# Count of Distinct Substrings occurring consecutively in a given String

Given a string str, the task is to find the number of distinct substrings that are placed consecutively in the given string.

Examples:

Input: str = “geeksgeeksforgeeks”
Output:
Explanation:
geeksgeeksforgeeks -> {“geeks”}
geeksgeeksforgeeks -> {“e”}
Only one consecutive occurrence of “e” is considered.
Therefore two distinct substrings {“geeks”, “e”} occur consecutively in the string.

Input: s = “geeksforgeeks”
Output:
Explanation:
geeksgeeksforgeeks -> {“e”, “e”}
Only one substring {“e”} occurs consecutively in the string.

Naive Approach:
The simplest approach is to generate all possible substrings of the given string, and for each substring, find the count of substrings in the given occurring consecutively in the string. Finally, print the count.

Time Complexity: O(N3)
Auxiliary Space: O(N)

Efficient Approach:
To optimize the above approach, the idea is to use Dynamic Programming
Follow the steps below to solve the problem:

1. If the length of the string does not exceed 1, then it is not possible to find any such consecutively placed similar substrings. So return 0 as the count.
2. Otherwise, initialize a memoization table dp[] of dimensions (N+1 * N+1) which is initialized to 0.
3. Initialize an unordered_set to store the distinct substrings placed consecutively.
4. Iterate from the end of the string.
5. While traversing the string if any repeating character is found, then dp[i][j] will be determined considering the previously computed dp value i.e., count of identical substrings up to dp[i+1][j+1] characters and including the current character.
6. If the character is not similar then, dp[i][j] will be filled with 0.
7. Similar substrings are consecutively placed together without any other characters and they will be the same for at most (j – i) characters. Hence, for valid substrings, dp[i][j] value must be greater than (j – i). Store those substrings in unordered_set which appears the maximum number of times consecutively.
8. Finally, return the size of the unordered_set as the count of distinct substrings placed consecutively.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement` `// the above approach` `#include ` `using` `namespace` `std;`   `// Function to count the distinct substrings` `// placed consecutively in the given string` `int` `distinctSimilarSubstrings(string str)` `{` `    ``// Length of the string` `    ``int` `n = str.size();`   `    ``// If length of the string` `    ``// does not exceed 1` `    ``if` `(n <= 1) {` `        ``return` `0;` `    ``}`   `    ``// Initialize a DP-table` `    ``vector > dp(` `        ``n + 1, vector<``int``>(n + 1, 0));`   `    ``// Stores the distinct substring` `    ``unordered_set substrings;`   `    ``// Iterate from end of the string` `    ``for` `(``int` `j = n - 1; j >= 0; j--) {`   `        ``// Iterate backward until` `        ``// dp table is all computed` `        ``for` `(``int` `i = j - 1; i >= 0; i--) {`   `            ``// If character at i-th index is` `            ``// same as character at j-th index` `            ``if` `(str[i] == str[j]) {`   `                ``// Update dp[i][j] based on` `                ``// previously computed value` `                ``dp[i][j] = dp[i + 1][j + 1] + 1;` `            ``}`   `            ``// Otherwise` `            ``else` `{`   `                ``dp[i][j] = 0;` `            ``}`   `            ``// Condition for consecutively` `            ``// placed similar substring` `            ``if` `(dp[i][j] >= j - i) {`   `                ``substrings.insert(` `                    ``str.substr(i, j - i));` `            ``}` `        ``}` `    ``}`   `    ``// Return the count` `    ``return` `substrings.size();` `}`   `// Driver Code` `int` `main()` `{` `    ``string str = ``"geeksgeeksforgeeks"``;`   `    ``cout << distinctSimilarSubstrings(str);` `    ``return` `0;` `}`

