Count of bitonic substrings from the given string

Last Updated : 31 Mar, 2023

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 decreasing 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++

// 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;
}

Java

// 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

Python3

# 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 point for 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

C#

// 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

Javascript

<script>
 
// Javascript program for the above approach
 
// Function to find all the bitonic
// sub strings
function subString(str, n)
{
    // Pick starting point
    var c = 0;
 
    // Iterate till length of the string
    for (var len = 1; len <= n; len++) {
 
        // Pick ending point for string
        for (var i = 0; i <= n - len; i++) {
 
            // Substring from i to j
            // is obtained
            var j = i + len - 1;
            var temp = str[i], f = 0;
 
            // Substrings of length 1
            if (j == i) {
 
                // Increase count
                c++;
                continue;
            }
 
            var 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
    document.write( c );
}
 
// Driver Code
 
// Given string
var str = "bade".split('');
 
// Function Call
subString(str, str.length);
 
// This code is contributed by itsok.
</script>

Output:

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

Closest pair in an Array such that one number is multiple of the other

grand_master

Improve

Article Tags :

Practice Tags :

Count of bitonic substrings from the given string

C++

Java

Python3

C#

Javascript

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