元音和辅音数量相等的最长子串

给定一个由小写英文字母组成的字符串S ，任务是从给定的字符串找出最长的子串的长度，它具有相同数量的元音和辅音。

例子：

Input: S = “geeksforgeeks” 
Output: 10 
Explanation: 
The substring “eeksforgee” consists of 5 vowels and 5 consonants. Remaining characters are only consonants. Therefore, any longer substring won’t have an equal number of vowels and consonants.

Input: S = “qwertyuiop” 
Output: 8 
Explanation: 
The substring “wertyuio” consists of 4 vowels and 4 consonants.

朴素的方法：最简单的解决方案是生成给定字符串的所有子串，并为每个子串检查元音和辅音的数量是否相等。最后，打印获得的元音和辅音数量相等的子串的最大长度。
时间复杂度： O(N ³ )
辅助空间： O(1)

高效方法：思路是考虑一个长度等于给定字符串的数组，分别存储元音和辅音对应的1和-1 ，使用HashMap打印出总和等于0的最长子数组的长度.

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to return the length of
// the longest substring having equal
// number of vowel and consonant
int maxsubstringLength(string S, int N)
{
    int arr[N];
 
    // Generate the array
    for (int i = 0; i < N; i++)
        if (S[i] == 'a' || S[i] == 'e' || S[i] == 'i'
            || S[i] == 'o' || S[i] == 'u')
            arr[i] = 1;
        else
            arr[i] = -1;
 
    // Initialize variable
    // to store result
    int maxLen = 0;
 
    // Stores the sum of subarray
    int curr_sum = 0;
 
    // Map to store indices of the sum
    unordered_map hash;
 
    // Loop through the array
    for (int i = 0; i < N; i++) {
        curr_sum += arr[i];
 
        // If sum is 0
        if (curr_sum == 0)
 
            // Count of vowels and consonants
            // are equal
            maxLen = max(maxLen, i + 1);
 
        // Update the maximum length
        // of substring in HashMap
        if (hash.find(curr_sum) != hash.end())
            maxLen = max(maxLen, i - hash[curr_sum]);
 
        // Store the index of the sum
        else
            hash[curr_sum] = i;
    }
 
    // Return the maximum
    // length of required substring
    return maxLen;
}
 
// Driver Code
int main()
{
    string S = "geeksforgeeks";
    int n = sizeof(S) / sizeof(S[0]);
    cout << maxsubstringLength(S, n);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
     
// Function to return the length of
// the longest subString having equal
// number of vowel and consonant
static int maxsubStringLength(char[] S, int N)
{
    int arr[] = new int[N];
     
    // Generate the array
    for(int i = 0; i < N; i++)
    if (S[i] == 'a' || S[i] == 'e' ||
        S[i] == 'i' || S[i] == 'o' ||
        S[i] == 'u')
        arr[i] = 1;
    else
        arr[i] = -1;
         
    // Initialize variable
    // to store result
    int maxLen = 0;
     
    // Stores the sum of subarray
    int curr_sum = 0;
     
    // Map to store indices of the sum
    HashMap hash = new HashMap<>();
     
    // Loop through the array
    for(int i = 0; i < N; i++)
    {
        curr_sum += arr[i];
         
        // If sum is 0
        if (curr_sum == 0)
         
            // Count of vowels and consonants
            // are equal
            maxLen = Math.max(maxLen, i + 1);
             
        // Update the maximum length
        // of subString in HashMap
        if (hash.containsKey(curr_sum))
            maxLen = Math.max(maxLen,
                              i - hash.get(curr_sum));
                               
        // Store the index of the sum
        else
         
            // hash[curr_sum] = i;
            hash.put(curr_sum, i);
    }
     
    // Return the maximum
    // length of required subString
    return maxLen;
}
 
// Driver Code
public static void main(String[] args)
{
    String S = "geeksforgeeks";
    int n = S.length();
     
    System.out.print(
        maxsubStringLength(S.toCharArray(), n));
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 program to implement
# the above approach
 
# Function to return the length of
# the longest substring having equal
# number of vowel and consonant
def maxsubstringLength(S, N):
 
    arr = [0] * N
 
    # Generate the array
    for i in range(N):
        if(S[i] == 'a' or S[i] == 'e' or
           S[i] == 'i' or S[i] == 'o' or
           S[i] == 'u'):
            arr[i] = 1
        else:
            arr[i] = -1
 
