给定长度为n且正整数为k的字符串S。任务是找到长度为k的回文子序列数，其中k <= 3。
例子：
输入：s =“ aabab”，k = 2输出：4 
输入：s =“ aaa”，k = 3输出：1
对于k = 1 ，我们可以轻松地说出字符串的字符数将是答案。
对于k = 2 ，我们可以轻松地制作成对的相同字符，因此我们必须维护字符串中每个字符的计数，然后计算
sum = 0
for character 'a' to 'z'
cnt = count(characater)
sum = sum + cnt*(cnt-1)/2
sum is the answer.现在，随着k的增加，变得很难找到。如何找到k = 3的答案?因此，我们的想法是看到长度为3的回文将采用TZT格式，因此我们必须维护两个矩阵，一个矩阵用于计算每个字符的前缀和，另一个矩阵用于计算字符串中每个字符的后缀和。
索引为i的字符T的前缀总和为L [T] [i]，即T在[0，i](索引)范围内出现的次数。
后缀总和在索引一个字符T I为R [T]已经发生在[I，N – 1]的范围内(指数)。
两个矩阵都将是26 * n，并且可以以复杂度O(26 * n)预先计算这两个矩阵，其中n是字符串的长度。
现在如何计算子序列?考虑一下：对于索引，我假设一个字符X在[0，i – 1]范围内出现n1次，在[i +1，n – 1]范围内出现n2次，那么该字符的答案将为n1 * n2即L [X] [i-1] * R [X] [i + 1]，这将给出格式Xs [i] -X的子序列数，其中s [i]是第i个字符指数。因此，对于每个指数，您都必须计算出
L[X][i-1] * R[X][i+1],
where i is the range [1, n-2] and
X will be from 'a' to 'z'以下是此方法的实现：
C++
// CPP program to count number of subsequences of
// given length.
#include
#define MAX 100
#define MAX_CHAR 26
using namespace std;
// Precompute the prefix and suffix array.
void precompute(string s, int n, int l[][MAX],
int r[][MAX])
{
l[s[0] - 'a'][0] = 1;
// Precompute the prefix 2D array
for (int i = 1; i < n; i++) {
for (int j = 0; j < MAX_CHAR; j++)
l[j][i] += l[j][i - 1];
l[s[i] - 'a'][i]++;
}
r[s[n - 1] - 'a'][n - 1] = 1;
// Precompute the Suffix 2D array.
for (int i = n - 2; i >= 0; i--) {
for (int j = 0; j < MAX_CHAR; j++)
r[j][i] += r[j][i + 1];
r[s[i] - 'a'][i]++;
}
}
// Find the number of palindromic subsequence of
// length k
int countPalindromes(int k, int n, int l[][MAX],
int r[][MAX])
{
int ans = 0;
// If k is 1.
if (k == 1) {
for (int i = 0; i < MAX_CHAR; i++)
ans += l[i][n - 1];
return ans;
}
// If k is 2
if (k == 2) {
// Adding all the products of prefix array
for (int i = 0; i < MAX_CHAR; i++)
ans += ((l[i][n - 1] * (l[i][n - 1] - 1)) / 2);
return ans;
}
// For k greater than 2. Adding all the products
// of value of prefix and suffix array.
for (int i = 1; i < n - 1; i++)
for (int j = 0; j < MAX_CHAR; j++)
ans += l[j][i - 1] * r[j][i + 1];
return ans;
}
// Driven Program
int main()
{
string s = "aabab";
int k = 2;
int n = s.length();
int l[MAX_CHAR][MAX] = { 0 }, r[MAX_CHAR][MAX] = { 0 };
precompute(s, n, l, r);
cout << countPalindromes(k, n, l, r) << endl;
return 0;
} Java
// Java program to count number of subsequences of
// given length.
class GFG
{
static final int MAX=100;
static final int MAX_CHAR=26;
// Precompute the prefix and suffix array.
static void precompute(String s, int n, int l[][],
int r[][])
{
l[s.charAt(0) - 'a'][0] = 1;
// Precompute the prefix 2D array
for (int i = 1; i < n; i++) {
for (int j = 0; j < MAX_CHAR; j++)
l[j][i] += l[j][i - 1];
l[s.charAt(i) - 'a'][i]++;
}
r[s.charAt(n - 1) - 'a'][n - 1] = 1;
// Precompute the Suffix 2D array.
for (int i = n - 2; i >= 0; i--) {
for (int j = 0; j < MAX_CHAR; j++)
r[j][i] += r[j][i + 1];
r[s.charAt(i) - 'a'][i]++;
}
}
// Find the number of palindromic subsequence of
// length k
static int countPalindromes(int k, int n, int l[][],
int r[][])
{
int ans = 0;
// If k is 1.
if (k == 1) {
for (int i = 0; i < MAX_CHAR; i++)
ans += l[i][n - 1];
return ans;
}
// If k is 2
if (k == 2) {
// Adding all the products of prefix array
for (int i = 0; i < MAX_CHAR; i++)
ans += ((l[i][n - 1] * (l[i][n - 1] - 1)) / 2);
return ans;
}
// For k greater than 2. Adding all the products
// of value of prefix and suffix array.
