Skip to content
Related Articles

Related Articles

Improve Article

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

  • Difficulty Level : Medium
  • Last Updated : 25 May, 2021

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++




// 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




// 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




# 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#




// 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

Javascript




<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>
Output: 
dd
ee
-1

 

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :