给定大小为N的字符串和一些查询，任务是为每个查询找到偶数长度在[L，R]范围内的偶数长度最小的词典学子序列。如果不存在这种回文子序列，则打印-1 。
例子：
Input: str = “dbdeke”, query[][] = {{0, 5}, {1, 5}, {1, 3}}
Output: dd
ee
-1
Explanation: dd
In the first query, possible palindromic subsequences are “dd”, “ee”, “ddee” and “dd” which are lexicographically smallest.
In the second query, only possible palindromic subsequence is “ee”.
In the third query, no such palindromic subsequence is possible.
Input: str = “abcd”, query[][] = {{0, 3}}
Output: -1
方法：对该问题的主要观察是，如果存在回文序列，则其长度必须大于2 。因此所得到的子序列将是长度的字符串> 2以相同字符。选择那些具有大于1的频率更大的范围在[L，R]和打印字符两次字符中的最小字符。如果不存在这样的字符，则打印-1 。
下面是上述方法的实现：
C++
// C++ program to find lexicographically smallest
// palindromic subsequence of even length
#include
using namespace std;
const int N = 100001;
// Frequency array for each character
int f[26][N];
// Preprocess the frequency array calculation
void precompute(string s, int n)
{
// Frequency array to track each character
// in position 'i'
for (int i = 0; i < n; i++) {
f[s[i] - 'a'][i]++;
}
// Calculating prefix sum
// over this frequency array
// to get frequency of a character
// in a range [L, R].
for (int i = 0; i < 26; i++) {
for (int j = 1; j < n; j++) {
f[i][j] += f[i][j - 1];
}
}
}
// Util function for palindromic subsequences
int palindromicSubsequencesUtil(int L, int R)
{
int c, ok = 0;
// Find frequency of all characters
for (int i = 0; i < 26; i++) {
// For each character
// find it's frequency
// in range [L, R]
int cnt = f[i][R];
if (L > 0)
cnt -= f[i][L - 1];
if (cnt > 1) {
// If frequency in this range is > 1,
// then we must take this character,
// as it will give
// lexicographically smallest one
ok = 1;
c = i;
break;
}
}
// There is no character
// in range [L, R] such
// that it's frequency is > 1.
if (ok == 0) {
return -1;
}
// Return the character's value
return c;
}
// Function to find lexicographically smallest
// palindromic subsequence of even length
void palindromicSubsequences(int Q[][2], int l)
{
for (int i = 0; i < l; i++) {
// Find in the palindromic subsequences
int x
= palindromicSubsequencesUtil(
Q[i][0], Q[i][1]);
// No such subsequence exists
if (x == -1) {
cout << -1 << "\n";
}
else {
char c = 'a' + x;
cout << c << c << "\n";
}
}
}
// Driver Code
int main()
{
string str = "dbdeke";
int Q[][2] = { { 0, 5 },
{ 1, 5 },
{ 1, 3 } };
int n = str.size();
int l = sizeof(Q) / sizeof(Q[0]);
// Function calls
precompute(str, n);
palindromicSubsequences(Q, l);
return 0;
} Java
// Java program to find lexicographically smallest
// palindromic subsequence of even length
import java.util.*;
class GFG{
static int N = 100001;
// Frequency array for each character
static int [][]f = new int[26][N];
// Preprocess the frequency array calculation
static void precompute(String s, int n)
{
// Frequency array to track each character
// in position 'i'
for (int i = 0; i < n; i++)
{
f[s.charAt(i) - 'a'][i]++;
}
// Calculating prefix sum
// over this frequency array
// to get frequency of a character
// in a range [L, R].
for (int i = 0; i < 26; i++)
{
for (int j = 1; j < n; j++)
{
f[i][j] += f[i][j - 1];
}
}
}
// Util function for palindromic subsequences
static int palindromicSubsequencesUtil(int L, int R)
{
int c = 0, ok = 0;
// Find frequency of all characters
for (int i = 0; i < 26; i++)
{
// For each character
// find it's frequency
// in range [L, R]
int cnt = f[i][R];
if (L > 0)
cnt -= f[i][L - 1];
if (cnt > 1)
{
// If frequency in this range is > 1,
// then we must take this character,
// as it will give
// lexicographically smallest one
ok = 1;
c = i;
break;
}
}
// There is no character
// in range [L, R] such
// that it's frequency is > 1.
