Count of K-length subarray with each element less than X times the next one

Given an array A[] of length N and two integers X and K, The task is to count the number of indices i (0 ≤ i < N−k) such that:
X⁰⋅a_i < X¹⋅a_{i + 1} < X²⋅a_i+2 < . . . < X^k⋅a_i+k.

Examples:

Input: A[] = {9, 5, 3, 4, 1}, X = 4, K = 1.
Output: 3.
Explanation: Three Subarrays satisfy the conditions:
i = 0: the subarray [a0, a1] = [9, 5] and 1.9 < 4.5.
i = 1: the subarray [a1, a2] = [5, 3] and 1.5 < 4.3.
i = 2: the subarray [a2, a3] = [3, 2] and 1.3 < 4.4.
i = 3: the subarray [a3, a4] = [2, 1] but 1.4 = 4.1, so this subarray doesn't satisfy the condition.

Input: A[] = {22, 12, 16, 4, 3, 22, 12}, X = 2, K = 3.
Output: 1
Explanation: There are total 4 subarray out of which 1 satisfy the given condition.
i = 0: the subarray [a0, a1, a2, a3] = [22, 12, 16, 4] and 1.22 < 2.12 < 4.16 > 8.4, so this subarray doesn't satisfy the condition.
i = 1: the subarray [a1, a2, a3, a4]=[12, 16, 4, 3] and 1.12 < 2.16 > 4.4 < 8.3, so this subarray doesn't satisfy the condition.
i = 2: the subarray [a2, a3, a4, a5]=[16, 4, 3, 22] and 1.16 > 2.4 < 4.8 < 8.22, so this subarray doesn't satisfy the condition.
i = 3: the subarray [a3, a4, a5, a6]=[4, 3, 22, 12] and 1.4 < 2.3 < 4.22 < 8.12, so this subarray satisfies the condition.

Naive Approach: To solve the problem follow the below idea.

Find all the subarray of length K+1 and such that every element in the subarray is X times smaller than its next element in the subarray .

Follow the steps below to implement the idea:

• Run a for loop from i = 0 to i < N-(K+1). In each iteration:
  • Run a for loop from i to i+K and trace that every element in this subarray is X times smaller than the next element.
  • After termination of the loop if every element in this loop is X times smaller than the next element increment ans variable by 1.
• Print the answer.

 Below is the implementation for the above approach:

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function performing Calculation
int solve(int a[], int n, int k, int X)
{
    int ans = 0;
    for (int i = 0; i < n - (k + 1); i++) {
        bool flag = true;
        for (int j = i; j < k; j++) {
            if (a[j] < X * a[j + 1])
                continue;
            else {
                flag = false;
            }
        }
        if (flag)
            ans++;
    }
    return ans;
}

// Driver Code
int main()
{
    int A[] = { 9, 5, 3, 4, 1 };
    int N = sizeof(A) / sizeof(A[0]);
    int K = 1, X = 4;

    // Function Call
    cout << solve(A, N, K, X);
    return 0;
}

// Java code to implement the approach
import java.io.*;

class GFG {

  // Function performing Calculation
  static int solve(int[] a, int n, int k, int X)
  {
    int ans = 0;
    for (int i = 0; i < n - (k + 1); i++) {
      boolean flag = true;
      for (int j = i; j < k; j++) {
        if (a[j] < X * a[j + 1]) {
          continue;
        }
        else {
          flag = false;
        }
      }
      if (flag) {
        ans++;
      }
    }
    return ans;
  }

  public static void main(String[] args)
  {
    int[] A = { 9, 5, 3, 4, 1 };
    int N = A.length;
    int K = 1, X = 4;

    // Function call
    System.out.print(solve(A, N, K, X));
  }
}

// This code is contributed by lokesh (lokeshmvs21).

# Python code to implement the approach

# Function performing Calculation
def solve(a, n, k, X):
    ans = 0
    for i in range(n - (k + 1)):
        flag = True
        for j in range(i, k):
            if a[j] < X * a[j + 1]:
                continue
            else:
                flag = False
        if flag:
            ans += 1
    return ans


# Driver Code
A = [9, 5, 3, 4, 1]
N = len(A)
K = 1
X = 4

# Function Call
print(solve(A, N, K, X))

# This code is contributed by Tapesh(tapeshdua420)

// C# code to implement the approach
using System;
using System.Collections.Generic;

public class GFG{

  // Function performing Calculation
  static int solve(int[] a, int n, int k, int X)
  {
    int ans = 0;
    for (int i = 0; i < n - (k + 1); i++) {
      bool flag = true;
      for (int j = i; j < k; j++) {
        if (a[j] < X * a[j + 1]) {
          continue;
        }
        else {
          flag = false;
        }
      }
      if (flag) {
        ans++;
      }
    }
    return ans;
  }

// Driver Code
static public void Main (){

    int[] A = { 9, 5, 3, 4, 1 };
    int N = A.Length;
    int K = 1, X = 4;

    // Function call
    Console.WriteLine(solve(A, N, K, X));
}
}

// This code is contributed by sanjoy_62.

