Dada una array arr[] de N enteros. La tarea es contar el número total de subarreglos de un arreglo dado de modo que la diferencia entre los elementos consecutivos en los subarreglos sea uno. Es decir, para cualquier índice 
en los subarreglos, arr[i+1] – arr[i] = 1 .
Nota : No considere subarreglos con un solo elemento.
Ejemplos:
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
Enfoque ingenuo : un enfoque simple es ejecutar dos bucles anidados y verificar cada subarreglo y calcular el recuento de subarreglos con elementos consecutivos que difieren en 1.
Enfoque eficiente : un enfoque eficiente es observar que en un arreglo de longitud digamos K , el número total de subarreglos de tamaño mayor que 1 = (K)*(K-1)/2.
Entonces, la idea es atravesar la array usando dos punteros para calcular subarreglos con elementos consecutivos en una ventana de longitud máxima y luego calcular todos los subarreglos en esa ventana usando la fórmula anterior.
A continuación se muestra el algoritmo paso a paso:
- Tome dos punteros para decir rápido y lento , para mantener una ventana de elementos consecutivos.
- Comience a atravesar la array.
- Si los elementos difieren en 1 incremento solo el puntero rápido.
- De lo contrario, calcule la longitud de la ventana actual entre los índices rápido y lento .
A continuación se muestra la implementación del enfoque dado:
C++
// 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;
}
Java
// 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
Python3
# 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 = i # 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
C#
// 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..
PHP
<?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
?>
Javascript
<script>
// Javascript 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
var result = 0;
// Take two pointers for maintaining a
// window of consecutive elements
var 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
var 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) {
var len = fast - slow + 1;
result += len * (len - 1) / 2;
}
return result;
}
// Driver Code
var arr = [ 1, 2, 3, 5, 6, 7 ];
var n = arr.length;
document.write(subarrayCount(arr, n));
// This code contributed by aashish1995
</script>
Producción:
6
Complejidad de tiempo : O(N), ya que estamos usando un bucle para recorrer N veces, por lo que la complejidad del programa será O(N).
Espacio auxiliar : O(1), ya que no estamos utilizando ningún espacio adicional.