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Count of bitonic substrings from the given string
  • Last Updated : 02 Aug, 2020

Given a string str, the task is to count all the bitonic substrings of the given string.

A bitonic substring is a substring of the given string in which elements are either strictly increasing or strictly decreasing, or first increasing and then decreasing.

Examples:

Input: str = “bade”
Output:  8
Explanation: 
Substrings of length 1 are always bitonic, “b”, “a”, “d”, “e” 
Substrings of length 2 are “ba”, “ad”, “de” and these all are also bitonic because they are increasing or decresing only. 
Substrings of length 3 are “bad “and “ade” in which “ade” is bitonic. 
Substring of length 4 “bade” is not bitonic because it decreases and then increases. 
So total 8 substrings are bitonic.

Input: str = “abc” 
Output:  6
Explanation: 
The given string is increasing, so all it’s substrings are also increasing and hence are bitonic. So total = 6.



 

Approach: The idea is to generate all possible substring of the given string and check if each substring is Bitonic or not. If the substring is Bitonic then increment the count for the answer. Finally, print the count of all the bitonic subarrays.

Below is the implementation of the above approach :

C++

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// C+++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find all the bitonic
// sub strings
void subString(char str[], int n)
{
    // Pick starting point
    int c = 0;
 
    // Iterate till length of the string
    for (int len = 1; len <= n; len++) {
 
        // Pick ending point for string
        for (int i = 0; i <= n - len; i++) {
 
            // Substring from i to j
            // is obtained
            int j = i + len - 1;
            char temp = str[i], f = 0;
 
            // Substrings of length 1
            if (j == i) {
 
                // Increase count
                c++;
                continue;
            }
 
            int k = i + 1;
 
            // For increasing sequence
            while (temp < str[k] && k <= j) {
                temp = str[k];
                k++;
                f = 2;
            }
 
            // Check for strictly increasing
            if (k > j) {
 
                // Increase count
                c++;
                f = 2;
            }
 
            // Check for decreasing sequence
            while (temp > str[k]
                   && k <= j
                   && f != 2) {
                k++;
                f = 0;
            }
 
            if (k > j && f != 2) {
 
                // Increase count
                c++;
                f = 0;
            }
        }
    }
 
    // Print the result
    cout << c << endl;
}
 
// Driver Code
int main()
{
    // Given string
    char str[] = "bade";
 
    // Function Call
    subString(str, strlen(str));
    return 0;
}

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Java

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// Java+ program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find all the bitonic
// sub Strings
static void subString(char str[], int n)
{
     
    // Pick starting point
    int c = 0;
 
    // Iterate till length of the String
    for(int len = 1; len <= n; len++)
    {
         
        // Pick ending point for String
        for(int i = 0; i <= n - len; i++)
        {
             
            // SubString from i to j
            // is obtained
            int j = i + len - 1;
            char temp = str[i], f = 0;
 
            // SubStrings of length 1
            if (j == i)
            {
                 
                // Increase count
                c++;
                continue;
            }
 
            int k = i + 1;
 
            // For increasing sequence
            while (k < n && temp < str[k])
            {
                temp = str[k];
                k++;
                f = 2;
            }
 
            // Check for strictly increasing
            if (k > j)
            {
 
                // Increase count
                c++;
                f = 2;
            }
 
            // Check for decreasing sequence
            while (k < n && temp > str[k] &&
                   f != 2)
            {
                k++;
                f = 0;
            }
 
            if (k > j && f != 2)
            {
 
                // Increase count
                c++;
                f = 0;
            }
        }
    }
 
    // Print the result
    System.out.print(c + "\n");
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given String
    char str[] = "bade".toCharArray();
 
    // Function call
    subString(str, str.length);
}
}
 
// This code is contributed by 29AjayKumar

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Python3

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# Python3 program for the above approach
 
# Function to find all the bitonic
# sub strings
def subString(str, n):
 
    # Pick starting po
    c = 0;
 
    # Iterate till length of the string
    for len in range(1, n + 1):
 
        # Pick ending pofor string
        for i in range(0, n - len + 1):
 
            # Substring from i to j
            # is obtained
            j = i + len - 1;
            temp = str[i]
            f = 0;
 
            # Substrings of length 1
            if (j == i):
 
                # Increase count
                c += 1
                continue;
            k = i + 1;
 
            # For increasing sequence
            while (k <= j and temp < str[k]):
                temp = str[k];
                k += 1;
                f = 2;
             
            # Check for strictly increasing
            if (k > j):
 
                # Increase count
                c += 1;
                f = 2;
             
            # Check for decreasing sequence
            while (k <= j and temp > str[k] and
                   f != 2):
                k += 1;
                f = 0;
            if (k > j and f != 2):
 
                # Increase count
                c += 1;
                f = 0;
                 
    # Print the result
    print(c)
 
# Driver code
 
# Given string
str = "bade";
 
# Function Call
subString(str, len(str))
     
# This code is contributed by grand_master

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C#

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// C# program for the above approach
using System;
class GFG{
 
// Function to find all the bitonic
// sub Strings
static void subString(char []str, int n)
{
     
    // Pick starting point
    int c = 0;
 
    // Iterate till length of the String
    for(int len = 1; len <= n; len++)
    {
         
        // Pick ending point for String
        for(int i = 0; i <= n - len; i++)
        {
             
            // SubString from i to j
            // is obtained
            int j = i + len - 1;
            char temp = str[i], f = (char)0;
 
            // SubStrings of length 1
            if (j == i)
            {
                 
                // Increase count
                c++;
                continue;
            }
 
            int k = i + 1;
 
            // For increasing sequence
            while (k < n && temp < str[k])
            {
                temp = str[k];
                k++;
                f = (char)2;
            }
 
            // Check for strictly increasing
            if (k > j)
            {
 
                // Increase count
                c++;
                f = (char)2;
            }
 
            // Check for decreasing sequence
            while (k < n && temp > str[k] &&
                f != 2)
            {
                k++;
                f = (char)0;
            }
 
            if (k > j && f != 2)
            {
 
                // Increase count
                c++;
                f = (char)0;
            }
        }
    }
 
    // Print the result
    Console.Write(c + "\n");
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given String
    char []str = "bade".ToCharArray();
 
    // Function call
    subString(str, str.Length);
}
}
 
// This code is contributed by 29AjayKumar

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Output: 

8



 

Time Complexity: O(N2)
Auxiliary Space: O(1)

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