<script>
// Javascript code to implement the approach

// Function performing Calculation
function solve(a, n, k, X)
{
    let ans = 0;
    for (let i = 0; i < n - (k + 1); i++) {
        let flag = true;
        for (let j = i; j < k; j++) {
            if (a[j] < X * a[j + 1])
                continue;
            else {
                flag = false;
            }
        }
        if (flag)
            ans++;
    }
    return ans;
}

// Driver Code
    let A = [ 9, 5, 3, 4, 1 ];
    let N = A.length;
    let K = 1, X = 4;

    // Function Call
    document.write(solve(A, N, K, X));
    
    // This code is contributed by satwik4409.
    </script>  

3

Time complexity: O(N * K)
Auxiliary Space: O(1).

Efficient Approach: To solve the problem follow the below idea.

Using two pointer approach and checking if every element in the subarray is 4 times smaller than its next element and then moving the window forward till the last element is reached.

Follow the steps below to implement the idea:

• Run a while loop from j = 0 to j < N - 1. In each iteration:
  • Check if  the length of window is equal to K + 1 (i.e j - i + 1 = k + 1) then increment answer by 1.
  • If j^th element is less than X times (j + 1)^th element (A[j] < X*A[j + 1]) then increment j by 1 otherwise move i pointer to j since no subarray that contains j^th and (j+1)^th element together can satisfy the given condition.
• Print the answer.

Below is the implementation for the above approach:

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function performing Calculation
int solve(int a[], int n, int k, int X)
{
    int ans = 0;
    int i = 0, j = 0;

    // Loop to utilize the sliding window concept
    while (j < n - 1) {
        if (j - i + 1 == k + 1) {
            i++;
            ans++;
        }
        if (a[j] < a[j + 1] * X) {
            j++;
        }
        else {
            i = j;
            j++;
        }
    }
    return ans;
}

// Driver Code
int main()
{
    int A[] = { 9, 5, 3, 4, 1 };
    int N = sizeof(A) / sizeof(A[0]);
    int K = 1, X = 4;

    // Function Call
    cout << solve(A, N, K, X);
    return 0;
}

// JAVA code to implement the approach

import java.util.*;
class GFG {
  // Function performing Calculation
  public static int solve(int a[], int n, int k, int X)
  {
    int ans = 0;
    int i = 0, j = 0;

    // Loop to utilize the sliding window concept
    while (j < n - 1) {
      if (j - i + 1 == k + 1) {
        i++;
        ans++;
      }
      if (a[j] < a[j + 1] * X) {
        j++;
      }
      else {
        i = j;
        j++;
      }
    }
    return ans;
  }

  // Driver Code
  public static void main(String[] args)
  {
    int A[] = { 9, 5, 3, 4, 1 };
    int N = A.length;
    int K = 1, X = 4;

    // Function Call
    System.out.print(solve(A, N, K, X));
  }
}

// This code is contributed by Taranpreet

# Python code to implement the approach

# Function performing Calculation
def solve(a, n, k, X):
    ans = 0
    for i in range(n - (k + 1)):
        flag = True
        for j in range(i, k):
            if a[j] < X * a[j + 1]:
                continue
            else:
                flag = False
        if flag:
            ans += 1
    return ans

if __name__ == '__main__':
    A = [9, 5, 3, 4, 1]
    N = len(A)
    K = 1
    X = 4

    # Function Call
    print(solve(A, N, K, X))

# This code is contributed by Tapesh(tapeshdua420)

// C# code to implement the approach

using System;
class GFG {
    // Function performing Calculation
    public static int solve(int[] a, int n, int k, int X)
    {
        int ans = 0;
        int i = 0, j = 0;

        // Loop to utilize the sliding window concept
        while (j < n - 1) {
            if (j - i + 1 == k + 1) {
                i++;
                ans++;
            }
            if (a[j] < a[j + 1] * X) {
                j++;
            }
            else {
                i = j;
                j++;
            }
        }
        return ans;
    }

    // Driver Code
    public static void Main(string[] args)
    {
        int[] A = { 9, 5, 3, 4, 1 };
        int N = A.Length;
        int K = 1, X = 4;

        // Function Call
        Console.WriteLine(solve(A, N, K, X));
    }
}

// This code is contributed by Tapesh(tapeshdua420)

    <script>
        // JavaScript code to implement the approach

        // Function performing Calculation
        const solve = (a, n, k, X) => {
            let ans = 0;
            let i = 0, j = 0;

            // Loop to utilize the sliding window concept
            while (j < n - 1) {
                if (j - i + 1 == k + 1) {
                    i++;
                    ans++;
                }
                if (a[j] < a[j + 1] * X) {
                    j++;
                }
                else {
                    i = j;
                    j++;
                }
            }
            return ans;
        }

        // Driver Code

        let A = [9, 5, 3, 4, 1];
        let N = A.length;
        let K = 1, X = 4;

        // Function Call
        document.write(solve(A, N, K, X));

        // This code is contributed by rakeshsahni

    </script>

3

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