Skip to content
Related Articles

Related Articles

Improve Article

Count pairs of non-overlapping palindromic sub-strings of the given string

  • Difficulty Level : Hard
  • Last Updated : 21 May, 2021

Given a string S. The task is to count the non-overlapping pairs of palindromic sub-strings S1 and S2 such that the strings should be S1[L1…R1] and S2[L2…R2] where 0 ≤ L1 ≤ R1 < L2 ≤ R2 < N. The task is to count the number of pairs of the non-overlapping palindromic sub-strings. 
Examples: 
 

Input: s = “aaa” 
Output:
All possible pairs are (s[0], s[1]), (s[0], s[2]), 
(s[0], s[1, 2]), (s[1], s[2]) and (s[0, 1], s[2])
Input: s = “abacaba” 
Output: 36 
 

 

Approach: We can 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. The following steps can be followed thereafter to get the number of pairs. 
 

  • Create a left[] array, where left[i] stores the count of the number of palindromes to the left on the index i including i.
  • Create a right[] array, where right[i] stores the count of the number of palindromes to the right on the index i including i.
  • Iterate from 0 to length-1 and add left[i]*right[i+1]. The summation of it for every index will be the required number of pairs.

Below is the implementation of the above approach: 
 



C++




// 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 return the number of pairs
int countPairs(string s)
{
 
    // Create the dp table initially
    bool dp[N][N];
    pre_process(dp, s);
    int n = s.length();
 
    // Declare the left array
    int left[n];
    memset(left, 0, sizeof left);
 
    // Declare the right array
    int right[n];
    memset(right, 0, sizeof right);
 
    // Initially left[0] is 1
    left[0] = 1;
 
    // Count the number of palindrome
    // pairs to the left
    for (int i = 1; i < n; i++) {
 
        for (int j = 0; j <= i; j++) {
 
            if (dp[j][i] == 1)
                left[i]++;
        }
    }
 
    // Initially right most as 1
    right[n - 1] = 1;
 
    // Count the number of palindrome
    // pairs to the right
    for (int i = n - 2; i >= 0; i--) {
 
        right[i] = right[i + 1];
 
        for (int j = n - 1; j >= i; j--) {
 
            if (dp[i][j] == 1)
                right[i]++;
        }
    }
 
    int ans = 0;
 
    // Count the number of pairs
    for (int i = 0; i < n - 1; i++)
        ans += left[i] * right[i + 1];
 
    return ans;
}
 
// Driver code
int main()
{
    string s = "abacaba";
    cout << countPairs(s);
 
    return 0;
}

Java




// 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 return the number of pairs
    static int countPairs(String s)
    {
 
        // Create the dp table initially
        boolean dp[][] = new boolean[N][N];
        pre_process(dp, s.toCharArray());
        int n = s.length();
 
        // Declare the left array
        int left[] = new int[n];
 
        // Declare the right array
        int right[] = new int[n];
 
        // Initially left[0] is 1
        left[0] = 1;
 
        // Count the number of palindrome
        // pairs to the left
        for (int i = 1; i < n; i++)
        {
 
            for (int j = 0; j <= i; j++)
            {
 
                if (dp[j][i] == true)
                {
                    left[i]++;
                }
            }
        }
 
        // Initially right most as 1
        right[n - 1] = 1;
 
        // Count the number of palindrome
        // pairs to the right
        for (int i = n - 2; i >= 0; i--)
        {
 
            right[i] = right[i + 1];
 
            for (int j = n - 1; j >= i; j--)
            {
 
                if (dp[i][j] == true)
                {
                    right[i]++;
                }
            }
        }
 
        int ans = 0;
 
        // Count the number of pairs
        for (int i = 0; i < n - 1; i++)
        {
            ans += left[i] * right[i + 1];
        }
 
        return ans;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String s = "abacaba";
        System.out.println(countPairs(s));
    }
}
 
// This code contributed by Rajput-Ji

Python3




# 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 return the number of pairs
def countPairs(s):
 
    # Create the dp table initially
    dp = [[False for i in range(N)]
                 for j in range(N)]
    pre_process(dp, s)
    n = len(s)
 
    # Declare the left array
    left = [0 for i in range(n)]
 
    # Declare the right array
    right = [0 for i in range(n)]
 
    # Initially left[0] is 1
    left[0] = 1
 
    # Count the number of palindrome
    # pairs to the left
    for i in range(1, n):
 
        for j in range(i + 1):
 
            if (dp[j][i] == 1):
                left[i] += 1
 
    # Initially right most as 1
    right[n - 1] = 1
 
    # Count the number of palindrome
    # pairs to the right
    for i in range(n - 2, -1, -1):
 
        right[i] = right[i + 1]
 
        for j in range(n - 1, i - 1, -1):
 
            if (dp[i][j] == 1):
                right[i] += 1
 
    ans = 0
 
    # Count the number of pairs
    for i in range(n - 1):
        ans += left[i] * right[i + 1]
 
    return ans
 
# Driver code
s = "abacaba"
print(countPairs(s))
 
# This code is contributed by mohit kumar

PHP




<?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 return the number of pairs
function countPairs($s)
{
 
