Count Subarrays with strictly decreasing consecutive elements

Given an array arr[] containing integers. The task is to find the number of decreasing subarrays with a difference of 1. 

Examples: 

Input: arr[] = {3, 2, 1, 4}
Output: 7
Explanation: Following are the possible decreasing subarrays with difference 1. 
[3], [2], [1], [4], [3,2], [2,1], and [3,2,1]
Therefore, the answer is 7. 

Input: arr[] = {5, 4, 3, 2, 1, 6}
Output: 16

Naive Approach: This problem can be solved by using Dynamic Programming. Follow the steps below to solve the given problem.

1. For every index i the task is to calculate number of subarrays ending at i which follows this pattern arr[i-2]==arr[i-1]+1, arr[i-1]==arr[i]+1.
2. Initialize a variable say ans = 0, to store the number of decreasing subarrays with a difference of 1.
3. We can make a dp[] array which stores the count of these continuous elements for every index.
4. dp[i] is the number of subarrays ending at i which follows this pattern.
5. Traverse dp[] and add each value in ans.
6. Return ans as the final result.

Below is the implementation of the above approach.

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

// Function to count number of
// decreasing subarrays with difference 1
long long getcount(vector<int>& p)
{
    int size = p.size(), cnt = 0;
    long long ans = 0;
    vector<int> dp(size, cnt);
    for (int i = 0; i < size; i++) {
        if (i == 0)
            cnt = 1;
        else if (p[i] + 1 == p[i - 1])
            cnt++;
        else
            cnt = 1;
        dp[i] = cnt;
    }
    for (int i = 0; i < size; i++)
        ans += dp[i];
    return ans;
}

// Driver Code
int main()
{
    vector<int> arr{ 5, 4, 3, 2, 1, 6 };

    // Function Call
    cout << getcount(arr);

    return 0;
}

// Java code to implement the above approach
import java.util.*;
public class GFG
{

  // Function to count number of
  // decreasing subarrays with difference 1
  static long getcount(int p[])
  {
    int size = p.length, cnt = 0;
    long ans = 0;

    int dp[] = new int[size];
    for(int i = 0; i < size; i++) {
      dp[i] = cnt;
    }

    for (int i = 0; i < size; i++) {
      if (i == 0)
        cnt = 1;
      else if (p[i] + 1 == p[i - 1])
        cnt++;
      else
        cnt = 1;
      dp[i] = cnt;
    }
    for (int i = 0; i < size; i++)
      ans += dp[i];
    return ans;
  }

  // Driver code
  public static void main(String args[])
  {
    int arr[] = { 5, 4, 3, 2, 1, 6 };

    // Function Call
    System.out.println(getcount(arr));
  }
}

// This code is contributed by Samim Hossain Mondal.

# Python code to implement the above approach

# Function to count number of
# decreasing subarrays with difference 1
def getcount(p):
    size = len(p)
    cnt = 0
    ans = 0
    dp = [cnt for i in range(size)]
    for i in range(size):
        if (i == 0):
            cnt = 1
        elif (p[i] + 1 == p[i - 1]):
            cnt += 1
        else:
            cnt = 1
        dp[i] = cnt
        
    for i in range(size):
        ans += dp[i]
    return ans

# Driver Code
arr = [5, 4, 3, 2, 1, 6]

# Function Call
print(getcount(arr))

# This code is contributed by Shubham Singh

// C# code to implement the above approach
using System;
class GFG
{
  
  // Function to count number of
  // decreasing subarrays with difference 1
  static long getcount(int []p)
  {
    int size = p.Length, cnt = 0;
    long ans = 0;

    int []dp = new int[size];
    for(int i = 0; i < size; i++) {
      dp[i] = cnt;
    }

    for (int i = 0; i < size; i++) {
      if (i == 0)
        cnt = 1;
      else if (p[i] + 1 == p[i - 1])
        cnt++;
      else
        cnt = 1;
      dp[i] = cnt;
    }
    for (int i = 0; i < size; i++)
      ans += dp[i];
    return ans;
  }

  // Driver code
  public static void Main()
  {
    int []arr = { 5, 4, 3, 2, 1, 6 };

    // Function Call
    Console.Write(getcount(arr));
  }
}

// This code is contributed by Samim Hossain Mondal.

