
# Function to find the length of
# recurring substring in range [l, r]
def recurringSubstring(s,  l, r):
 
    # Length of the string
    N = len(s)
 
    # Length of resultant string
    ans = 0
    for i in range(l-1, r):
 
        # Add lexicographic value of s[i]
        ans = ans + (ord(s[i]) - ord('a') + 1)
 
    return ans
 
 
# Driver Code
s = "cbbde"
l = 2
r = 5
 
print(recurringSubstring(s, l, r))
 
 
# This code is contributed by Potta Lokesh

C＃

// C# program for the above approach
using System;
class GFG
{
// Function to find the length of
// recurring substring in range [l, r]
static int recurringSubstring(string s, int l, int r)
{
    // Length of the string
    int N = s.Length;
 
    // Length of resultant string
    int ans = 0;
    for (int i = l - 1; i <= r - 1; i++) {
 
        // Add lexicographic value of s[i]
        ans += (s[i] - 'a' + 1);
    }
    return ans;
}
 
// Driver Code
public static void Main()
{
    string s = "cbbde";
    int l = 2, r = 5;
 
    Console.Write(recurringSubstring(s, l, r));
}
}
// This code is contributed by Samim Hossain Mondal.

Javascript

// Javascript program for the above approach
 
// Function to find the length of
// recurring substring in range [l, r]
function recurringSubstring(s, l, r)
{
    // Length of the string
    let N = s.length;
 
    // Length of resultant string
    let ans = 0;
    for (let i = l - 1; i <= r - 1; i++) {
 
        // Add lexicographic value of s[i]
        ans += (s.charCodeAt(s[i]) - s.charCodeAt('a') + 1);
    }
    return ans;
}
 
// Driver Code
let s = "cbbde";
let l = 2, r = 5;
 
document.write(recurringSubstring(s, l, r));
 
// This code is contributed by Samim Hossain Mondal

输出

13

时间复杂度： O(N)，其中 N 是字符串的长度
辅助空间： O(1)