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++
// 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
// 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
# 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// 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 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 */ |
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
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.
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