Total number of odd length palindrome sub-sequence around each center

Given a string str, the task is to find the number of odd length palindromic sub-sequences around of str with str[i] as center i.e. every index will be considered as the center one by one.

Examples:

Input: str = “xyzx”
Output: 1 2 2 1
For index 0: There is only a single sub-sequence possible i.e. “x”
For index 1: Two sub-sequences are possible i.e. “y” and “xyx”
For index 2: “z” and “xzx”
For index 3: “x”

Input: str = “aaaa”
Output: 1 3 3 1

Approach: We will use dynamic programming to solve this problem. Let’s denote length of the string str be N. Now, Let dp[i][j] denote the number of palindromic sub-sequences from 0 to i – 1 and number of palindromic sub-sequences from j to N – 1
Let len be the distance between i and j. For each length len, we will fix our i and j, and check whether characters str[i] and str[j] are equal or not. Then according to it, we will make our dp transitions.

If str[i] != str[j] then dp[i][j] = dp[i – 1][j] + dp[i][j + 1] – dp[i – 1][j + 1]
If str[i] == str[j] then dp[i][j] = dp[i – 1][j] + dp[i][j + 1]
Base case:
If i == 0 and j == n – 1 then dp[i][j] = 2 if str[i] == str[j] else dp[i][j] = 1

Below is the implementation of the above approach:

C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function to find the total palindromic` `// odd length sub-sequences` `void` `solve(string& s)` `{` `    ``int` `n = s.length();`   `    ``// dp array to store the number of palindromic` `    ``// subsequences for 0 to i-1 and j+1 to n-1` `    ``int` `dp[n][n];` `    ``memset``(dp, 0, ``sizeof` `dp);`   `    ``// We will start with the largest` `    ``// distance between i and j` `    ``for` `(``int` `len = n - 1; len >= 0; --len) {`   `        ``// For each len, we fix our i` `        ``for` `(``int` `i = 0; i + len < n; ++i) {`   `            ``// For this i we will find our j` `            ``int` `j = i + len;`   `            ``// Base cases` `            ``if` `(i == 0 and j == n - 1) {` `                ``if` `(s[i] == s[j])` `                    ``dp[i][j] = 2;` `                ``else` `if` `(s[i] != s[j])` `                    ``dp[i][j] = 1;` `            ``}` `            ``else` `{` `                ``if` `(s[i] == s[j]) {`   `                    ``// If the characters are equal` `                    ``// then look for out of bound index` `                    ``if` `(i - 1 >= 0) {` `                        ``dp[i][j] += dp[i - 1][j];` `                    ``}` `                    ``if` `(j + 1 <= n - 1) {` `                        ``dp[i][j] += dp[i][j + 1];` `                    ``}` `                    ``if` `(i - 1 < 0 or j + 1 >= n) {`   `                        ``// We have only 1 way that is to` `                        ``// just pick these characters` `                        ``dp[i][j] += 1;` `                    ``}` `                ``}` `                ``else` `if` `(s[i] != s[j]) {`   `                    ``// If the characters are not equal` `                    ``if` `(i - 1 >= 0) {` `                        ``dp[i][j] += dp[i - 1][j];` `                    ``}` `                    ``if` `(j + 1 <= n - 1) {` `                        ``dp[i][j] += dp[i][j + 1];` `                    ``}` `                    ``if` `(i - 1 >= 0 and j + 1 <= n - 1) {`   `                        ``// Subtract it as we have` `                        ``// counted it twice` `                        ``dp[i][j] -= dp[i - 1][j + 1];` `                    ``}` `                ``}` `            ``}` `        ``}` `    ``}` `    ``vector<``int``> ways;` `    ``for` `(``int` `i = 0; i < n; ++i) {` `        ``if` `(i == 0 or i == n - 1) {`   `            ``// We have just 1 palindrome` `            ``// sequence of length 1` `            ``ways.push_back(1);` `        ``}` `        ``else` `{`   `            ``// Else total ways would be sum of dp[i-1][i+1],` `            ``// that is number of palindrome sub-sequences` `            ``// from 1 to i-1 + number of palindrome` `            ``// sub-sequences from i+1 to n-1` `            ``int` `total = dp[i - 1][i + 1];` `            ``ways.push_back(total);` `        ``}` `    ``}` `    ``for` `(``int` `i = 0; i < ways.size(); ++i) {` `        ``cout << ways[i] << ``" "``;` `    ``}` `}`   `// Driver code` `int` `main()` `{` `    ``string s = ``"xyxyx"``;` `    ``solve(s);`   `    ``return` `0;` `}`

