给定字符串S和整数K。任务是找到字典上S的最大子序列，例如T，以使T中的每个字符必须出现至少K次。
例子：
输入：S =“ banana”，K =2。输出：nn每个字符至少存在两次的可能子序列是： 
在以上子序列中，“ nn”在字典上最大。
这个想法是贪婪地解决上述问题。如果要按字典顺序使子序列最大，则必须优先考虑按字典顺序较大的字符。 ‘z’是最大的字符，假设z在S中出现f z次。如果f z > = K，则在字符串T中追加’z’z k次，并继续从S的左侧删除字符，直到所有z都为删除。应用带有“ y”，“ w”，…..，“ a”的策略。最后，您会找到答案。
让我们来看一个例子。假设S =“ zzwzawa”且K =2。从最大字符“ z”开始。这里f z = 3> =K。因此T将变为“ zzz”，我们将删除S左侧的字母，直到所有z都被删除为止。因此，现在S将变为“ awa”。下一个最大的是“ y”，但在k中出现0次，因此我们将其跳过。我们也将跳过“ w”，“ v”等，直到进入出现两次的“ a”。现在，T将成为“ zzzaa”，S将成为一个空字符串。我们的答案是“ zzzaa”。
以下是此方法的实现：
C++
// C++ program to find lexicographically largest
// subsequence where every character appears at
// least k times.
#include
using namespace std;
// Find lexicographically largest subsequence of
// s[0..n-1] such that every character appears
// at least k times. The result is filled in t[]
void subsequence(char s[], char t[], int n, int k)
{
int last = 0, cnt = 0, new_last = 0, size = 0;
// Starting from largest charter 'z' to 'a'
for (char ch = 'z'; ch >= 'a'; ch--) {
cnt = 0;
// Counting the frequency of the character
for (int i = last; i < n; i++) {
if (s[i] == ch)
cnt++;
}
// If frequency is greater than k
if (cnt >= k) {
// From the last point we leave
for (int i = last; i < n; i++) {
// check if string contain ch
if (s[i] == ch) {
// If yes, append to output string
t[size++] = ch;
new_last = i;
}
}
// Update the last point.
last = new_last;
}
}
t[size] = '\0';
}
// Driver code
int main()
{
char s[] = "banana";
int n = sizeof(s);
int k = 2;
char t[n];
subsequence(s, t, n - 1, k);
cout << t << endl;
return 0;
} Java
import java.util.Arrays;
// Java program to find lexicographically largest
// subsequence where every character appears at
// least k times.
class GFG {
// Find lexicographically largest subsequence of
// s[0..n-1] such that every character appears
// at least k times. The result is filled in t[]
static void subsequence(char s[], char t[], int n, int k)
{
int last = 0, cnt = 0, new_last = 0, size = 0;
// Starting from largest charter 'z' to 'a'
for (char ch = 'z'; ch >= 'a'; ch--) {
cnt = 0;
// Counting the frequency of the character
for (int i = last; i < n; i++) {
if (s[i] == ch)
cnt++;
}
// If frequency is greater than k
if (cnt >= k) {
// From the last point we leave
for (int i = last; i < n; i++) {
// check if string contain ch
if (s[i] == ch) {
// If yes, append to output string
t[size++] = ch;
new_last = i;
}
}
// Update the last point.
last = new_last;
}
}
t[size] = '\0';
}
// Driver code
public static void main(String[] args) {
char s[] = {'b','a','n','a','n','a'};
int n = s.length;
int k = 2;
char t[] = new char[n];
subsequence(s, t, n - 1, k);
for(int i = 0;iPython3
# Python3 program to find lexicographically largest
# subsequence where every character appears at
# least k times.
# Find lexicographically largest subsequence of
# s[0..n-1] such that every character appears
# at least k times. The result is filled in t[]
def subsequence(s, t, n, k):
last = 0
cnt = 0
new_last = 0
size = 0
string = 'zyxwvutsrqponmlkjihgfedcba'
# Starting from largest charter 'z' to 'a'
for ch in string:
cnt = 0
for i in range(last, n):
if s[i] == ch:
cnt += 1
# If frequency is greater than k
if cnt >= k:
# From the last point we leave
for i in range(last, n):
# check if string contain ch
if s[i] == ch:
# If yes, append to output string
t[size] = ch
new_last = i
size += 1
# Update the last point.
last = new_last
# Driver Code
if __name__ == "__main__":
s = ['b', 'a', 'n', 'a', 'n', 'a']
n = len(s)
k = 2
t = [''] * n
subsequence(s, t, n - 1, k)
t = ''.join(t)
print(t)
# This code is contributed by
# sanjeev2552C#
// C# program to find lexicographically
// largest subsequence where every
// character appears at least k times.
using System;
class GFG
{
// Find lexicographically largest subsequence
// of s[0..n-1] such that every character
// appears at least k times. The result is
// filled in t[]
static void subsequence(char []s, char []t,
int n, int k)
{
int last = 0, cnt = 0,
new_last = 0, size = 0;
// Starting from largest character
// 'z' to 'a'
for (char ch = 'z'; ch >= 'a'; ch--)
{
cnt = 0;
// Counting the frequency of
// the character
for (int i = last; i < n; i++)
{
if (s[i] == ch)
cnt++;
}
// If frequency is greater than k
if (cnt >= k)
{
// From the last point we leave
for (int i = last; i < n; i++)
{
// check if string contain ch
if (s[i] == ch)
{
// If yes, append to output string
t[size++] = ch;
new_last = i;
}
}
// Update the last point.
last = new_last;
}
}
t[size] = '\0';
}
// Driver code
public static void Main()
{
char []s = {'b','a','n','a','n','a'};
int n = s.Length;
int k = 2;
char []t = new char[n];
subsequence(s, t, n - 1, k);
for(int i = 0; i < t.Length; i++)
if(t[i] != 0)
Console.Write(t[i]);
}
}
// This code contributed by Rajput-Ji输出：
nn