<script>

// JavaScript program for above approach 

// Function to count number of decreasing
// subarrays with difference 1
function getcount(p)
{
    let size = p.length, cnt = 0;
    let ans = 0;
    let dp = new Array(size).fill(cnt);
    
    for(let i = 0; i < size; i++) 
    {
        if (i == 0)
            cnt = 1;
        else if (p[i] + 1 == p[i - 1])
            cnt++;
        else
            cnt = 1;
            
        dp[i] = cnt;
    }
    for(let i = 0; i < size; i++)
        ans += dp[i];
        
    return ans;
}

// Driver Code
let arr = [ 5, 4, 3, 2, 1, 6 ];

// Function Call
document.write(getcount(arr));

// This code is contributed by Potta Lokesh

</script>

16

Time complexity: O(N) 
Auxiliary Space: O(N)

Efficient Approach: In the above approach the auxiliary space complexity can be further optimized to constant space by replacing dp[] array with a variable to keep track of the current number of subarrays. Follow the steps below to solve the given problem.

• Initialize a variable say count = 0.
• Start traversing the array when arr[i]-arr[i-1 ]==1 to make a chain of numbers that are decreasing by 1, then count++.
• Add the count to the ans.
• When the chain breaks that means, arr[i]-arr[i-1] !=1 then reset the count.
• Return ans as the final result.

Below is the implementation of the above approach.

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

// Function to count the number of
// decreasing subarrays with difference 1
long long getcount(vector<int>& arr)
{
    long long int ans = arr.size();
    long long int count = 0;
    for (int i = 1; i < arr.size(); i++) {
        if (arr[i - 1] - arr[i] == 1)
            count++;
        else
            count = 0;
        ans = ans + count;
    }
    return ans;
}

// Driver Code
int main()
{
    vector<int> arr{ 5, 4, 3, 2, 1, 6 };

    // Function Call
    cout << getcount(arr);

    return 0;
}

// Java program for above approach
class GFG
{

  // Function to count the number of
  // decreasing subarrays with difference 1
  static long getcount(int[] arr)
  {
    int ans = arr.length;
    int count = 0;
    for (int i = 1; i < arr.length; i++) {
      if (arr[i - 1] - arr[i] == 1)
        count++;
      else
        count = 0;
      ans = ans + count;
    }
    return ans;
  }

  // Driver Code
  public static void main(String[] args)
  {
    int[] arr = { 5, 4, 3, 2, 1, 6 };

    // Function Call
    System.out.print(getcount(arr));

  }
}

// This code is contributed by 29AjayKumar 

#Python program for the above approach

# Function to count the number of
# decreasing subarrays with difference 1
def getcount(arr):
    ans = len(arr)
    count = 0
    for i in range(1, len(arr)):
        if (arr[i - 1] - arr[i] == 1):
            count+=1
        else:
            count = 0
        ans = ans + count
        
    return ans

  # Driver Code
arr = [ 5, 4, 3, 2, 1, 6 ]

# Function Call
print(getcount(arr))

# This code is contributed by Shubham Singh

// C# program for above approach
using System;

public class GFG
{

  // Function to count the number of
  // decreasing subarrays with difference 1
  static long getcount(int[] arr)
  {
    int ans = arr.Length;
    int count = 0;
    for (int i = 1; i < arr.Length; i++) {
      if (arr[i - 1] - arr[i] == 1)
        count++;
      else
        count = 0;
      ans = ans + count;
    }
    return ans;
  }

  // Driver Code
  public static void Main(String[] args)
  {
    int[] arr = { 5, 4, 3, 2, 1, 6 };

    // Function Call
    Console.Write(getcount(arr));

  }
}

// This code is contributed by 29AjayKumar 

<script>
// javascript program for above approach 

// Function to count the number of
  // decreasing subarrays with difference 1
  function getcount(arr)
  {
    var ans = arr.length;
    var count = 0;
    for (var i = 1; i < arr.length; i++) {
      if (arr[i - 1] - arr[i] == 1)
        count++;
      else
        count = 0;
      ans = ans + count;
    }
    return ans;
  }

  // Driver Code
var arr = [ 5, 4, 3, 2, 1, 6 ];

// Function Call
document.write(getcount(arr));

// This code is contributed by 29AjayKumar 
</script>

16

Time complexity: O(N) 
Auxiliary Space: O(1)