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Count of subarrays with largest element at least twice the largest of remaining elements

  • Last Updated : 13 Jul, 2021

Given an array arr[] consisting of N positive integers, the task is to find the count of subarrays such that the maximum element of the subarray is greater than twice the maximum of all other elements of the array.

Examples:

Input: arr[] = {1, 6, 10, 9, 7, 3}
Output: 4
Explanation:
Below are the subarrays satisfying the given condition:

  1. Consider the subarray {6, 10, 9, 7}. Now the maximum element of this subarray is 10 which is greater than twice the maximum elements of the remaining array elements i.e., 2*max{1, 3} = 2*3 = 6.
  2. Consider the subarray {6, 10, 9, 7, 3}. Now the maximum element of this subarray is 10 which is greater than twice the maximum elements of the remaining array elements i.e., 2*max{1} = 2*1 = 2.
  3. Consider the subarray {1, 6, 10, 9, 7}. Now the maximum element of this subarray is 10 which is greater than twice the maximum elements of the remaining array elements i.e., 2*max{3} = 2*3 = 6.
  4. Consider the subarray {1, 6, 10, 9, 7, 3}. Now the maximum element of this subarray is 10 which is greater than twice the maximum elements of the remaining array elements i.e., 2*max{} = 2*0 = 0.

Therefore, the total number of subarrays is 4.

Input: arr[] = {1, 10, 2, 3}
Output: 6



Naive Approach: The simplest approach to solve the given problem is to generate all possible subarrays of the given array arr[] and then count the number of subarrays having a maximum element greater than twice the maximum of all other elements. After checking for all the subarrays, print the count of the subarray obtained.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <iostream>
using namespace std;
 
// Function to find count of subarrays
// which have max element greater than
// twice maximum of all other elements
void countSubarray(int arr[], int n)
{
    // Stores the count of subarrays
    int count = 0;
 
    // Generate all possible subarrays
    for (int i = 0; i < n; i++) {
 
        for (int j = i; j < n; j++) {
 
            // Stores the maximum element
            // of the subarray
            int mxSubarray = 0;
 
            // Stores the maximum of all
            // other elements
            int mxOther = 0;
 
            // Find the maximum element
            // in the subarray [i, j]
            for (int k = i; k <= j; k++) {
                mxSubarray = max(mxSubarray,
                                 arr[k]);
            }
 
            // Find the maximum of all
            // other elements
            for (int k = 0; k < i; k++) {
                mxOther = max(
                    mxOther, arr[k]);
            }
 
            for (int k = j + 1; k < n; k++) {
                mxOther = max(
                    mxOther, arr[k]);
            }
 
            // If the maximum of subarray
            // is greater than twice the
            // maximum of other elements
            if (mxSubarray > (2 * mxOther))
                count++;
        }
    }
 
    // Print the maximum value obtained
    cout << count;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 6, 10, 9, 7, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);
    countSubarray(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
 
class GFG{
     
// Function to find count of subarrays
// which have max element greater than
// twice maximum of all other elements
public static void countSubarray(int arr[], int n)
{
     
    // Stores the count of subarrays
    int count = 0;
 
    // Generate all possible subarrays
    for(int i = 0; i < n; i++)
    {
        for(int j = i; j < n; j++)
        {
             
            // Stores the maximum element
            // of the subarray
            int mxSubarray = 0;
 
            // Stores the maximum of all
            // other elements
            int mxOther = 0;
 
            // Find the maximum element
            // in the subarray [i, j]
            for(int k = i; k <= j; k++)
            {
                mxSubarray = Math.max(
                    mxSubarray, arr[k]);
            }
 
            // Find the maximum of all
            // other elements
            for(int k = 0; k < i; k++)
            {
                mxOther = Math.max(mxOther, arr[k]);
            }
 
            for(int k = j + 1; k < n; k++)
            {
                mxOther = Math.max(mxOther, arr[k]);
            }
 
            // If the maximum of subarray
            // is greater than twice the
            // maximum of other elements
            if (mxSubarray > (2 * mxOther))
                count++;
        }
    }
 
    // Print the maximum value obtained
    System.out.println(count);
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 6, 10, 9, 7, 3 };
    int N = arr.length;
     
    countSubarray(arr, N);
}
}
 
// This code is contributed by Potta Lokesh

Python3




# Python 3 program for the above approach
 
# Function to find count of subarrays
# which have max element greater than
# twice maximum of all other elements
def countSubarray(arr, n):
    # Stores the count of subarrays
    count = 0
 
    # Generate all possible subarrays
    for i in range(n):
        for j in range(i, n, 1):
            # Stores the maximum element
            # of the subarray
            mxSubarray = 0
 
            # Stores the maximum of all
            # other elements
            mxOther = 0
 
            # Find the maximum element
            # in the subarray [i, j]
            for k in range(i, j + 1, 1):
                mxSubarray = max(mxSubarray, arr[k])
 
            # Find the maximum of all
            # other elements
            for k in range(0, i, 1):
                mxOther = max(mxOther, arr[k])
 
            for k in range(j + 1,n,1):
                mxOther = max(mxOther, arr[k])
 
            # If the maximum of subarray
            # is greater than twice the
            # maximum of other elements
            if (mxSubarray > (2 * mxOther)):
                count += 1
 
