Recuento de subarreglos que contiene un número dado exactamente K veces

Dada una array A[] de N elementos que consta de valores de 1 a N con duplicados, la tarea es encontrar el número total de subarreglos que contienen un número dado num exactamente K veces.

Ejemplos: 

Entrada: A[] = {1, 2, 1, 5, 1}, num = 1, K = 2 
Salida: 2 
Explicación: 
Subarreglos {1, 2, 1, 5}, {1, 2, 1}, { 2, 1, 5, 1} y {1, 5, 1} contiene 1 exactamente dos veces.

Entrada: A[] = {1, 5, 3, 5, 7, 5, 6, 5, 10, 5, 12, 5}, num = 5, K = 3 
Salida: 14 

Enfoque ingenuo: una solución simple es generar todos los subarreglos de la array dada y contar el número de subarreglos en los que el número dado aparece exactamente K veces.

Complejidad temporal: O(N ² ) donde N es el tamaño de la array dada.

Enfoque eficiente: 

• Almacene los índices que contienen el número dado num .
• Recorra la array indices[] y calcule el número de subarreglos posibles para cada índice K.
• El número de subarreglos posibles para cualquier índice K de num es igual al

Producto de (i ^th index – (i-1) ^th index) y ((i + K) ^th index – (i+(K – 1)) ^th index) 
 

• El conteo de todos esos subarreglos da el conteo del total de subarreglos posibles en el arreglo dado.

A continuación se muestra la implementación del enfoque anterior:

// C++ program to count subarrays
// which contains a given number
// exactly K times
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to return
// the count of subarrays
// which contains given
// number exactly K times
int countSubarrays(int A[], int num,
                   int K, int size)
{
    // Store the indices
    // containing num
    vector<int> indices;
    for (int i = 0; i < size; i++) {
        if (A[i] == num)
            indices.push_back(i);
    }
 
    // If the occurrence of num
    // in the entire array
    // is less than K
    if (indices.size() < K)
 
        // No such subarrays are possible
        return 0;
 
    // Store the previous
    // index of num
    int prev = -1;
 
    // Store the count of
    // total subarrays
    int ans = 0;
 
    // Store the count of
    // subarrays for current
    // K occurrences
    int ctr = 0;
 
    for (int i = 0;
         i <= indices.size() - K;
         i++) {
 
        ctr = indices[i] - prev;
 
        if (i < indices.size() - K) {
 
            ctr *= (indices[i + K]
                    - indices[i + K - 1]);
        }
        else {
            ctr *= ((size - 1)
                    - indices[i + K - 1] + 1);
        }
 
        ans += ctr;
        prev = indices[i];
    }
 
    return ans;
}
 
// Driver code
int main()
{
    int A[] = { 1, 5, 3, 5, 7, 5, 6,
                5, 10, 5, 12, 5 };
 
    int num = 5;
 
    int K = 3;
 
    int size = sizeof(A) / sizeof(int);
 
    cout << countSubarrays(A, num, K, size);
 
    return 0;
}

// Java program to count subarrays
// which contains a given number
// exactly K times
 
import java.util.*;
public class Main {
 
    // Function to return
    // the count of subarrays
    // which contains given
    // number exactly K times
    public static int countSubarrays(
        int A[], int num,
        int K, int size)
    {
        // Store the indices
        // containing num
        ArrayList<Integer> indices
            = new ArrayList<Integer>();
 
        for (int i = 0; i < size; i++) {
            if (A[i] == num) {
                indices.add(i);
            }
        }
 
        if (indices.size() < K) {
            return 0;
        }
 
        // Store the previous
        // index of num
        int prev = -1;
 
        // Store the count of
        // total subarrays
        int ans = 0;
 
        // Store the count of
        // subarrays for current
        // K occurrences
        int ctr = 0;
 
        for (int i = 0;
             i <= indices.size() - K;
             i++) {
 
            ctr = indices.get(i) - prev;
 
            if (i < indices.size() - K) {
 
                ctr *= (indices.get(i + K)
                        - indices.get(i + K - 1));
            }
            else {
                ctr *= ((size - 1)
                        - indices.get(i + K - 1) + 1);
            }
 
            ans += ctr;
            prev = indices.get(i);
        }
 
        return ans;
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        int A[] = { 1, 5, 3, 5, 7, 5,
                    6, 5, 10, 5, 12, 5 };
 
        int num = 5;
 
        int K = 3;
 
        int size = A.length;
 
