Smallest Palindromic Subsequence of Even Length in Range [L, R]

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++

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// 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;
}

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Java

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// 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

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Output:

dd
ee
-1

Time Complexity: O(26 * N + 26 * Q), where N is the length of string

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