## Java

 `// Java program to implement ` `// the above approach ` `import` `java.io.*;` `import` `java.util.ArrayList;`   `class` `GFG{`   `// Function to count the distinct substrings ` `// placed consecutively in the given string     ` `static` `int` `distinctSimilarSubstrings(String str)` `{` `    `  `    ``// Length of the string ` `    ``int` `n = str.length();` `    `  `    ``// If length of the string ` `    ``// does not exceed 1 ` `    ``if` `(n <= ``1``)` `        ``return` `0``;` `        `  `    ``// Initialize a DP-table ` `    ``long` `dp[][] = ``new` `long``[n + ``1``][n + ``1``];` `    `  `    ``// Declaring ArrayList to store strings` `    ``ArrayList list = ``new` `ArrayList();`   `    ``// Iterate from end of the string ` `    ``for``(``int` `j = n - ``1``; j >= ``0``; j--) ` `    ``{` `        `  `        ``// Iterate backward until` `        ``// dp table is all computed` `        ``for``(``int` `i = j - ``1``; i >= ``0``; i--) ` `        ``{` `            `  `            ``// If character at i-th index is` `            ``// same as character at j-th index` `            ``if` `(str.charAt(i) == str.charAt(j)) ` `            ``{` `                `  `                ``// Update dp[i][j] based on` `                ``// previously computed value` `                ``dp[i][j] = dp[i + ``1``][j + ``1``] + ``1``;` `            ``}` `            `  `            ``// Otherwise` `            ``else` `            ``{` `                ``dp[i][j] = ``0``;` `            ``}`   `            ``// Condition for consecutively` `            ``// placed similar substring` `            ``if` `(dp[i][j] >= j - i)` `            ``{` `                ``list.add(str.substring(j - i, i));` `            ``}` `        ``}` `    ``}` `    `  `    ``// Return the count` `    ``return` `list.size();` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``String str = ``"geeksforgeeks"``;` `    `  `    ``System.out.println(distinctSimilarSubstrings(str));` `}` `}`   `// This code is contributed by user_00`

## Python3

 `# Python3 program to implement ` `# the above approach `   `# Function to count the distinct substrings` `# placed consecutively in the given string` `def` `distinctSimilarSubstrings(``str``):`   `    ``# Length of the string` `    ``n ``=` `len``(``str``)`   `    ``# If length of the string` `    ``# does not exceed 1` `    ``if``(n <``=` `1``):` `        ``return` `0`   `    ``# Initialize a DP-table` `    ``dp ``=` `[[``0` `for` `x ``in` `range``(n ``+` `1``)]` `             ``for` `y ``in` `range``(n ``+` `1``)]`   `    ``# Stores the distinct substring` `    ``substrings ``=` `set``()`   `    ``# Iterate from end of the string` `    ``for` `j ``in` `range``(n ``-` `1``, ``-``1``, ``-``1``):`   `        ``# Iterate backward until` `        ``# dp table is all computed` `        ``for` `i ``in` `range``(j ``-` `1``, ``-``1``, ``-``1``):`   `            ``# If character at i-th index is` `            ``# same as character at j-th index` `            ``if``(``str``[i] ``=``=` `str``[j]):`   `                ``# Update dp[i][j] based on` `                ``# previously computed value` `                ``dp[i][j] ``=` `dp[i ``+` `1``][j ``+` `1``] ``+` `1`   `            ``# Otherwise` `            ``else``:` `                ``dp[i][j] ``=` `0`   `            ``# Condition for consecutively` `            ``# placed similar substring` `            ``if``(dp[i][j] >``=` `j ``-` `i):` `                ``substrings.add(``str``[i : j ``-` `i])`   `    ``# Return the count ` `    ``return` `len``(substrings)`   `# Driver Code` `str` `=` `"geeksgeeksforgeeks"`   `# Function call` `print``(distinctSimilarSubstrings(``str``))`   `# This code is contributed by Shivam Singh`

Output:

```2

```

Time Complexity: O(N)
Auxiliary Space: O(N) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : SHIVAMSINGH67, user_00