Queries to check if substring[L…R] is palindrome or not

Given a string str and Q queries. Every query consists of two numbers L and R. The task is to print if the sub-string[L…R] is palindrome or not.

Examples:

Input: str = “abacccde”, Q[][] = {{0, 2}, {1, 2}, {2, 4}, {3, 5}}
Output:
Yes
No
No
Yes

Input: str = “abaaab”, Q[][] = {{0, 1}, {1, 5}}
Output:
No
Yes

Simple Approach: A naive approach is to check for every sub-string if it is palindrome or not. In the worst case, every query can take up to O(Q).

Efficient Approach: An efficient approach is to use Dynamic Programming to solve the above problem. We can initially create the DP table which stores if substring[i….j] is palindrome or not. We maintain a boolean dp[n][n] that is filled in a bottom-up manner. The value of dp[i][j] is true if the substring is a palindrome, otherwise false. To calculate dp[i][j], we first check the value of dp[i+1][j-1], if the value is true and s[i] is same as s[j], then we make dp[i][j] true. Otherwise, the value of dp[i][j] is made false.

Now for every query, check if dp[l][r] is true or not.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define N 100
  
// Pre-processing function
void pre_process(bool dp[N][N], string s)
{
  
    // Get the size of the string
    int n = s.size();
  
    // Initially mark every
    // position as false
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++)
            dp[i][j] = false;
    }
  
    // For the length
    for (int j = 1; j <= n; j++) {
  
        // Iterate for every index with
        // length j
        for (int i = 0; i <= n - j; i++) {
  
            // If the length is less than 2
            if (j <= 2) {
  
                // If characters are equal
                if (s[i] == s[i + j - 1])
                    dp[i][i + j - 1] = true;
            }
  
            // Check for equal
            else if (s[i] == s[i + j - 1])
                dp[i][i + j - 1] = dp[i + 1][i + j - 2];
        }
    }
}
  
// Function to answer every query in O(1)
void answerQuery(int l, int r, bool dp[N][N])
{
    if (dp[l][r])
        cout << "Yes\n";
    else
        cout << "No\n";
}
  
// Driver code
int main()
{
    string s = "abaaab";
    bool dp[N][N];
    pre_process(dp, s);
  
    int queries[][2] = { { 0, 1 }, { 1, 5 } };
    int q = sizeof(queries) / sizeof(queries[0]);
  
    for (int i = 0; i < q; i++)
        answerQuery(queries[i][0], queries[i][1], dp);
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG {
  
    static int N = 100;
  
    // Pre-processing function
    static void pre_process(boolean dp[][], char[] s)
    {
  
        // Get the size of the string
        int n = s.length;
  
        // Initially mark every
        // position as false
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                dp[i][j] = false;
            }
        }
  
        // For the length
        for (int j = 1; j <= n; j++) {
  
            // Iterate for every index with
            // length j
            for (int i = 0; i <= n - j; i++) {
  
                // If the length is less than 2
                if (j <= 2) {
  
                    // If characters are equal
                    if (s[i] == s[i + j - 1]) {
                        dp[i][i + j - 1] = true;
                    }
                }
  
                // Check for equal
                else if (s[i] == s[i + j - 1]) {
                    dp[i][i + j - 1] = dp[i + 1][i + j - 2];
                }
            }
        }
    }
  
    // Function to answer every query in O(1)
    static void answerQuery(int l, int r, boolean dp[][])
    {
        if (dp[l][r]) {
            System.out.println("Yes");
        }
        else {
            System.out.println("No");
        }
    }
  
    // Driver code
    public static void main(String[] args)
    {
        String s = "abaaab";
        boolean[][] dp = new boolean[N][N];
        pre_process(dp, s.toCharArray());
  
        int queries[][] = { { 0, 1 }, { 1, 5 } };
        int q = queries.length;
  
        for (int i = 0; i < q; i++) {
            answerQuery(queries[i][0], queries[i][1], dp);
        }
    }
}
  
// This code contributed by Rajput-Ji

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Python3

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# Python3 implementation of the approach
N = 100
  
# Pre-processing function
def pre_process(dp, s):
  
    # Get the size of the string
    n = len(s)
  
    # Initially mark every
    # position as false
    for i in range(n):
        for j in range(n):
            dp[i][j] = False
  
