Count Subarrays with Consecutive elements differing by 1

Given an array arr[] of N integers. The task is to count the total number of subarrays of the given array such that difference between the consecutive elements in the subarrays is one. That is, for any index i in the subarrays, arr[i+1] – arr[i] = 1.

Note: Do not consider subarrays with single element.

Examples:

Input : arr[] = {1, 2, 3}
Output : 3
The subarrays are {1, 2}. {2, 3} and {1, 2, 3}

Input : arr[] = {1, 2, 3, 5, 6, 7}
Output : 6


Naive Approach: A simple approach is to run two nested loops and check every subarray and calculate the count of subarrays with consecutive elements differing by 1.

Efficient Approach: An efficient approach is to observe that in an array of length say K, total number of subarrays of size greater than 1 = (K)*(K-1)/2.

So, the idea is to traverse the array by using two pointers to calculate subarrays with consecutive elements in a window of maximum length and then calculate all subarrays in that window using the above formula.

Below is the step by step algorithm:

  • Take two pointers say fast and slow, for maintaining a window of consecutive elements.
  • Start traversing the array.
  • If elements differ by 1 increment only the fast pointer.
  • Else, calculate the length of current window between the indexes fast and slow.

Below is the implementation of the given approach:

C++

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// C++ program to count Subarrays with
// Consecutive elements differing by 1
  
#include <iostream>
using namespace std;
  
// Function to count Subarrays with
// Consecutive elements differing by 1
int subarrayCount(int arr[], int n)
{
    // Variable to store count of subarrays
    // whose consecutive elements differ by 1
    int result = 0;
  
    // Take two pointers for maintaining a
    // window of consecutive elements
    int fast = 0, slow = 0;
  
    // Traverse the array
    for (int i = 1; i < n; i++) {
  
        // If elements differ by 1
        // increment only the fast pointer
        if (arr[i] - arr[i - 1] == 1) {
            fast++;
        }
        else {
  
            // Calculate length of subarray
            int len = fast - slow + 1;
  
            // Calculate total subarrays except 
            // Subarrays with single element
            result += len * (len - 1) / 2;
  
            // Update fast and slow
            fast = i;
            slow = i;
        }
    }
  
    // For last iteration. That is if array is
    // traversed and fast > slow
    if (fast != slow) {
        int len = fast - slow + 1;
        result += len * (len - 1) / 2;
    }
  
    return result;
}
  
// Driver Code
int main()
{
  
    int arr[] = { 1, 2, 3, 5, 6, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << subarrayCount(arr, n);
  
    return 0;
}

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Java

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// Java program to count Subarrays with
// Consecutive elements differing by 1
class cfg 
{
  
// Function to count Subarrays with
// Consecutive elements differing by 1
static int subarrayCount(int arr[], int n)
{
    // Variable to store count of subarrays
    // whose consecutive elements differ by 1
    int result = 0;
  
    // Take two pointers for maintaining a
    // window of consecutive elements
    int fast = 0, slow = 0;
  
    // Traverse the array
    for (int i = 1; i < n; i++) {
  
        // If elements differ by 1
        // increment only the fast pointer
        if (arr[i] - arr[i - 1] == 1) {
            fast++;
        }
        else {
  
            // Calculate length of subarray
            int len = fast - slow + 1;
  
            // Calculate total subarrays except 
            // Subarrays with single element
            result += len * (len - 1) / 2;
  
            // Update fast and slow
            fast = i;
            slow = i;
        }
    }
  
    // For last iteration. That is if array is
    // traversed and fast > slow
    if (fast != slow) {
        int len = fast - slow + 1;
        result += len * (len - 1) / 2;
    }
  
    return result;
}
  
// Driver Code
public static void main(String[] args)
{
  
    int arr[] = { 1, 2, 3, 5, 6, 7 };
    int n = arr.length;
  
    System.out.println(subarrayCount(arr, n));
  
}
}
//This code is contributed by Mukul Singh

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Python3

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# Python3 program to count Subarrays with 
# Consecutive elements differing by 1 
  
