Count Strictly Increasing Subarrays

Given an array of integers, count number of subarrays (of size more than one) that are strictly increasing.
Expected Time Complexity : O(n)
Expected Extra Space: O(1)

Examples:

Input: arr[] = {1, 4, 3}
Output: 1
There is only one subarray {1, 4}

Input: arr[] = {1, 2, 3, 4}
Output: 6
There are 6 subarrays {1, 2}, {1, 2, 3}, {1, 2, 3, 4}
                      {2, 3}, {2, 3, 4} and {3, 4}

Input: arr[] = {1, 2, 2, 4}
Output: 2
There are 2 subarrays {1, 2} and {2, 4}

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A Simple Solution is to generate all possible subarrays, and for every subarray check if subarray is strictly increasing or not. Worst case time complexity of this solution would be O(n3).

A Better Solution is to use the fact that if subarray arr[i:j] is not strictly increasing, then subarrays arr[i:j+1], arr[i:j+2], .. arr[i:n-1] cannot be strictly increasing. Below is the program based on above idea.

C++

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// C++ program to count number of strictly
// increasing subarrays
#include<bits/stdc++.h>
using namespace std;
  
int countIncreasing(int arr[], int n)
{
    // Initialize count of subarrays as 0
    int cnt = 0;
  
    // Pick starting point
    for (int i=0; i<n; i++)
    {
        // Pick ending point
        for (int j=i+1; j<n; j++)
        {
            if (arr[j] > arr[j-1])
                cnt++;
  
            // If subarray arr[i..j] is not strictly 
            // increasing, then subarrays after it , i.e., 
            // arr[i..j+1], arr[i..j+2], .... cannot
            // be strictly increasing
            else
                break;
        }
    }
    return cnt;
}
  
// Driver program
int main()
{
  int arr[] = {1, 2, 2, 4};
  int n = sizeof(arr)/sizeof(arr[0]);
  cout << "Count of strictly increasing subarrays is "
       << countIncreasing(arr, n);
  return 0;
}

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Java

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// Java program to count number of strictly
// increasing subarrays
  
  
class Test
{
    static int arr[] = new int[]{1, 2, 2, 4};
      
    static int countIncreasing(int n)
    {
        // Initialize count of subarrays as 0
        int cnt = 0;
       
        // Pick starting point
        for (int i=0; i<n; i++)
        {
            // Pick ending point
            for (int j=i+1; j<n; j++)
            {
                if (arr[j] > arr[j-1])
                    cnt++;
       
                // If subarray arr[i..j] is not strictly 
                // increasing, then subarrays after it , i.e., 
                // arr[i..j+1], arr[i..j+2], .... cannot
                // be strictly increasing
                else
                    break;
            }
        }
        return cnt;
    }
    // Driver method to test the above function
    public static void main(String[] args) 
    {
        System.out.println("Count of strictly increasing subarrays is "
                                               countIncreasing(arr.length));
    }
}

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Python3

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# Python3 program to count number
# of strictly increasing subarrays
  
def countIncreasing(arr, n):
      
    # Initialize count of subarrays as 0
    cnt = 0
  
    # Pick starting point
    for i in range(0, n) :
          
        # Pick ending point
        for j in range(i + 1, n) :
            if arr[j] > arr[j - 1] :
                cnt += 1
  
            # If subarray arr[i..j] is not strictly 
            # increasing, then subarrays after it , i.e., 
            # arr[i..j+1], arr[i..j+2], .... cannot
            # be strictly increasing
            else:
                break
    return cnt
  
  
# Driver code
arr = [1, 2, 2, 4]
n = len(arr)
print ("Count of strictly increasing subarrays is",
                            countIncreasing(arr, n))
  
# This code is contributed by Shreyanshi Arun.

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C#

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// C# program to count number of 
// strictly increasing subarrays
using System;
  
class Test
{
    static int []arr = new int[]{1, 2, 2, 4};
      
    static int countIncreasing(int n)
    {
        // Initialize count of subarrays as 0
        int cnt = 0;
      
        // Pick starting point
        for (int i = 0; i < n; i++)
        {
            // Pick ending point
            for (int j = i + 1; j < n; j++)
            {
                if (arr[j] > arr[j - 1])
                    cnt++;
      
                // If subarray arr[i..j] is not strictly 
                // increasing, then subarrays after it , 
                // i.e.,  arr[i..j+1], arr[i..j+2], .... 
                // cannot be strictly increasing
                else
                    break;
            }
        }
        return cnt;
    }
      
    // Driver Code
    public static void Main(String[] args) 
    {
        Console.Write("Count of strictly increasing"
            "subarrays is " + countIncreasing(arr.Length));
    }
}
  
// This code is contribute by parashar.

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PHP

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<?php
// PHP program to count number of
// strictly increasing subarrays
  
function countIncreasing( $arr, $n)
{
      
    // Initialize count of subarrays 
    // as 0
    $cnt = 0;
  
    // Pick starting point
    for ( $i = 0; $i < $n; $i++)
    {
        // Pick ending point
        for ( $j = $i+1; $j < $n; $j++)
        {
            if ($arr[$j] > $arr[$j-1])
                $cnt++;
  
            // If subarray arr[i..j] is
            // not strictly increasing,
            // then subarrays after it,
            // i.e., arr[i..j+1], 
            // arr[i..j+2], .... cannot
            // be strictly increasing
            else
                break;
        }
    }
    return $cnt;
}
  
// Driver program
  
$arr = array(1, 2, 2, 4);
$n = count($arr);
echo "Count of strictly increasing ",
                     "subarrays is ",
            countIncreasing($arr, $n);
  
// This code is contribute by anuj_67.
?>

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Output :

Count of strictly increasing subarrays is 2

Time complexity of the above solution is O(m) where m is number of subarrays in output

This problem and solution are contributed by Rahul Agrawal.

