给定字符串str仅包含小写英文字母和整数K ，任务是找到一个K长度的子字符串，该子字符串包含最大数量的元音(即‘a’，’e’，’i’，’o’，’u’ )。如果有多个这样的子字符串，则返回字典上最小的子字符串。
例子：
Input: str = “geeksforgeeks”, K = 4
Output: eeks
Explanation:
The substrings with maximum count of vowels are “geek”, “eeks” which contains 2 vowels. But “eeks” is lexicographically smallest.
Input: str = “ceebbaceeffo”, K = 3
Output: ace
Explanation:
Lexicographically substrings with maximum count of vowels are “ace”.
天真的方法：
为了解决上述问题，我们必须生成所有长度为K的子串，并在字典上存储所有包含最大元音数量的子串中最小的子串。
时间复杂度： O(N 2 )
高效方法：
可以通过创建前缀和数组来优化上述过程 元音的pref [] ，其中ith索引包含从0到ith索引的元音计数。任何子串str [l：r]的元音计数可以由pref [r] -pref [l-1]给出。然后，找到在词典上最小的子串，并带有最大数量的元音。
下面是上述方法的实现：
C++
// C++ program to find
// lexicographically smallest
// K-length substring containing
// maximum number of vowels
#include
using namespace std;
// Function that prints the
// lexicographically smallest
// K-length substring containing
// maximum number of vowels
string maxVowelSubString(
string str, int K)
{
// Store the length of the string
int N = str.length();
// Initialize a prefix sum array
int pref[N];
// Loop through the string to
// create the prefix sum array
for (int i = 0; i < N; i++) {
// Store 1 at the index
// if it is a vowel
if (str[i] == 'a'
or str[i] == 'e'
or str[i] == 'i'
or str[i] == 'o'
or str[i] == 'u')
pref[i] = 1;
// Otherwise, store 0
else
pref[i] = 0;
// Process the prefix array
if (i)
pref[i] += pref[i - 1];
}
// Initialize the variable to store
// maximum count of vowels
int maxCount = pref[K - 1];
// Initialize the variable
// to store substring
// with maximum count of vowels
string res = str.substr(0, K);
// Loop through the prefix array
for (int i = K; i < N; i++) {
// Store the current
// count of vowels
int currCount
= pref[i]
- pref[i - K];
// Update the result if current count
// is greater than maximum count
if (currCount > maxCount) {
maxCount = currCount;
res = str.substr(i - K + 1, K);
}
// Update lexicographically smallest
// substring if the current count
// is equal to the maximum count
else if (currCount == maxCount) {
string temp
= str.substr(
i - K + 1, K);
if (temp < res)
res = temp;
}
}
// Return the result
return res;
}
// Driver Program
int main()
{
string str = "ceebbaceeffo";
int K = 3;
cout << maxVowelSubString(str, K);
return 0;
} Java
// Java program to find
// lexicographically smallest
// K-length substring containing
// maximum number of vowels
class GFG{
// Function that prints the
// lexicographically smallest
// K-length substring containing
// maximum number of vowels
static String maxVowelSubString(String str,
int K)
{
// Store the length of the string
int N = str.length();
// Initialize a prefix sum array
int []pref = new int[N];
// Loop through the string to
// create the prefix sum array
for (int i = 0; i < N; i++)
{
// Store 1 at the index
// if it is a vowel
if (str.charAt(i) == 'a' ||
str.charAt(i) == 'e' ||
str.charAt(i) == 'i' ||
str.charAt(i) == 'o' ||
str.charAt(i) == 'u')
pref[i] = 1;
// Otherwise, store 0
else
pref[i] = 0;
// Process the prefix array
if (i != 0)
pref[i] += pref[i - 1];
}
// Initialize the variable to store
// maximum count of vowels
int maxCount = pref[K - 1];
// Initialize the variable
// to store substring
// with maximum count of vowels
String res = str.substring(0, K);
// Loop through the prefix array
for (int i = K; i < N; i++)
{
// Store the current
// count of vowels
int currCount = pref[i] -
pref[i - K];
// Update the result if current count
// is greater than maximum count
if (currCount > maxCount)
{
maxCount = currCount;
res = str.substring(i - K + 1,
i + 1);
}
// Update lexicographically smallest
// substring if the current count
// is equal to the maximum count