        System.out.println(
            countSubarrays(A, num, K, size));
    }
}

# Python3 program to
# count subarrays which
# contains a given number
# exactly K times
 
# Function to return
# the count of subarrays
# which contains given
# number exactly K times
def countSubarrays(A, num,
                   K, size):
  # Store the indices
  # containing num
  indices = []
  for i in range (size):
    if (A[i] == num):
      indices.append(i)
       
  # If the occurrence of num
  # in the entire array
  # is less than K
  if (len(indices) < K):
     
    # No such subarrays are possible
    return 0
   
  # Store the previous
  # index of num
  prev = -1
 
  # Store the count of
  # total subarrays
  ans = 0
 
  # Store the count of
  # subarrays for current
  # K occurrences
  ctr = 0
   
  for i in range (len(indices) - K + 1):
    ctr = indices[i] - prev
    if (i < len(indices) - K):
      ctr *= (indices[i + K] -
              indices[i + K - 1])       
    else:
      ctr *= ((size - 1) -
               indices[i + K - 1] + 1)
    ans += ctr
    prev = indices[i]
  return ans
 
# Driver code
if __name__ == "__main__":
  A = [1, 5, 3, 5, 7, 5,
       6, 5, 10, 5, 12, 5]
  num = 5
  K = 3
  size = len(A)
  print(countSubarrays(A, num, K, size))
     
# This code is contributed by Chitranayal

// C# program to count subarrays
// which contains a given number
// exactly K times
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG{
 
// Function to return the count of subarrays
// which contains given number exactly K times
public static int countSubarrays(int[] A, int num,
                                 int K, int size)
{
     
    // Store the indices
    // containing num
    ArrayList indices = new ArrayList();
 
    for(int i = 0; i < size; i++)
    {
        if (A[i] == num)
        {
            indices.Add(i);
        }
    }
 
    if (indices.Count < K)
    {
        return 0;
    }
 
    // Store the previous
    // index of num
    int prev = -1;
 
    // Store the count of
    // total subarrays
    int ans = 0;
 
    // Store the count of
    // subarrays for current
    // K occurrences
    int ctr = 0;
 
    for(int i = 0;
            i <= indices.Count - K;
            i++)
    {
        ctr = (int)indices[i] - prev;
        if (i < indices.Count - K)
        {
            ctr *= ((int)indices[i + K] -
                    (int)indices[i + K - 1]);
        }
        else
        {
            ctr *= ((size - 1) -
                    (int)indices[i + K - 1] + 1);
        }
        ans += ctr;
        prev = (int)indices[i];
    }
    return ans;
}
 
// Driver code
static public void Main()
{
    int[] A = { 1, 5, 3, 5, 7, 5,
                6, 5, 10, 5, 12, 5 };
 
    int num = 5;
    int K = 3;
    int size = A.Length;
 
    Console.WriteLine(countSubarrays(A, num, K, size));
}
}
 
// This code is contributed by akhilsaini

<script>
 
// JavaScript program to count subarrays
// which contains a given number
// exactly K times
 
// Function to return the count of subarrays
// which contains given number exactly K times
function countSubarrays(A, num, K, size)
{
      
    // Store the indices
    // containing num
    let indices = [];
  
    for(let i = 0; i < size; i++)
    {
        if (A[i] == num)
        {
            indices.push(i);
        }
    }
  
    if (indices.length < K)
    {
        return 0;
    }
  
    // Store the previous
    // index of num
    let prev = -1;
  
    // Store the count of
    // total subarrays
    let ans = 0;
  
    // Store the count of
    // subarrays for current
    // K occurrences
    let ctr = 0;
  
    for(let i = 0;
            i <= indices.length - K;
            i++)
    {
        ctr = indices[i] - prev;
        if (i < indices.length - K)
        {
            ctr *= (indices[i + K] -
                    indices[i + K - 1]);
        }
        else
        {
            ctr *= ((size - 1) -
                    indices[i + K - 1] + 1);
        }
        ans += ctr;
        prev = indices[i];
    }
    return ans;
}
  
  // Driver Code
     
    let A = [ 1, 5, 3, 5, 7, 5,
                6, 5, 10, 5, 12, 5 ];
  
    let num = 5;
    let K = 3;
    let size = A.length;
  
    document.write(countSubarrays(A, num, K, size));
            
</script>

14

Complejidad de tiempo: O(N) , donde N es el tamaño de la array. 
Complejidad espacial: O(N)