    # Initialize variable
    # to store result
    maxLen = 0
 
    # Stores the sum of subarray
    curr_sum = 0
 
    # Map to store indices of the sum
    hash = {}
 
    # Loop through the array
    for i in range(N):
        curr_sum += arr[i]
 
        # If sum is 0
        if(curr_sum == 0):
 
            # Count of vowels and consonants
            # are equal
            maxLen = max(maxLen, i + 1)
 
        # Update the maximum length
        # of substring in HashMap
        if(curr_sum in hash.keys()):
            maxLen = max(maxLen, i - hash[curr_sum])
 
        # Store the index of the sum
        else:
            hash[curr_sum] = i
 
    # Return the maximum
    # length of required substring
    return maxLen
 
# Driver Code
S = "geeksforgeeks"
n = len(S)
 
# Function call
print(maxsubstringLength(S, n))
 
# This code is contributed by Shivam Singh

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to return the length of
// the longest subString having equal
// number of vowel and consonant
static int maxsubStringLength(char[] S, int N)
{
    int []arr = new int[N];
     
    // Generate the array
    for(int i = 0; i < N; i++)
    if (S[i] == 'a' || S[i] == 'e' ||
        S[i] == 'i' || S[i] == 'o' ||
        S[i] == 'u')
        arr[i] = 1;
    else
        arr[i] = -1;
         
    // Initialize variable
    // to store result
    int maxLen = 0;
     
    // Stores the sum of subarray
    int curr_sum = 0;
     
    // Map to store indices of the sum
    Dictionary hash = new Dictionary();
     
    // Loop through the array
    for(int i = 0; i < N; i++)
    {
        curr_sum += arr[i];
         
        // If sum is 0
        if (curr_sum == 0)
         
            // Count of vowels and consonants
            // are equal
            maxLen = Math.Max(maxLen, i + 1);
             
        // Update the maximum length
        // of subString in Dictionary
        if (hash.ContainsKey(curr_sum))
            maxLen = Math.Max(maxLen,
                              i - hash[curr_sum]);
                               
        // Store the index of the sum
        else
         
            // hash[curr_sum] = i;
            hash.Add(curr_sum, i);
    }
     
    // Return the maximum
    // length of required subString
    return maxLen;
}
 
// Driver Code
public static void Main(String[] args)
{
    String S = "geeksforgeeks";
    int n = S.Length;
     
    Console.Write(maxsubStringLength(
                    S.ToCharArray(), n));
}
}
 
// This code is contributed by Princi Singh

Javascript

 
// Javascript program to implement
// the above approach
 
// Function to return the length of
// the longest subString having equal
// number of vowel and consonant
function maxsubStringLength(S, N)
{
    let arr = Array.from({length: N}, (_, i) => 0);
      
    // Generate the array
    for(let i = 0; i < N; i++)
    if (S[i] == 'a' || S[i] == 'e' ||
        S[i] == 'i' || S[i] == 'o' ||
        S[i] == 'u')
        arr[i] = 1;
    else
        arr[i] = -1;
          
    // Initialize variable
    // to store result
    let maxLen = 0;
      
    // Stores the sum of subarray
    let curr_sum = 0;
      
    // Map to store indices of the sum
    let hash = new Map();
      
    // Loop through the array
    for(let i = 0; i < N; i++)
    {
        curr_sum += arr[i];
          
        // If sum is 0
        if (curr_sum == 0)
          
            // Count of vowels and consonants
            // are equal
            maxLen = Math.max(maxLen, i + 1);
              
        // Update the maximum length
        // of subString in HashMap
        if (hash.has(curr_sum))
            maxLen = Math.max(maxLen,
                              i - hash.get(curr_sum));
                                
        // Store the index of the sum
        else
          
            // hash[curr_sum] = i;
            hash.set(curr_sum, i);
    }
      
    // Return the maximum
    // length of required subString
    return maxLen;
}
 
// Driver code
 
    let S = "geeksforgeeks";
    let n = S.length;
      
    document.write(maxsubStringLength(S.split(''), n));
 

输出：

10

时间复杂度： O(NlogN)
辅助空间： O(N)

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