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

21 Jun, 2022
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 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++
Java
Python3
PHP
C#
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define N 100 // Pre-processing functionvoid 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 pairsint 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 codeint main(){    string s = "abacaba";    cout << countPairs(s);     return 0;}
// Java implementation of the approachclass 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 implementation of the approachN = 100 # Pre-processing functiondef 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 pairsdef 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 codes = "abacaba"print(countPairs(s)) # This code is contributed by mohit kumar
<?php// PHP implementation of the approach$N = 100; // Pre-processing functionfunction pre_process($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 $dp;} // Function to return the number of pairsfunction countPairs($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# implementation of the approachusing 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 */
<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>
36
 
Time Complexity: O(N*N), as we are using nested loops for traversing N*N times. Where N is the length of the string.
Auxiliary Space: O(N*N), as we are using extra space for the dp matrix.
ankthon
mohit kumar 29
Rajput-Ji
princiraj1992
ManasChhabra2
todaysgaurav
rohitsingh57
palindrome
substring
Combinatorial
Dynamic Programming
Strings
Strings
Dynamic Programming
Combinatorial
palindrome
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