给定一个字符串S和两个整数N和K ，任务是找到最大长度的后缀，使得后缀字符串中每个字符的出现次数小于N并且最多可以从输入字符串删除K 个元素以获得最大长度的后缀.
例子：
Input: S = “iahagafedcba”, N = 1, K = 0
Output: fedcba
Explanation:
Maximum length suffix in the given string
Such that occurence of each character is 1 is “fedcba”,
Because if we take the string “afedcba”,
then the occurence of character “a” will be 2.
Input: S = “iahagafedcba”, N = 1, K = 2
Output: hgfedcba
Explanation:
Maximum length suffix in the given string
Such that occurrence of each character is 1 is “hgfedcba”,
After deleting character “a” two times.
做法：思路是用hash-map来存储字符串的字符出现的频率。
- 初始化一个空字符串存储字符串的最长后缀。
- 使用循环从最后一个迭代字符串-
- 将哈希映射中字符的出现次数增加 1
- 如果当前字符出现次数小于N，则将该字符添加到后缀字符串
- 否则，如果 K 的值大于 0，则将 K 的值减 1
- 打印后缀字符串。
下面是上述方法的实现：
C++
// C++ implementation to find
// longest suffix of the string
// such that occurrence of each
// character is less than K
#include
using namespace std;
// Function to find the maximum
// length suffix in the string
void maximumSuffix(string s,
int n, int k){
// Length of the string
int i = s.length() - 1;
// Map to store the number
// of occurrence of character
int arr[26] = { 0 };
string suffix = "";
// Loop to iterate string
// from the last character
while (i > -1) {
int index = s[i] - 'a';
// Condition to check if the
// occurrence of each character
// is less than given number
if (arr[index] < n) {
arr[index]++;
suffix += s[i];
i--;
continue;
}
// Condition when character
// cannot be deleted
if (k == 0)
break;
k--;
i--;
}
reverse(suffix.begin(), suffix.end());
// Longest suffix
cout << suffix;
}
// Driver Code
int main()
{
string str = "iahagafedcba";
int n = 1, k = 2;
// Function call
maximumSuffix(str, n, k);
return 0;
} Java
// Java implementation to find
// longest suffix of the String
// such that occurrence of each
// character is less than K
class GFG{
// Function to find the maximum
// length suffix in the String
static void maximumSuffix(String s,
int n, int k){
// Length of the String
int i = s.length() - 1;
// Map to store the number
// of occurrence of character
int arr[] = new int[26];
String suffix = "";
// Loop to iterate String
// from the last character
while (i > -1) {
int index = s.charAt(i) - 'a';
// Condition to check if the
// occurrence of each character
// is less than given number
if (arr[index] < n) {
arr[index]++;
suffix += s.charAt(i);
i--;
continue;
}
// Condition when character
// cannot be deleted
if (k == 0)
break;
k--;
i--;
}
suffix = reverse(suffix);
// Longest suffix
System.out.print(suffix);
}
static String reverse(String input) {
char[] a = input.toCharArray();
int l, r = a.length - 1;
for (l = 0; l < r; l++, r--) {
char temp = a[l];
a[l] = a[r];
a[r] = temp;
}
return String.valueOf(a);
}
// Driver Code
public static void main(String[] args)
{
String str = "iahagafedcba";
int n = 1, k = 2;
// Function call
maximumSuffix(str, n, k);
}
}
// This code is contributed by 29AjayKumarPython 3
# Python 3 implementation to find
# longest suffix of the string
# such that occurrence of each
# character is less than K
# Function to find the maximum
# length suffix in the string
def maximumSuffix(s, n, k):
# Length of the string
i = len(s)- 1
# Map to store the number
# of occurrence of character
arr = [0 for i in range(26)]
suffix = ""
# Loop to iterate string
# from the last character
while (i > -1):
index = ord(s[i]) - ord('a');
# Condition to check if the
# occurrence of each character
# is less than given number
if (arr[index] < n):
arr[index] += 1
suffix += s[i]
i -= 1
continue
# Condition when character
# cannot be deleted
if (k == 0):
break
k -= 1
i -= 1
suffix = suffix[::-1]
# Longest suffix
print(suffix)
# Driver Code
if __name__ == '__main__':
str = "iahagafedcba"
n = 1
k = 2
# Function call
maximumSuffix(str, n, k)
# This code is contributed by Surendra_GangwarC#
// C# implementation to find
// longest suffix of the String
// such that occurrence of each
// character is less than K
using System;
class GFG{
// Function to find the maximum
// length suffix in the String
static void maximumSuffix(String s,
int n, int k){
// Length of the String
int i = s.Length - 1;
// Map to store the number
// of occurrence of character
int []arr = new int[26];
String suffix = "";
// Loop to iterate String
// from the last character
while (i > -1) {
int index = s[i] - 'a';
// Condition to check if the
// occurrence of each character
// is less than given number
if (arr[index] < n) {
arr[index]++;
suffix += s[i];
i--;
continue;
}
// Condition when character
// cannot be deleted
if (k == 0)
break;
k--;
i--;
}
suffix = reverse(suffix);
// longest suffix
Console.Write(suffix);
}
static String reverse(String input) {
char[] a = input.ToCharArray();
int l, r = a.Length - 1;
for (l = 0; l < r; l++, r--) {
char temp = a[l];
a[l] = a[r];
a[r] = temp;
}
return String.Join("",a);
}
// Driver Code
public static void Main(String[] args)
{
String str = "iahagafedcba";
int n = 1, k = 2;
// Function call
maximumSuffix(str, n, k);
}
}
// This code is contributed by Rajput-Ji输出：
hgfedcba性能分析：
- 时间复杂度：在上面给出的方法中，有一个循环用于迭代字符串，在最坏的情况下需要 O(L) 时间。因此，这种方法的时间复杂度为O(L) 。
- 辅助空间复杂度：在上面给出的方法中，有额外的空间用于存储字符的频率。因此，上述方法的辅助空间复杂度将为O(L)
如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live