else if (currCount == maxCount)
{
String temp = str.substring(i - K + 1,
i + 1);
if (temp.compareTo(res) < 0)
res = temp;
}
}
// Return the result
return res;
}
// Driver Code
public static void main(String []args)
{
String str = "ceebbaceeffo";
int K = 3;
System.out.print(maxVowelSubString(str, K));
}
}
// This code is contributed by ChitranayalPython3
# Python3 program to find
# lexicographically smallest
# K-length substring containing
# maximum number of vowels
# Function that prints the
# lexicographically smallest
# K-length substring containing
# maximum number of vowels
def maxVowelSubString(str1, K):
# Store the length of the string
N = len(str1)
# Initialize a prefix sum array
pref = [0 for i in range(N)]
# Loop through the string to
# create the prefix sum array
for i in range(N):
# Store 1 at the index
# if it is a vowel
if (str1[i] == 'a' or
str1[i] == 'e' or
str1[i] == 'i' or
str1[i] == 'o' or
str1[i] == 'u'):
pref[i] = 1
# Otherwise, store 0
else:
pref[i] = 0
# Process the prefix array
if (i):
pref[i] += pref[i - 1]
# Initialize the variable to
# store maximum count of vowels
maxCount = pref[K - 1]
# Initialize the variable
# to store substring with
# maximum count of vowels
res = str1[0:K]
# Loop through the prefix array
for i in range(K, N):
# Store the current
# count of vowels
currCount = pref[i] - pref[i - K]
# Update the result if current count
# is greater than maximum count
if (currCount > maxCount):
maxCount = currCount
res = str1[i - K + 1 : i + 1]
# Update lexicographically smallest
# substring if the current count
# is equal to the maximum count
elif (currCount == maxCount):
temp = str1[i - K + 1 : i + 1]
if (temp < res):
res = temp
# Return the result
return res
# Driver code
if __name__ == '__main__':
str1 = "ceebbaceeffo"
K = 3
print(maxVowelSubString(str1, K))
# This code is contributed by Surendra_GangwarC#
// C# program to find
// lexicographically smallest
// K-length substring containing
// maximum number of vowels
using System;
class GFG{
// Function that prints the
// lexicographically smallest
// K-length substring containing
// maximum number of vowels
static string maxVowelSubString(string str,
int K)
{
// Store the length of the string
int N = str.Length;
// Initialize a prefix sum array
int []pref = new int[N];
// Loop through the string to
// create the prefix sum array
for (int i = 0; i < N; i++)
{
// Store 1 at the index
// if it is a vowel
if (str[i] == 'a' ||
str[i] == 'e' ||
str[i] == 'i' ||
str[i] == 'o' ||
str[i] == 'u')
pref[i] = 1;
// Otherwise, store 0
else
pref[i] = 0;
// Process the prefix array
if (i != 0)
pref[i] += pref[i - 1];
}
// Initialize the variable to store
// maximum count of vowels
int maxCount = pref[K - 1];
// Initialize the variable
// to store substring
// with maximum count of vowels
string res = str.Substring(0, K);
// Loop through the prefix array
for (int i = K; i < N; i++)
{
// Store the current
// count of vowels
int currCount = pref[i] -
pref[i - K];
// Update the result if current count
// is greater than maximum count
if (currCount > maxCount)
{
maxCount = currCount;
res = str.Substring(i - K + 1, K);
}
// Update lexicographically smallest
// substring if the current count
// is equal to the maximum count
else if (currCount == maxCount)
{
string temp = str.Substring(i - K + 1, K);
if (string.Compare(temp, res) == -1)
res = temp;
}
}
// Return the result
return res;
}
// Driver Code
public static void Main()
{
string str = "ceebbaceeffo";
int K = 3;
Console.Write(maxVowelSubString(str, K));
}
}
// This code is contributed by Code_Mech输出：
ace
时间复杂度： O(N)
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