Given a string of size N and some queries, the task is to find the lexicographically smallest palindromic subsequence of even length in range [L, R] for each query. If no such palindromic subsequence exists then the print -1.
Examples:
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
Approach: The main observation of this problem is if there exists a palindromic subsequence, then it must be of length > 2. Therefore the resultant subsequence will be a string of length > 2 with the same characters. Choose the smallest character among those characters which have a frequency greater than 1 in the range [L, R] and print that character twice. If there is no such character exists then print -1.
Below is the implementation of above approach:
// C++ program to find lexicographically smallest // palindromic subsequence of even length #include <bits/stdc++.h> 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 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 PrinciRaj1992 |
# 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 chitranayal |
// 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 |
<script> // Javascript program to find lexicographically smallest // palindromic subsequence of even length var N = 100001;
// Frequency array for each character var f = Array.from(Array(26), ()=> Array(N).fill(0));
// Preprocess the frequency array calculation function precompute(s, n)
{ // Frequency array to track each character
// in position 'i'
for ( var i = 0; i < n; i++) {
f[s[i].charCodeAt(0) - 'a' .charCodeAt(0)][i]++;
}
// Calculating prefix sum
// over this frequency array
// to get frequency of a character
// in a range [L, R].
for ( var i = 0; i < 26; i++) {
for ( var j = 1; j < n; j++) {
f[i][j] += f[i][j - 1];
}
}
} // Util function for palindromic subsequences function palindromicSubsequencesUtil(L, R)
{ var c, ok = 0;
// Find frequency of all characters
for ( var i = 0; i < 26; i++) {
// For each character
// find it's frequency
// in range [L, R]
var 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 function palindromicSubsequences(Q, l)
{ for ( var i = 0; i < l; i++) {
// Find in the palindromic subsequences
var x
= palindromicSubsequencesUtil(
Q[i][0], Q[i][1]);
// No such subsequence exists
if (x == -1) {
document.write( -1 + "<br>" );
}
else {
var c = String.fromCharCode('a'.charCodeAt(0) + x);
document.write( c + c + "<br>" );
}
}
} // Driver Code var str = "dbdeke" ;
var Q = [ [ 0, 5 ],
[ 1, 5 ],
[ 1, 3 ] ];
var n = str.length;
var l = Q.length;
// Function calls precompute(str, n); palindromicSubsequences(Q, l); // This code is contributed by itsok. </script> |
dd ee -1
Time Complexity: O(26 * N + 26 * Q), where N is the length of string
Auxiliary Space: O(26*N)