if (ok == 0)
{
return -1;
}
// Return the character's value
return c;
}
// Function to find lexicographically smallest
// palindromic subsequence of even length
static void palindromicSubsequences(int Q[][], int l)
{
for (int i = 0; i < l; i++)
{
// Find in the palindromic subsequences
int x = palindromicSubsequencesUtil(
Q[i][0], Q[i][1]);
// No such subsequence exists
if (x == -1)
{
System.out.print(-1 + "\n");
}
else
{
char c = (char) ('a' + x);
System.out.print((char) c + "" +
(char) c + "\n");
}
}
}
// Driver Code
public static void main(String[] args)
{
String str = "dbdeke";
int Q[][] = { { 0, 5 },
{ 1, 5 },
{ 1, 3 } };
int n = str.length();
int l = Q.length;
// Function calls
precompute(str, n);
palindromicSubsequences(Q, l);
}
}
// This code is contributed by PrinciRaj1992Python3
# Python3 program to find lexicographically
# smallest palindromic subsequence of even length
N = 100001
# Frequency array for each character
f = [[ 0 for x in range (N)]
for y in range (26)]
# Preprocess the frequency array calculation
def precompute(s, n):
# Frequency array to track each character
# in position 'i'
for i in range(n):
f[ord(s[i]) - ord('a')][i] += 1
# Calculating prefix sum
# over this frequency array
# to get frequency of a character
# in a range [L, R].
for i in range(26):
for j in range(1, n):
f[i][j] += f[i][j - 1]
# Util function for palindromic subsequences
def palindromicSubsequencesUtil(L, R):
ok = 0
# Find frequency of all characters
for i in range(26):
# For each character
# find it's frequency
# in range [L, R]
cnt = f[i][R]
if (L > 0):
cnt -= f[i][L - 1]
if (cnt > 1):
# If frequency in this range is > 1,
# then we must take this character,
# as it will give
# lexicographically smallest one
ok = 1
c = i
break
# There is no character
# in range [L, R] such
# that it's frequency is > 1.
if (ok == 0):
return -1
# Return the character's value
return c
# Function to find lexicographically smallest
# palindromic subsequence of even length
def palindromicSubsequences(Q, l):
for i in range(l):
# Find in the palindromic subsequences
x = palindromicSubsequencesUtil(Q[i][0],
Q[i][1])
# No such subsequence exists
if (x == -1):
print(-1)
else :
c = ord('a') + x
print(2 * chr(c))
# Driver Code
if __name__ == "__main__":
st = "dbdeke"
Q = [ [ 0, 5 ],
[ 1, 5 ],
[ 1, 3 ] ]
n = len(st)
l = len(Q)
# Function calls
precompute(st, n)
palindromicSubsequences(Q, l)
# This code is contributed by chitranayalC#
// C# program to find lexicographically smallest
// palindromic subsequence of even length
using System;
class GFG{
static int N = 100001;
// Frequency array for each character
static int [,]f = new int[26, N];
// Preprocess the frequency array calculation
static void precompute(String s, int n)
{
// Frequency array to track each character
// in position 'i'
for(int i = 0; i < n; i++)
{
f[s[i] - 'a', i]++;
}
// Calculating prefix sum
// over this frequency array
// to get frequency of a character
// in a range [L, R].
for(int i = 0; i < 26; i++)
{
for(int j = 1; j < n; j++)
{
f[i, j] += f[i, j - 1];
}
}
}
// Util function for palindromic subsequences
static int palindromicSubsequencesUtil(int L, int R)
{
int c = 0, ok = 0;
// Find frequency of all characters
for(int i = 0; i < 26; i++)
{
// For each character
// find it's frequency
// in range [L, R]
int cnt = f[i, R];
if (L > 0)
cnt -= f[i, L - 1];
if (cnt > 1)
{
// If frequency in this range is > 1,
// then we must take this character,
// as it will give
// lexicographically smallest one
ok = 1;
c = i;
break;
}
}
// There is no character
// in range [L, R] such
// that it's frequency is > 1.
if (ok == 0)
{
return -1;
}
// Return the character's value
return c;
}
// Function to find lexicographically smallest
// palindromic subsequence of even length
static void palindromicSubsequences(int [,]Q, int l)
{
for(int i = 0; i < l; i++)
{
// Find in the palindromic subsequences
int x = palindromicSubsequencesUtil(Q[i, 0],
Q[i, 1]);
// No such subsequence exists
if (x == -1)
{
Console.Write(-1 + "\n");
}
else
{
char c = (char)('a' + x);
Console.Write((char) c + "" +
(char) c + "\n");
}
}
}
// Driver Code
public static void Main(String[] args)
{
String str = "dbdeke";
int [,]Q = { { 0, 5 },
{ 1, 5 },
{ 1, 3 } };
int n = str.Length;
int l = Q.GetLength(0);
// Function calls
precompute(str, n);
palindromicSubsequences(Q, l);
}
}
// This code is contributed by amal kumar choubey输出：
dd
ee
-1
时间复杂度： O(26 * N + 26 * Q)，其中N是字符串的长度