Java

 `// Java implementation of above approach` `import` `java.util.*;`   `class` `GFG ` `{`   `// Function to find the total palindromic` `// odd length sub-sequences` `static` `void` `solve(``char``[] s)` `{` `    ``int` `n = s.length;`   `    ``// dp array to store the number of palindromic` `    ``// subsequences for 0 to i-1 and j+1 to n-1` `    ``int` `[][]dp = ``new` `int``[n][n];`   `    ``// We will start with the largest` `    ``// distance between i and j` `    ``for` `(``int` `len = n - ``1``; len >= ``0``; --len) ` `    ``{`   `        ``// For each len, we fix our i` `        ``for` `(``int` `i = ``0``; i + len < n; ++i)` `        ``{`   `            ``// For this i we will find our j` `            ``int` `j = i + len;`   `            ``// Base cases` `            ``if` `(i == ``0` `&& j == n - ``1``) ` `            ``{` `                ``if` `(s[i] == s[j])` `                    ``dp[i][j] = ``2``;` `                ``else` `if` `(s[i] != s[j])` `                    ``dp[i][j] = ``1``;` `            ``}` `            ``else` `            ``{` `                ``if` `(s[i] == s[j]) ` `                ``{`   `                    ``// If the characters are equal` `                    ``// then look for out of bound index` `                    ``if` `(i - ``1` `>= ``0``) ` `                    ``{` `                        ``dp[i][j] += dp[i - ``1``][j];` `                    ``}` `                    ``if` `(j + ``1` `<= n - ``1``) ` `                    ``{` `                        ``dp[i][j] += dp[i][j + ``1``];` `                    ``}` `                    ``if` `(i - ``1` `< ``0` `|| j + ``1` `>= n) ` `                    ``{`   `                        ``// We have only 1 way that is to` `                        ``// just pick these characters` `                        ``dp[i][j] += ``1``;` `                    ``}` `                ``}` `                ``else` `if` `(s[i] != s[j])` `                ``{`   `                    ``// If the characters are not equal` `                    ``if` `(i - ``1` `>= ``0``) ` `                    ``{` `                        ``dp[i][j] += dp[i - ``1``][j];` `                    ``}` `                    ``if` `(j + ``1` `<= n - ``1``) ` `                    ``{` `                        ``dp[i][j] += dp[i][j + ``1``];` `                    ``}` `                    ``if` `(i - ``1` `>= ``0` `&& j + ``1` `<= n - ``1``) ` `                    ``{`   `                        ``// Subtract it as we have` `                        ``// counted it twice` `                        ``dp[i][j] -= dp[i - ``1``][j + ``1``];` `                    ``}` `                ``}` `            ``}` `        ``}` `    ``}` `    `  `    ``Vector ways = ``new` `Vector<>();` `    ``for` `(``int` `i = ``0``; i < n; ++i) ` `    ``{` `        ``if` `(i == ``0` `|| i == n - ``1``) ` `        ``{`   `            ``// We have just 1 palindrome` `            ``// sequence of length 1` `            ``ways.add(``1``);` `        ``}` `        ``else` `        ``{`   `            ``// Else total ways would be sum of dp[i-1][i+1],` `            ``// that is number of palindrome sub-sequences` `            ``// from 1 to i-1 + number of palindrome` `            ``// sub-sequences from i+1 to n-1` `            ``int` `total = dp[i - ``1``][i + ``1``];` `            ``ways.add(total);` `        ``}` `    ``}` `    ``for` `(``int` `i = ``0``; i < ways.size(); ++i) ` `    ``{` `        ``System.out.print(ways.get(i) + ``" "``);` `    ``}` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``char``[] s = ``"xyxyx"``.toCharArray();` `    ``solve(s);` `}` `}`   `// This code has been contributed by 29AjayKumar`

