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+ 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 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# 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
<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>
Time Complexity: O(N2)
Auxiliary Space: O(1)
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