for (int i = 1; i < n - 1; i++)
for (int j = 0; j < MAX_CHAR; j++)
ans += l[j][i - 1] * r[j][i + 1];
return ans;
}
// Driver code
public static void main (String[] args)
{
String s = "aabab";
int k = 2;
int n = s.length();
int l[][]=new int[MAX_CHAR][MAX];
int r[][]=new int[MAX_CHAR][MAX];
precompute(s, n, l, r);
System.out.println(countPalindromes(k, n, l, r));
}
}
// This code is contributed by Anant Agarwal.Python3
# Python3 program to count number of
# subsequences of given length.
MAX = 100
MAX_CHAR = 26
# Precompute the prefix and suffix array.
def precompute(s, n, l, r):
l[ord(s[0]) - ord('a')][0] = 1
# Precompute the prefix 2D array
for i in range(1, n):
for j in range(MAX_CHAR):
l[j][i] += l[j][i - 1]
l[ord(s[i]) - ord('a')][i] += 1
r[ord(s[n - 1]) - ord('a')][n - 1] = 1
# Precompute the Suffix 2D array.
k = n - 2
while(k >= 0):
for j in range(MAX_CHAR):
r[j][k] += r[j][k + 1]
r[ord(s[k]) - ord('a')][k] += 1
k -= 1
# Find the number of palindromic
# subsequence of length k
def countPalindromes(k, n, l, r):
ans = 0
# If k is 1.
if (k == 1):
for i in range(MAX_CHAR):
ans += l[i][n - 1]
return ans
# If k is 2
if (k == 2):
# Adding all the products of
# prefix array
for i in range(MAX_CHAR):
ans += ((l[i][n - 1] * (l[i][n - 1] - 1)) / 2)
return ans
# For k greater than 2. Adding all
# the products of value of prefix
# and suffix array.
for i in range(1, n - 1):
for j in range(MAX_CHAR):
ans += l[j][i - 1] * r[j][i + 1]
return ans
# Driven Program
s = "aabab"
k = 2
n = len(s)
l = [[0 for x in range(MAX)] for y in range(MAX_CHAR)]
r = [[0 for x in range(MAX)] for y in range(MAX_CHAR)]
precompute(s, n, l, r)
print (countPalindromes(k, n, l, r))
# This code is written by Sachin BishtC#
// C# program to count number of
// subsequences of given length.
using System;
class GFG {
static int MAX=100;
static int MAX_CHAR=26;
// Precompute the prefix
// and suffix array.
static void precompute(string s, int n,
int [,]l, int [,]r)
{
l[s[0] - 'a',0] = 1;
// Precompute the
// prefix 2D array
for (int i = 1; i < n; i++)
{
for (int j = 0; j < MAX_CHAR; j++)
l[j, i] += l[j,i - 1];
l[s[i] - 'a',i]++;
}
r[s[n - 1] - 'a',n - 1] = 1;
// Precompute the Suffix 2D array.
for (int i = n - 2; i >= 0; i--)
{
for (int j = 0; j < MAX_CHAR; j++)
r[j, i] += r[j,i + 1];
r[s[i] - 'a',i]++;
}
}
// Find the number of palindromic
// subsequence of length k
static int countPalindromes(int k, int n,
int [,]l, int [,]r)
{
int ans = 0;
// If k is 1.
if (k == 1)
{
for (int i = 0; i < MAX_CHAR; i++)
ans += l[i,n - 1];
return ans;
}
// If k is 2
if (k == 2) {
// Adding all the products
// of prefix array
for (int i = 0; i < MAX_CHAR; i++)
ans += ((l[i,n - 1] *
(l[i,n - 1] - 1)) / 2);
return ans;
}
// For k greater than 2.
// Adding all the products
// of value of prefix and
// suffix array.
for (int i = 1; i < n - 1; i++)
for (int j = 0; j < MAX_CHAR; j++)
ans += l[j,i - 1] * r[j, i + 1];
return ans;
}
// Driver code
public static void Main ()
{
string s = "aabab";
int k = 2;
int n = s.Length;
int [,]l=new int[MAX_CHAR,MAX];
int [,]r=new int[MAX_CHAR,MAX];
precompute(s, n, l, r);
Console.Write(countPalindromes(k, n, l, r));
}
}
// This code is contributed by Nitin Mittal.PHP
= 0; $i--)
{
for ($j = 0; $j < $MAX_CHAR; $j++)
$r[$j][$i] += $r[$j][$i + 1];
$r[ord($s[$i]) - ord('a')][$i]++;
}
}
// Find the number of palindromic
// subsequence of length k
function countPalindromes($k, $n, &$l, &$r)
{
global $MAX, $MAX_CHAR;
$ans = 0;
// If k is 1.
if ($k == 1)
{
for ($i = 0; $i < $MAX_CHAR; $i++)
$ans += $l[$i][$n - 1];
return $ans;
}
// If k is 2
if ($k == 2)
{
// Adding all the products of
// prefix array
for ($i = 0; $i < $MAX_CHAR; $i++)
$ans += (($l[$i][$n - 1] *
($l[$i][$n - 1] - 1)) / 2);
return $ans;
}
// For k greater than 2. Adding all
// the products of value of prefix
// and suffix array.
for ($i = 1; $i < $n - 1; $i++)
for ($j = 0; $j < $MAX_CHAR; $j++)
$ans += $l[$j][$i - 1] *
$r[$j][$i + 1];
return $ans;
}
// Driver Code
$s = "aabab";
$k = 2;
$n = strlen($s);
$l = array_fill(0, $MAX_CHAR,
array_fill(0, $MAX, NULL));
$r = array_fill(0, $MAX_CHAR,
array_fill(0, $MAX, NULL));
precompute($s, $n, $l, $r);
echo countPalindromes($k, $n, $l, $r) . "\n";
// This code is contributed by ita_c
?>输出：
4