Python3

 `# Function to find the total palindromic` `# odd Length sub-sequences` `def` `solve(s):` `    ``n ``=` `len``(s)`   `    ``# dp array to store the number of palindromic` `    ``# subsequences for 0 to i-1 and j+1 to n-1` `    ``dp``=``[[``0` `for` `i ``in` `range``(n)] ``for` `i ``in` `range``(n)]`   `    ``# We will start with the largest` `    ``# distance between i and j` `    ``for` `Len` `in` `range``(n``-``1``,``-``1``,``-``1``):`   `        ``# For each Len, we fix our i` `        ``for` `i ``in` `range``(n):`   `            ``if` `i ``+` `Len` `>``=` `n:` `                ``break`   `            ``# For this i we will find our j` `            ``j ``=` `i ``+` `Len`   `            ``# Base cases` `            ``if` `(i ``=``=` `0` `and` `j ``=``=` `n ``-` `1``):` `                ``if` `(s[i] ``=``=` `s[j]):` `                    ``dp[i][j] ``=` `2` `                ``elif` `(s[i] !``=` `s[j]):` `                    ``dp[i][j] ``=` `1` `            ``else``:` `                ``if` `(s[i] ``=``=` `s[j]):` `                    ``# If the characters are equal` `                    ``# then look for out of bound index` `                    ``if` `(i ``-` `1` `>``=` `0``):` `                        ``dp[i][j] ``+``=` `dp[i ``-` `1``][j]`   `                    ``if` `(j ``+` `1` `<``=` `n ``-` `1``):` `                        ``dp[i][j] ``+``=` `dp[i][j ``+` `1``]`   `                    ``if` `(i ``-` `1` `< ``0` `or` `j ``+` `1` `>``=` `n):`   `                        ``# We have only 1 way that is to` `                        ``# just pick these characters` `                        ``dp[i][j] ``+``=` `1`   `                ``elif` `(s[i] !``=` `s[j]):`   `                    ``# If the characters are not equal` `                    ``if` `(i ``-` `1` `>``=` `0``):` `                        ``dp[i][j] ``+``=` `dp[i ``-` `1``][j]`   `                    ``if` `(j ``+` `1` `<``=` `n ``-` `1``):` `                        ``dp[i][j] ``+``=` `dp[i][j ``+` `1``]`   `                    ``if` `(i ``-` `1` `>``=` `0` `and` `j ``+` `1` `<``=` `n ``-` `1``):`   `                        ``# Subtract it as we have` `                        ``# counted it twice` `                        ``dp[i][j] ``-``=` `dp[i ``-` `1``][j ``+` `1``]`   `    ``ways ``=` `[]` `    ``for` `i ``in` `range``(n):` `        ``if` `(i ``=``=` `0` `or` `i ``=``=` `n ``-` `1``):`   `            ``# We have just 1 palindrome` `            ``# sequence of Length 1` `            ``ways.append(``1``)` `        ``else``:`   `            ``# Else total ways would be sum of dp[i-1][i+1],` `            ``# that is number of palindrome sub-sequences` `            ``# from 1 to i-1 + number of palindrome` `            ``# sub-sequences from i+1 to n-1` `            ``total ``=` `dp[i ``-` `1``][i ``+` `1``]` `            ``ways.append(total)`   `    ``for` `i ``in` `ways:` `        ``print``(i,end``=``" "``)`   `# Driver code`   `s ``=` `"xyxyx"` `solve(s)`   `# This code is contributed by mohit kumar 29`