    # For the length
    for j in range(1, n + 1):
  
        # Iterate for every index with
        # length j
        for i in range(n - j + 1):
  
            # If the length is less than 2
            if (j <= 2):
  
                # If characters are equal
                if (s[i] == s[i + j - 1]):
                    dp[i][i + j - 1] = True
  
            # Check for equal
            elif (s[i] == s[i + j - 1]):
                dp[i][i + j - 1] = dp[i + 1][i + j - 2]
  
# Function to answer every query in O(1)
def answerQuery(l, r, dp):
  
    if (dp[l][r]):
        print("Yes")
    else:
        print("No")
  
# Driver code
s = "abaaab"
dp = [[0 for i in range(N)]
         for i in range(N)]
pre_process(dp, s)
  
queries = [[0, 1], [1, 5]]
q = len(queries)
  
for i in range(q):
    answerQuery(queries[i][0], 
                queries[i][1], dp)
  
# This code is contributed by mohit kumar

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C#

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// C# implementation of the approach
using System;
class GFG {
  
    static int N = 100;
  
    // Pre-processing function
    static void pre_process(bool[, ] dp, char[] s)
    {
  
        // Get the size of the string
        int n = s.Length;
  
        // Initially mark every
        // position as false
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                dp[i, j] = false;
            }
        }
  
        // For the length
        for (int j = 1; j <= n; j++) {
  
            // Iterate for every index with
            // length j
            for (int i = 0; i <= n - j; i++) {
  
                // If the length is less than 2
                if (j <= 2) {
  
                    // If characters are equal
                    if (s[i] == s[i + j - 1]) {
                        dp[i, i + j - 1] = true;
                    }
                }
  
                // Check for equal
                else if (s[i] == s[i + j - 1]) {
                    dp[i, i + j - 1] = dp[i + 1, i + j - 2];
                }
            }
        }
    }
  
    // Function to answer every query in O(1)
    static void answerQuery(int l, int r, bool[, ] dp)
    {
        if (dp[l, r]) {
            Console.WriteLine("Yes");
        }
        else {
            Console.WriteLine("No");
        }
    }
  
    // Driver code
    public static void Main()
    {
        string s = "abaaab";
        bool[, ] dp = new bool[N, N];
        pre_process(dp, s.ToCharArray());
  
        int[, ] queries = { { 0, 1 }, { 1, 5 } };
        int q = queries.Length;
  
        for (int i = 0; i < q; i++) {
            answerQuery(queries[i, 0], queries[i, 1], dp);
        }
    }
}
  
// This code is contributed by Ankit.

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PHP

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<?php
// PHP implementation of the approach 
$N = 100;
  
// Pre-processing function 
function pre_process($dp, $s
  
    // Get the size of the string 
    $n = strlen($s); 
  
    // Initially mark every 
    // position as false 
    for ($i = 0; $i < $n; $i++) 
    
        for ($j = 0; $j < $n; $j++) 
            $dp[$i][$j] = false; 
    
  
    // For the length 
    for ($j = 1; $j <= $n; $j++)
    
          
        // Iterate for every index with 
        // length j 
        for ($i = 0; $i <= $n - $j; $i++) 
        
  
            // If the length is less than 2 
            if ($j <= 2) 
            
  
                // If characters are equal 
                if ($s[$i] == $s[$i + $j - 1]) 
                    $dp[$i][$i + $j - 1] = true; 
            
  
            // Check for equal 
            else if ($s[$i] == $s[$i + $j - 1]) 
                $dp[$i][$i + $j - 1] = $dp[$i + 1][$i + $j - 2]; 
        
    
    return $dp;
  
// Function to answer every query in O(1) 
function answerQuery($l, $r, $dp
    if ($dp[$l][$r]) 
        echo "Yes\n"
    else
        echo "No\n"
  
// Driver code 
$s = "abaaab"
$dp = array(array());
$dp = pre_process($dp, $s); 
  
$queries = array(array(0, 1), 
                 array(1, 5)); 
$q = count($queries); 
  
for ($i = 0; $i < $q; $i++) 
    answerQuery($queries[$i][0], 
                $queries[$i][1], $dp); 
  
// This code is contributed by Ryuga
?>

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Output:

No
Yes


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Striver(underscore)79 at Codechef and codeforces D

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