Create Find all distinct palindromic substrings of a given string.cpp

Author: MohammedHamzaMalik

Competitive-Programming/String II/Find all distinct palindromic substrings of a given string.cpp

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+/*
+Find all distinct palindromic substrings of a given string
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+Ninja did not do homework. As a penalty, the teacher has given a string ‘S’ to ninja and asked him to print all distinct palindromic substrings of the given string. A string is said to be palindrome if the reverse of the string is the same as the string itself. For example, the string “bccb” is a palindrome, but the string “def” is not a palindrome.
+Input Format:
+The first line of input contains an integer ‘T’, denoting the number of test cases.
+The first and only line of each test case contains the string ‘S’. 
+Output Format:
+For each test case, print all distinct palindromic substrings of the given string. Print the substrings in sorted manner and they should be space-separated.    
+
+Print the output of each test case in a separate line.
+Constraints:
+1 <= T <= 10
+1 <= |S| <= 1000
+
+String S contains lowercase English letters only.
+
+Where ‘T’ represents the number of test cases and ‘S’ represents the given string.
+
+Time limit: 1 sec
+Sample Input 1:
+2
+aba
+aabb
+Sample Output 1:
+a aba b
+a aa b bb
+Explanation of sample input 1:
+For the first test case, 
+All the possible substrings are [ ‘a’, ‘b’, ‘ab’, ‘ba’, ‘aba’ ] out of which [ ‘a’, ‘aba', 'b’ ] are 
+palindromic substrings.
+
+For the second test case,
+All the possible substrings are [ ‘a’, ‘b’, ‘aa’, ‘ab’, ‘bb’, ‘aab’, ‘abb’ , ‘aabb’ ] out of which [ 
+‘a’, ‘b’, ‘aa’, ‘bb’ ] are palindromic substrings.
+Sample Input 2:
+2
+babbb
+abcdc
+Sample Output 2:
+a b bab bb bbb 
+a b c cdc d
+*/
+
+
+
+#include<bits/stdc++.h>
+
+vector<string>distinctPalindrome(string &s)
+{
+    int n = s.size();
+    
+    vector<vector<bool> > dp(n, vector<bool>(n, false));
+    for (int i = 0; i < n; i++)
+    {    
+        dp[i][i] = 1;
+        
+        if (i < n && s[i] == s[i + 1]) {
+            dp[i][i + 1] = 1;
+        }
+    }
+    
+    for (int len = 3; len <= n; len++)
+    {
+        for (int i = 0; i + len - 1 < n; i++)
+        {
+            if (s[i] == s[i + (len - 1)] && dp[i + 1][i + (len - 1) - 1])
+            {
+                dp[i][i + (len - 1)] = true;
+            }
+        }
+    }
+ 
+    vector<int> kmp(n, 0);
+    
+    for (int i = 0; i < n; i++)
+    {
+        int j = 0, k = 1;
+        while (k + i < n)
+        {
+            if (s[j + i] == s[k + i])
+            {
+                dp[k + i - j][k + i] = false;
+                kmp[k++] = ++j;
+            }
+            
+            else if (j > 0)
+            {
+                j = kmp[j - 1];
+            }
+            
+            else
+            {
+                kmp[k++] = 0;
+            }
+        }
+    }
+    
+    vector<string>ans;
+    
+    for (int i = 0; i < n; i++)
+    {
+        string str;
+        for (int j = i; j < n; j++)
+        {
+            str += s[j];
+            if (dp[i][j])
+            {
+                ans.push_back(str);
+            }
+        }
+    }
+    
+    sort(ans.begin(),ans.end());
+    
+    return ans;
+}