C#

 `// C# implementation of above approach ` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG ` `{ `   `    ``// Function to find the total palindromic ` `    ``// odd length sub-sequences ` `    ``static` `void` `solve(``char``[] s) ` `    ``{ ` `        ``int` `n = s.Length; ` `    `  `        ``// dp array to store the number of palindromic ` `        ``// subsequences for 0 to i-1 and j+1 to n-1 ` `        ``int` `[,]dp = ``new` `int``[n, n]; ` `    `  `        ``// We will start with the largest ` `        ``// distance between i and j ` `        ``for` `(``int` `len = n - 1; len >= 0; --len) ` `        ``{ ` `    `  `            ``// For each len, we fix our i ` `            ``for` `(``int` `i = 0; i + len < n; ++i) ` `            ``{ ` `    `  `                ``// For this i we will find our j ` `                ``int` `j = i + len; ` `    `  `                ``// Base cases ` `                ``if` `(i == 0 && j == n - 1) ` `                ``{ ` `                    ``if` `(s[i] == s[j]) ` `                        ``dp[i, j] = 2; ` `                    ``else` `if` `(s[i] != s[j]) ` `                        ``dp[i, j] = 1; ` `                ``} ` `                ``else` `                ``{ ` `                    ``if` `(s[i] == s[j]) ` `                    ``{ ` `    `  `                        ``// If the characters are equal ` `                        ``// then look for out of bound index ` `                        ``if` `(i - 1 >= 0) ` `                        ``{ ` `                            ``dp[i, j] += dp[i - 1, j]; ` `                        ``} ` `                        ``if` `(j + 1 <= n - 1) ` `                        ``{ ` `                            ``dp[i, j] += dp[i, j + 1]; ` `                        ``} ` `                        ``if` `(i - 1 < 0 || j + 1 >= n) ` `                        ``{ ` `    `  `                            ``// We have only 1 way that is to ` `                            ``// just pick these characters ` `                            ``dp[i, j] += 1; ` `                        ``} ` `                    ``} ` `                    ``else` `if` `(s[i] != s[j]) ` `                    ``{ ` `    `  `                        ``// If the characters are not equal ` `                        ``if` `(i - 1 >= 0) ` `                        ``{ ` `                            ``dp[i, j] += dp[i - 1, j]; ` `                        ``} ` `                        ``if` `(j + 1 <= n - 1) ` `                        ``{ ` `                            ``dp[i, j] += dp[i, j + 1]; ` `                        ``} ` `                        ``if` `(i - 1 >= 0 && j + 1 <= n - 1) ` `                        ``{ ` `    `  `                            ``// Subtract it as we have ` `                            ``// counted it twice ` `                            ``dp[i, j] -= dp[i - 1, j + 1]; ` `                        ``} ` `                    ``} ` `                ``} ` `            ``} ` `        ``} ` `        `  `        ``List<``int``> ways = ``new` `List<``int``>();`   `        ``for` `(``int` `i = 0; i < n; ++i) ` `        ``{ ` `            ``if` `(i == 0 || i == n - 1) ` `            ``{ ` `    `  `                ``// We have just 1 palindrome ` `                ``// sequence of length 1 ` `                ``ways.Add(1); ` `            ``} ` `            ``else` `            ``{ ` `    `  `                ``// Else total ways would be sum of dp[i-1][i+1], ` `                ``// that is number of palindrome sub-sequences ` `                ``// from 1 to i-1 + number of palindrome ` `                ``// sub-sequences from i+1 to n-1 ` `                ``int` `total = dp[i - 1,i + 1]; ` `                ``ways.Add(total); ` `            ``} ` `        ``} ` `        ``for` `(``int` `i = 0; i < ways.Capacity; ++i) ` `        ``{ ` `            ``Console.Write(ways[i] + ``" "``); ` `        ``} ` `    ``} ` `    `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``char``[] s = ``"xyxyx"``.ToCharArray(); ` `        ``solve(s); ` `    ``} ` `} `   `// This code is contributed by AnkitRai01`

Javascript

 ``

Output:

`1 3 4 3 1`

Time Complexity: O(n2)
Auxiliary Space: O(n2), where n is the length of the given string.

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