给定字符串S ，任务是通过重新排列或删除元素来查找不超过K个连续出现元素的最大词典字符串。
例子：
Input: S = “baccc”
K = 2
Output: Result = “ccbca”
Explanation: Since K=2, a maximum of 2 same characters can be placed consecutively.
No. of ‘c’ = 3.
No. of ‘b’ = 1.
No. of ‘a’ = 1.
Since the largest lexicographical string has to be printed, therefore, the answer is “ccbca”.
Input: S = “xxxxzaz”
K = 3
Output: result = “zzxxxax”
方法：
- 形成面积为26，其中，索引i被使用选择的频率阵列(在一个字符串的字符- “A”)。
- 初始化一个空字符串以存储相应的更改。
- 对于i = 25到0，请执行以下操作：
- 如果索引i处的频率大于k，则附加(i +’a’)K次。将索引i处的频率降低K，找到下一个优先级最高的元素，然后追加回答，并将相应索引处的频率降低1。
- 如果索引i处的频率大于0但小于k，则将其频率附加(i +’a’)倍。
- 如果索引i处的频率为0，则该索引不能用于形成元素，因此请检查下一个可能的最高优先级元素。
C++
// C++ code for the above approach
#include
using namespace std;
#define ll long long int
// Function to find the
// largest lexicographical
// string with given constraints.
string getLargestString(string s,
ll k)
{
// vector containing frequency
// of each character.
vector frequency_array(26, 0);
// assigning frequency to
for (int i = 0;
i < s.length(); i++) {
frequency_array[s[i] - 'a']++;
}
// empty string of string class type
string ans = "";
// loop to iterate over
// maximum priority first.
for (int i = 25; i >= 0;) {
// if frequency is greater than
// or equal to k.
if (frequency_array[i] > k) {
// temporary variable to operate
// in-place of k.
int temp = k;
string st(1, i + 'a');
while (temp > 0) {
// concatenating with the
// resultant string ans.
ans += st;
temp--;
}
frequency_array[i] -= k;
// handling k case by adjusting
// with just smaller priority
// element.
int j = i - 1;
while (frequency_array[j]
<= 0
&& j >= 0) {
j--;
}
// condition to verify if index
// j does have frequency
// greater than 0;
if (frequency_array[j] > 0
&& j >= 0) {
string str(1, j + 'a');
ans += str;
frequency_array[j] -= 1;
}
else {
// if no such element is found
// than string can not be
// processed further.
break;
}
}
// if frequency is greater than 0
// and less than k.
else if (frequency_array[i] > 0) {
// here we don't need to fix K
// consecutive element criteria.
int temp = frequency_array[i];
frequency_array[i] -= temp;
string st(1, i + 'a');
while (temp > 0) {
ans += st;
temp--;
}
}
// otherwise check for next
// possible element.
else {
i--;
}
}
return ans;
}
// Driver program
int main()
{
string S = "xxxxzza";
int k = 3;
cout << getLargestString(S, k)
<< endl;
return 0;
} Java
// Java code for
// the above approach
import java.util.*;
class GFG{
// Function to find the
// largest lexicographical
// String with given constraints.
static String getLargestString(String s,
int k)
{
// Vector containing frequency
// of each character.
int []frequency_array = new int[26];
// Assigning frequency
for (int i = 0;
i < s.length(); i++)
{
frequency_array[s.charAt(i) - 'a']++;
}
// Empty String of
// String class type
String ans = "";
// Loop to iterate over
// maximum priority first.
for (int i = 25; i >= 0😉
{
// If frequency is greater than
// or equal to k.
if (frequency_array[i] > k)
{
// Temporary variable to
// operate in-place of k.
int temp = k;
String st = String.valueOf((char)(i + 'a'));
while (temp > 0)
{
// Concatenating with the
// resultant String ans.
ans += st;
temp--;
}
frequency_array[i] -= k;
// Handling k case by adjusting
// with just smaller priority
// element.
int j = i - 1;
while (frequency_array[j] <= 0 &&
j >= 0)
{
j--;
}
// Condition to verify if index
// j does have frequency
// greater than 0;
if (frequency_array[j] > 0 &&
j >= 0)
{
String str = String.valueOf((char)(j + 'a'));
ans += str;
frequency_array[j] -= 1;
}
else
{
// If no such element is found
// than String can not be
// processed further.