An Efficient Solution can count subarrays in O(n) time. The idea is based on fact that a sorted subarray of length ‘len’ adds len*(len-1)/2 to result. For example, {10, 20, 30, 40} adds 6 to the result.

C++

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// C++ program to count number of strictly
// increasing subarrays in O(n) time.
#include<bits/stdc++.h>
using namespace std;
  
int countIncreasing(int arr[], int n)
{
    int cnt = 0;  // Initialize result
  
    // Initialize length of current increasing
    // subarray
    int len = 1;
  
    // Traverse through the array
    for (int i=0; i < n-1; ++i)
    {
        // If arr[i+1] is greater than arr[i],
        // then increment length
        if (arr[i + 1] > arr[i])
            len++;
              
        // Else Update count and reset length
        else
        {
            cnt += (((len - 1) * len) / 2);
            len = 1;
        }
    }
      
    // If last length is more than 1
    if (len > 1)
        cnt += (((len - 1) * len) / 2);
  
    return cnt;
}
  
// Driver program
int main()
{
  int arr[] = {1, 2, 2, 4};
  int n = sizeof(arr)/sizeof(arr[0]);
  cout << "Count of strictly increasing subarrays is "
       << countIncreasing(arr, n);
  return 0;
}

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Java

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// Java program to count number of strictly
// increasing subarrays
  
  
class Test
{
    static int arr[] = new int[]{1, 2, 2, 4};
      
    static int countIncreasing(int n)
    {
        int cnt = 0// Initialize result
           
        // Initialize length of current increasing
        // subarray
        int len = 1;
       
        // Traverse through the array
        for (int i=0; i < n-1; ++i)
        {
            // If arr[i+1] is greater than arr[i],
            // then increment length
            if (arr[i + 1] > arr[i])
                len++;
                   
            // Else Update count and reset length
            else
            {
                cnt += (((len - 1) * len) / 2);
                len = 1;
            }
        }
           
        // If last length is more than 1
        if (len > 1)
            cnt += (((len - 1) * len) / 2);
       
        return cnt;
    }
    // Driver method to test the above function
    public static void main(String[] args) 
    {
        System.out.println("Count of strictly increasing subarrays is "
                                               countIncreasing(arr.length));
    }
}

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Python3

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# Python3 program to count number of 
# strictlyincreasing subarrays in O(n) time.
  
def countIncreasing(arr, n):
    cnt = 0 # Initialize result
  
    # Initialize length of current 
    # increasing subarray
    len = 1
  
    # Traverse through the array
    for i in range(0, n - 1) :
          
        # If arr[i+1] is greater than arr[i],
        # then increment length
        if arr[i + 1] > arr[i] :
            len += 1
              
        # Else Update count and reset length
        else:
            cnt += (((len - 1) * len) / 2)
            len = 1
      
    # If last length is more than 1
    if len > 1:
        cnt += (((len - 1) * len) / 2)
  
    return cnt
  
  
# Driver program
arr = [1, 2, 2, 4]
n = len(arr)
  
print ("Count of strictly increasing subarrays is",
                        int(countIncreasing(arr, n)))
  
  
# This code is contributed by Shreyanshi Arun.

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C#

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// C# program to count number of strictly
// increasing subarrays
using System;
  
class GFG {
      
    static int []arr = new int[]{1, 2, 2, 4};
      
    static int countIncreasing(int n)
    {
        int cnt = 0; // Initialize result
          
        // Initialize length of current 
        // increasing subarray
        int len = 1;
      
        // Traverse through the array
        for (int i = 0; i < n-1; ++i)
        {
              
            // If arr[i+1] is greater than
            // arr[i], then increment length
            if (arr[i + 1] > arr[i])
                len++;
                  
            // Else Update count and reset
            // length
            else
            {
                cnt += (((len - 1) * len) / 2);
                len = 1;
            }
        }
          
        // If last length is more than 1
        if (len > 1)
            cnt += (((len - 1) * len) / 2);
      
        return cnt;
    }
      
    // Driver method to test the
    // above function
    public static void Main() 
    {
        Console.WriteLine("Count of strictly "
                  + "increasing subarrays is "
               + countIncreasing(arr.Length));
    }
}
  
// This code is contribute by anuj_67.

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PHP

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<?php
// PHP program to count number of strictly
// increasing subarrays in O(n) time.
  
function countIncreasing( $arr, $n)
{
      
    // Initialize result
    $cnt = 0; 
  
    // Initialize length of
    // current increasing
    // subarray
    $len = 1;
  
    // Traverse through the array
    for($i = 0; $i < $n - 1; ++$i)
    {
          
        // If arr[i+1] is greater than arr[i],
        // then increment length
        if ($arr[$i + 1] > $arr[$i])
            $len++;
              
        // Else Update count and
        // reset length
        else
        {
            $cnt += ((($len - 1) * $len) / 2);
            $len = 1;
        }
    }
      
    // If last length is
    // more than 1
    if ($len > 1)
        $cnt += ((($len - 1) * $len) / 2);
  
    return $cnt;
}
  
// Driver Code
$arr = array(1, 2, 2, 4);
$n = count($arr);
echo "Count of strictly increasing subarrays is "
                     , countIncreasing($arr, $n);
  
// This code is contribute by anuj_67
?>

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Output :

Count of strictly increasing subarrays is 2

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Improved By : parashar, vt_m



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