    // Create the dp table initially
    $dp = array(array());
    $dp = pre_process($dp, $s);
     
    $n = strlen($s);
 
    // Declare the left array
    $left = array_fill(0, $n, 0);
 
    // Declare the right array
    $right = array_fill(0, $n, 0);
 
    // Initially left[0] is 1
    $left[0] = 1;
 
    // Count the number of palindrome
    // pairs to the left
    for ($i = 1; $i < $n; $i++)
    {
        for ($j = 0; $j <= $i; $j++)
        {
            if ($dp[$j][$i] == 1)
                $left[$i]++;
        }
    }
 
    // Initially right most as 1
    $right[$n - 1] = 1;
 
    // Count the number of palindrome
    // pairs to the right
    for ($i = $n - 2; $i >= 0; $i--)
    {
        $right[$i] = $right[$i + 1];
 
        for ($j = $n - 1; $j >= $i; $j--)
        {
            if ($dp[$i][$j] == 1)
                $right[$i]++;
        }
    }
 
    $ans = 0;
 
    // Count the number of pairs
    for ($i = 0; $i < $n - 1; $i++)
        $ans += $left[$i] * $right[$i + 1];
 
    return $ans;
}
 
// Driver code
$s = "abacaba";
echo countPairs($s);
 
// This code is contributed by Ryuga
?>

C#




// 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 return the number of pairs
    static int countPairs(String s)
    {
 
        // Create the dp table initially
        bool [,]dp = new bool[N,N];
        pre_process(dp, s.ToCharArray());
        int n = s.Length;
 
        // Declare the left array
        int []left = new int[n];
 
        // Declare the right array
        int []right = new int[n];
 
        // Initially left[0] is 1
        left[0] = 1;
 
        // Count the number of palindrome
        // pairs to the left
        for (int i = 1; i < n; i++)
        {
 
            for (int j = 0; j <= i; j++)
            {
 
                if (dp[j,i] == true)
                {
                    left[i]++;
                }
            }
        }
 
        // Initially right most as 1
        right[n - 1] = 1;
 
        // Count the number of palindrome
        // pairs to the right
        for (int i = n - 2; i >= 0; i--)
        {
 
            right[i] = right[i + 1];
 
            for (int j = n - 1; j >= i; j--)
            {
 
                if (dp[i,j] == true)
                {
                    right[i]++;
                }
            }
        }
 
        int ans = 0;
 
        // Count the number of pairs
        for (int i = 0; i < n - 1; i++)
        {
            ans += left[i] * right[i + 1];
        }
 
        return ans;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        String s = "abacaba";
        Console.Write(countPairs(s));
    }
}
 
/* This code contributed by PrinciRaj1992 */

Javascript




<script>
// javascript implementation of the approach   
var N = 100;
 
    // Pre-processing function
    function pre_process( dp,  s) {
 
        // Get the size of the string
        var n = s.length;
 
        // 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];
                }
            }
        }
    }
 
    // Function to return the number of pairs
    function countPairs(s) {
 
        // Create the dp table initially
        var dp = Array(N).fill().map(()=>Array(N).fill(false));
        pre_process(dp, s);
        var n = s.length;
 
        // Declare the left array
        var left = Array(n).fill(0);
 
        // Declare the right array
        var right = Array(n).fill(0);
 
        // Initially left[0] is 1
        left[0] = 1;
 
        // Count the number of palindrome
        // pairs to the left
        for (i = 1; i < n; i++) {
 
            for (j = 0; j <= i; j++) {
 
                if (dp[j][i] == true) {
                    left[i]++;
                }
            }
        }
 
        // Initially right most as 1
        right[n - 1] = 1;
 
        // Count the number of palindrome
        // pairs to the right
        for (i = n - 2; i >= 0; i--) {
 
            right[i] = right[i + 1];
 
            for (j = n - 1; j >= i; j--) {
 
                if (dp[i][j] == true) {
                    right[i]++;
                }
            }
        }
 
        var ans = 0;
 
        // Count the number of pairs
        for (i = 0; i < n - 1; i++) {
            ans += left[i] * right[i + 1];
        }
 
        return ans;
    }
 
    // Driver code
     
        var s = "abacaba";
        document.write(countPairs(s));
 
// This code is contributed by todaysgaurav
</script>
Output: 
36

 

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 :