break;
}
}
// If frequency is greater than 0
// and less than k.
else if (frequency_array[i] > 0)
{
// Here we don't need to fix K
// consecutive element criteria.
int temp = frequency_array[i];
frequency_array[i] -= temp;
String st = String.valueOf((char)(i + 'a'));
while (temp > 0)
{
ans += st;
temp--;
}
}
// Otherwise check for next
// possible element.
else
{
i--;
}
}
return ans;
}
// Driver code
public static void main(String[] args)
{
String S = "xxxxzza";
int k = 3;
System.out.print(getLargestString(S, k));
}
}
// This code is contributed by shikhasingrajputPython3
# Python3 code for the above approach
# Function to find the
# largest lexicographical
# string with given constraints.
def getLargestString(s, k):
# Vector containing frequency
# of each character.
frequency_array = [0] * 26
# Assigning frequency to
for i in range(len(s)):
frequency_array[ord(s[i]) -
ord('a')] += 1
# Empty string of
# string class type
ans = ""
# Loop to iterate over
# maximum priority first.
i = 25
while i >= 0:
# If frequency is greater than
# or equal to k.
if (frequency_array[i] > k):
# Temporary variable to
# operate in-place of k.
temp = k
st = chr( i + ord('a'))
while (temp > 0):
# concatenating with the
# resultant string ans.
ans += st
temp -= 1
frequency_array[i] -= k
# Handling k case by adjusting
# with just smaller priority
# element.
j = i - 1
while (frequency_array[j] <= 0 and
j >= 0):
j -= 1
# Condition to verify if index
# j does have frequency
# greater than 0;
if (frequency_array[j] > 0 and
j >= 0):
str1 = chr(j + ord( 'a'))
ans += str1
frequency_array[j] -= 1
else:
# if no such element is found
# than string can not be
# processed further.
break
# If frequency is greater than 0
#and less than k.
elif (frequency_array[i] > 0):
# Here we don't need to fix K
# consecutive element criteria.
temp = frequency_array[i]
frequency_array[i] -= temp
st = chr(i + ord('a'))
while (temp > 0):
ans += st
temp -= 1
# Otherwise check for next
# possible element.
else:
i -= 1
return ans
# Driver code
if __name__ == "__main__":
S = "xxxxzza"
k = 3
print (getLargestString(S, k))
# This code is contributed by ChitranayalC#
// C# code for
// the above approach
using System;
class GFG{
// Function to find the
// largest lexicographical
// String with given constraints.
static String getLargestString(String s,
int k)
{
// List containing frequency
// of each character.
int []frequency_array = new int[26];
// Assigning frequency
for (int i = 0; i < s.Length; i++)
{
frequency_array[s[i] - 'a']++;
}
// Empty String of
// String class type
String ans = "";
// Loop to iterate over
// maximum priority first.
for (int i = 25; i >= 0;)
{
// If frequency is greater than
// or equal to k.
if (frequency_array[i] > k)
{
// Temporary variable to
// operate in-place of k.
int temp = k;
String st = String.Join("",
(char)(i + 'a'));
while (temp > 0)
{
// Concatenating with the
// resultant String ans.
ans += st;
temp--;
}
frequency_array[i] -= k;
// Handling k case by adjusting
// with just smaller priority
// element.
int j = i - 1;
while (frequency_array[j] <= 0 &&
j >= 0)
{
j--;
}
// Condition to verify if index
// j does have frequency
// greater than 0;
if (frequency_array[j] > 0 &&
j >= 0)
{
String str = String.Join("",
(char)(j + 'a'));
ans += str;
frequency_array[j] -= 1;
}
else
{
// If no such element is found
// than String can not be
// processed further.
break;
}
}
// If frequency is greater than 0
// and less than k.
else if (frequency_array[i] > 0)
{
// Here we don't need to fix K
// consecutive element criteria.
int temp = frequency_array[i];
frequency_array[i] -= temp;
String st = String.Join("",
(char)(i + 'a'));
while (temp > 0)
{
ans += st;
temp--;
}
}
// Otherwise check for next
// possible element.
else
{
i--;
}
}
return ans;
}
// Driver code
public static void Main(String[] args)
{
String S = "xxxxzza";
int k = 3;
Console.Write(getLargestString(S, k));
}
}
// This code is contributed by Princi Singh输出
zzxxxax
时间复杂度： O(N)