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

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: 5
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 lenth-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++

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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 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;
}

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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 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
?>

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Output:

36


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Striver(underscore)79 at Codechef and codeforces D

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Improved By : Ryuga