Given a string S, the task is to count the number of Reverse Bitonic Substrings in the given string.
Reverse bitonic substring: A string in which the ASCII values of the characters of the string follow any of the following patterns:
- Strictly Increasing
- Strictly decreasing
- Decreasing and then increasing
Examples:
Input: S = “bade”
Output: 10
Explanation:
All possible substrings of length 1, {“b”, “a”, “d”, “e”} are always reverse bitonic.
Substrings of length 2 which are reverse bitonic are {“ba”, “ad”, “de”}.
Substrings of length 3 which are reverse bitonic are {“bad “, “ade”}.
Only substring of length 4 which is reverse bitonic is “bade”.
Therefore, the count of reverse bitonic substrings is 10.Input: S = “abc”
Output: 6
Approach :
The approach is to generate all possible substrings of the given string and follow the steps below for each substring to solve the problem:
- Traverse the string and for each character, check if the ASCII value of the next character is smaller than the ASCII value of the current character or not.
- If at any point, the ASCII value of the next character is greater than the ASCII value of the current character, traverse from that index and for each character from now onwards, check if the ASCII value of the next character is greater than the ASCII value of the current character or not.
- If at any point, the ASCII value of the next character is smaller than the ASCII value of the current character before the end of the substring is reached, then ignore the substring.
- If the end of the substring is reached, increase count.
- After completing all the above steps for all substrings, print the final value of count.
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 calculate the number // of reverse bitonic substrings int CountsubString(char str[], int n) { // Stores the count int c = 0; // All possible lengths of substrings for (int len = 1; len <= n; len++) { // Starting point of a substring for (int i = 0; i <= n - len; i++) { // Ending point of a substring int j = i + len - 1; char temp = str[i], f = 0; // Condition for reverse // bitonic substrings of // length 1 if (j == i) { c++; continue; } int k = i + 1; // Check for decreasing sequence while (temp > str[k] && k <= j) { temp = str[k]; k++; } // If end of substring // is reached if (k > j) { c++; f = 2; } // For increasing sequence while (temp < str[k] && k <= j && f != 2) { temp = str[k]; k++; } // If end of substring // is reached if (k > j && f != 2) { c++; f = 0; } } } // Return the number // of bitonic substrings return c; } // Driver Code int main() { char str[] = "bade"; cout << CountsubString(str, strlen(str)); return 0; } |
Java
// Java program to implement // the above approach class GFG{ // Function to calculate the number // of reverse bitonic substrings public static int CountsubString(char[] str, int n) { // Stores the count int c = 0; // All possible lengths of substrings for(int len = 1; len <= n; len++) { // Starting point of a substring for(int i = 0; i <= n - len; i++) { // Ending point of a substring int j = i + len - 1; char temp = str[i], f = 0; // Condition for reverse // bitonic substrings of // length 1 if (j == i) { c++; continue; } int k = i + 1; // Check for decreasing sequence while (temp > str[k] && k <= j) { temp = str[k]; k++; } // If end of substring // is reached if (k > j) { c++; f = 2; } // For increasing sequence while (k <= j && temp < str[k] && f != 2) { temp = str[k]; k++; } // If end of substring // is reached if (k > j && f != 2) { c++; f = 0; } } } // Return the number // of bitonic substrings return c; } // Driver codepublic static void main(String[] args){ char str[] = { 'b', 'a', 'd', 'e' }; System.out.println(CountsubString( str, str.length));}}// This code is contributed by divyeshrabadiya07 |
Python3
# Python3 program to implement # the above approach # Function to calculate the number # of reverse bitonic substrings def CountsubString(strr, n): # Stores the count c = 0 # All possible lengths of substrings for len in range(n + 1): # Starting point of a substring for i in range(n - len): # Ending point of a substring j = i + len - 1 temp = strr[i] f = 0 # Condition for reverse # bitonic substrings of # length 1 if (j == i): c += 1 continue k = i + 1 # Check for decreasing sequence while (k <= j and temp > strr[k]): temp = strr[k] k += 1 # If end of substring # is reache if (k > j): c += 1 f = 2 # For increasing sequence while (k <= j and f != 2 and temp < strr[k]): temp = strr[k] k += 1 # If end of substring # is reached if (k > j and f != 2): c += 1 f = 0 # Return the number # of bitonic substrings return c # Driver Code if __name__ == '__main__': strr = "bade" print(CountsubString(strr, len(strr))) # This code is contributed by mohit kumar 29 |
C#
// C# program to implement // the above approach using System;class GFG{ // Function to calculate the number // of reverse bitonic substrings public static int CountsubString(char[] str, int n) { // Stores the count int c = 0; // All possible lengths of substrings for(int len = 1; len <= n; len++) { // Starting point of a substring for(int i = 0; i <= n - len; i++) { // Ending point of a substring int j = i + len - 1; char temp = str[i], f = '0'; // Condition for reverse // bitonic substrings of // length 1 if (j == i) { c++; continue; } int k = i + 1; // Check for decreasing sequence while (temp > str[k] && k <= j) { temp = str[k]; k++; } // If end of substring // is reached if (k > j) { c++; f = '2'; } // For increasing sequence while (k <= j && temp < str[k] && f != '2') { temp = str[k]; k++; } // If end of substring // is reached if (k > j && f != 2) { c++; f = '0'; } } } // Return the number // of bitonic substrings return c; } // Driver codepublic static void Main(String[] args){ char []str = { 'b', 'a', 'd', 'e' }; Console.WriteLine(CountsubString( str, str.Length) - 1);}}// This code is contributed by amal kumar choubey |
Javascript
<script>// Javascript Program to implement // the above approach // Function to calculate the number // of reverse bitonic substrings function CountsubString(str, n) { // Stores the count var c = 0; // All possible lengths of substrings for (var len = 1; len <= n; len++) { // Starting point of a substring for (var i = 0; i <= n - len; i++) { // Ending point of a substring var j = i + len - 1; var temp = str[i], f = 0; // Condition for reverse // bitonic substrings of // length 1 if (j == i) { c++; continue; } var k = i + 1; // Check for decreasing sequence while (temp > str[k] && k <= j) { temp = str[k]; k++; } // If end of substring // is reached if (k > j) { c++; f = 2; } // For increasing sequence while (temp < str[k] && k <= j && f != 2) { temp = str[k]; k++; } // If end of substring // is reached if (k > j && f != 2) { c++; f = 0; } } } // Return the number // of bitonic substrings return c; } // Driver Code var str = "bade".split(''); document.write( CountsubString(str, str.length)); // This code is contributed by importantly.</script> |
Output:
10
Time Complexity: O(N^3)
Auxiliary Space: O(1)
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