Create Count Special Palindromic Substrings.cpp

Author: MohammedHamzaMalik

Competitive-Programming/String II/Count Special Palindromic Substrings.cpp

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+/*
+Count Special Palindromic Substrings
+Send Feedback
+You are given a string 'STR'. Your task is to count the number of special palindromic substrings of size greater than 1 that the given string contains. A substring is said to be a special palindrome in the following two cases:
+If all the characters in the substring are the same.
+
+If the length of the substring is odd and only the middle element is different, while all the other characters are the same. 
+Example:
+“aba” is a special palindrome, while “abaa” is not
+Input Format:
+The first line contains an integer ‘T’, which denotes the number of test cases or queries to be run. Then the test cases are as follows.
+
+The first and the only line of each test case contains the string 'STR'.
+Output Format:
+For each test case, print the count of special palindromic substrings.
+
+Print the output of each test case in a separate line.
+Note:
+You don’t need to print anything; It has already been taken care of. Just implement the given function.
+Constraints:
+1 <= T <= 100
+1 <= |STR| <= 10000
+
+Time limit: 1 sec
+Sample Input 1:
+2
+bcbc
+ccddd
+Sample Output 1:
+2
+4
+Explanation For Sample Input 1:
+In the first test case, 
+The special palindromic substrings in the given string are: bcb and cbc. Hence, the answer is 2 in this case.
+
+In the second test case, 
+The special palindromic substrings in the string are: cc, dd, ddd and dd. Hence, the answer is 4 in this case. 
+Sample Input 2:
+2
+abccdcdf
+baabaab
+Sample Output 2:
+3
+4
+*/
+
+
+
+/*
+    Time complexity: O(N) 
+    Space complexity: O(N)
+	
+    Where N is the size of the string.
+*/
+
+
+int specialPalindromes(string &str)
+{
+
+    // Initialise ans to store count of special palindromes
+    int ans = 0;
+
+    // Initialise variable to store length of string
+    int n = str.length();
+
+    // Create an array with all values 0 to store count of continuous same characters
+    vector<int> countSame(n, 0);
+
+    // First pointer
+    int a = 0;
+
+    // Start traversing through the string
+    while(a < n)
+    {
+
+        // Initialise a variable to store temporary count of continuous same characters
+        int countContinuous = 1;
+
+        // Initialise a variable to store next character
+        int b = a + 1;
+
+        // Count same characters
+        while(b < n and str[a] == str[b] && (b < n))
+        {
+            countContinuous++;
+            b++;
+        }
+            
+        // CASE: 1
+        // Count total substrings from the count of  continuous same characters
+        ans = ans + (countContinuous * (countContinuous - 1) / 2);
+
+        // Store this count in the array 
+        countSame[a] = countContinuous;
+
+        a = b;
+    }
+
+    // CASE: 2
+    for(int i = 1; i < n; i++)
+    {
+
+        // If current character is equal to previous character
+        if(str[i] == str[i - 1])
+        {
+
+            // Update values of array
+            countSame[i] = countSame[i - 1];
+        }
+
+        if(i > 0 && i < (n - 1) && (str[i - 1] == str[i + 1]) && (str[i] != str[i + 1]))
+        {
+            ans += min(countSame[i - 1], countSame[i + 1]);
+        }
+    }
+
+    // Return count
+    return ans;
+}