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:2Explanation:geeksgeeksforgeeks -> {“geeks”}

geeksgeeksforgeeks -> {“e”}

Only one consecutive occurrence of “e” is considered.

Therefore two distinct substrings {“geeks”, “e”} occur consecutively in the string.

Therefore, the answer is 2.

Input:s = “geeksforgeeks”Output:1Explanation:

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(N^{3})**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:

- 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**. - Otherwise, initialize a memoization table
**dp[]**of dimensions**(N+1 * N+1)**which is initialized to**0**. - Initialize an
**unordered_set**to store the distinct substrings placed consecutively. - Iterate from the end of the string.
- 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. - If the character is not similar then,
**dp[i][j]**will be filled with 0. - 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. - 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 <bits/stdc++.h> ` `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<vector<` `int` `> > dp( ` ` ` `n + 1, vector<` `int` `>(n + 1, 0)); ` ` ` ` ` `// Stores the distinct substring ` ` ` `unordered_set<string> 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; ` `} ` |

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## 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 ` |

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**Output:**

2

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

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