# Function to count Subarrays with 
# Consecutive elements differing by 1 
def subarrayCount(arr, n) :
      
    # Variable to store count of subarrays 
    # whose consecutive elements differ by 1 
    result = 0
  
    # Take two pointers for maintaining a 
    # window of consecutive elements 
    fast, slow = 0, 0
  
    # Traverse the array 
    for i in range(1, n) : 
  
        # If elements differ by 1 
        # increment only the fast pointer 
        if (arr[i] - arr[i - 1] == 1) : 
            fast += 1
          
        else
  
            # Calculate length of subarray 
            length = fast - slow + 1
  
            # Calculate total subarrays except 
            # Subarrays with single element 
            result += length * (length - 1) // 2
  
            # Update fast and slow 
            fast = i
            slow =
  
    # For last iteration. That is if array is 
    # traversed and fast > slow 
    if (fast != slow) : 
        length = fast - slow + 1
        result += length * (length - 1) // 2
      
    return result
  
# Driver Code 
if __name__ == "__main__" :
  
    arr = [ 1, 2, 3, 5, 6, 7 ]
    n = len(arr)
  
    print(subarrayCount(arr, n))
  
# This code is contributed by Ryuga

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C#

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// C# program to count Subarrays with
// Consecutive elements differing by 1
using System;
class cfg 
{
  
// Function to count Subarrays with
// Consecutive elements differing by 1
static int subarrayCount(int []arr, int n)
{
    // Variable to store count of subarrays
    // whose consecutive elements differ by 1
    int result = 0;
  
    // Take two pointers for maintaining a
    // window of consecutive elements
    int fast = 0, slow = 0;
  
    // Traverse the array
    for (int i = 1; i < n; i++) {
  
        // If elements differ by 1
        // increment only the fast pointer
        if (arr[i] - arr[i - 1] == 1) {
            fast++;
        }
        else {
  
            // Calculate length of subarray
            int len = fast - slow + 1;
  
            // Calculate total subarrays except 
            // Subarrays with single element
            result += len * (len - 1) / 2;
  
            // Update fast and slow
            fast = i;
            slow = i;
        }
    }
  
    // For last iteration. That is if array is
    // traversed and fast > slow
    if (fast != slow) {
        int len = fast - slow + 1;
        result += len * (len - 1) / 2;
    }
  
    return result;
}
  
// Driver Code
public static void Main()
{
  
    int []arr = { 1, 2, 3, 5, 6, 7 };
    int n = arr.Length;
  
    Console.WriteLine(subarrayCount(arr, n));
  
}
}
//This code is contributed by inder_verma..

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PHP

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<?php
// PHP program to count Subarrays with
// Consecutive elements differing by 1
  
// Function to count Subarrays with
// Consecutive elements differing by 1
function subarrayCount($arr, $n)
{
    // Variable to store count of subarrays
    // whose consecutive elements differ by 1
    $result = 0;
  
    // Take two pointers for maintaining a
    // window of consecutive elements
    $fast = 0; $slow = 0;
  
    // Traverse the array
    for ($i = 1; $i < $n; $i++)
    {
  
        // If elements differ by 1
        // increment only the fast pointer
        if ($arr[$i] - $arr[$i - 1] == 1) 
        {
            $fast++;
        }
        else
        {
  
            // Calculate length of subarray
            $len = $fast - $slow + 1;
  
            // Calculate total subarrays except 
            // Subarrays with single element
            $result += $len * ($len - 1) / 2;
  
            // Update fast and slow
            $fast = $i;
            $slow = $i;
        }
    }
  
    // For last iteration. That is if array 
    // is traversed and fast > slow
    if ($fast != $slow)
    {
        $len = $fast - $slow + 1;
        $result += $len * ($len - 1) / 2;
    }
  
    return $result;
}
  
// Driver Code
$arr = array(1, 2, 3, 5, 6, 7);
$n = sizeof($arr);
  
echo subarrayCount($arr, $n);
  
// This code is contributed 
// by Akanksha Rai
?>

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Output:

6

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



My Personal Notes arrow_drop_up

Striver(underscore)79 at Codechef and codeforces D

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