    # Print the maximum value obtained
    print(count)
 
# Driver Code
if __name__ == '__main__':
    arr = [1, 6, 10, 9, 7, 3]
    N = len(arr)
    countSubarray(arr, N)
     
    # This code is contributed by bgangwar59.

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find count of subarrays
// which have max element greater than
// twice maximum of all other elements
static void countSubarray(int []arr, int n)
{
     
    // Stores the count of subarrays
    int count = 0;
 
    // Generate all possible subarrays
    for(int i = 0; i < n; i++)
    {
        for(int j = i; j < n; j++)
        {
             
            // Stores the maximum element
            // of the subarray
            int mxSubarray = 0;
 
            // Stores the maximum of all
            // other elements
            int mxOther = 0;
 
            // Find the maximum element
            // in the subarray [i, j]
            for(int k = i; k <= j; k++)
            {
                mxSubarray = Math.Max(mxSubarray,
                                      arr[k]);
            }
 
            // Find the maximum of all
            // other elements
            for(int k = 0; k < i; k++)
            {
                mxOther = Math.Max(mxOther, arr[k]);
            }
 
            for(int k = j + 1; k < n; k++)
            {
                mxOther = Math.Max(mxOther, arr[k]);
            }
 
            // If the maximum of subarray
            // is greater than twice the
            // maximum of other elements
            if (mxSubarray > (2 * mxOther))
                count++;
        }
    }
 
    // Print the maximum value obtained
    Console.Write(count);
}
 
// Driver Code
public static void Main()
{
    int []arr = { 1, 6, 10, 9, 7, 3 };
    int N = arr.Length;
     
    countSubarray(arr, N);
}
}
 
// This code is contributed by SURENDRA_GANGWAR

Javascript




<script>
 
// JavaScript program for the above approach
 
// Function to find count of subarrays
// which have max element greater than
// twice maximum of all other elements
function countSubarray(arr, n)
{
     
    // Stores the count of subarrays
    let count = 0;
 
    // Generate all possible subarrays
    for(let i = 0; i < n; i++)
    {
        for(let j = i; j < n; j++)
        {
             
            // Stores the maximum element
            // of the subarray
            let mxSubarray = 0;
             
            // Stores the maximum of all
            // other elements
            let mxOther = 0;
 
            // Find the maximum element
            // in the subarray [i, j]
            for(let k = i; k <= j; k++)
            {
                mxSubarray = Math.max(mxSubarray,
                                      arr[k]);
            }
 
            // Find the maximum of all
            // other elements
            for(let k = 0; k < i; k++)
            {
                mxOther = Math.max(mxOther, arr[k]);
            }
 
            for(let k = j + 1; k < n; k++)
            {
                mxOther = Math.max(
                    mxOther, arr[k]);
            }
 
            // If the maximum of subarray
            // is greater than twice the
            // maximum of other elements
            if (mxSubarray > (2 * mxOther))
                count++;
        }
    }
 
    // Print the maximum value obtained
    document.write(count);
}
 
// Driver Code
let arr = [ 1, 6, 10, 9, 7, 3 ];
let N = arr.length;
 
countSubarray(arr, N);
 
// This code is contributed by Potta Lokesh
   
</script>
Output: 
4

 

Time Complexity: O(N3)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized using the observation that the maximum element of the array will always be a part of the subarray and all elements having a value greater than half of the maximum element is also included in the subarray. Follow the steps below to solve the problem:

  • Initialize a variable, say mx that stores the maximum element of the array.
  • Initialize two variables L and R to store the left and the right endpoints of the subarray.
  • Iterate over the range [0, N – 1] using the variable i and if the value of 2*arr[i] is greater than mx then initialize L to i and break from the loop.
  • Iterate over the range [N – 1, 0] in a reverse manner using the variable i and if the value of 2*arr[i] is greater than mx then initialize R to i and break from the loop.
  • After completing the above steps, print the value of (L + 1)*(N – R) as the result.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find count of subarrays
// which have max element greater than
// twice maximum of all other elements
void countSubarray(int arr[], int n)
{
    int count = 0, L = 0, R = 0;
 
    // Stores the maximum element of
    // the array
    int mx = *max_element(arr, arr + n);
 
    // Traverse the given array
    for (int i = 0; i < n; i++) {
 
        // If the value of 2*arr[i] is
        // greater than mx
        if (arr[i] * 2 > mx) {
 
            // Update the value of L
            // and break out of loop
            L = i;
            break;
        }
    }
 
    for (int i = n - 1; i >= 0; i--) {
 
        // If the value 2*arr[i] is
        // greater than mx
        if (arr[i] * 2 > mx) {
 
            // Update the value of R
            // and break out of loop
            R = i;
            break;
        }
    }
 
    // Print the final answer
    cout << (L + 1) * (n - R);
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 6, 10, 9, 7, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);
    countSubarray(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG {
 
    // Function to find count of subarrays
    // which have max element greater than
    // twice maximum of all other elements
    static void countSubarray(int[] arr, int n)
    {
        int L = 0, R = 0;
 
        // Stores the maximum element of
        // the array
 
        int mx = Integer.MIN_VALUE;
        for (int i = 0; i < n; i++)
            mx = Math.max(mx, arr[i]);
 
        // Traverse the given array
        for (int i = 0; i < n; i++) {
 
            // If the value of 2*arr[i] is
            // greater than mx
            if (arr[i] * 2 > mx) {
 
                // Update the value of L
                // and break out of loop
                L = i;
                break;
            }
        }
 
        for (int i = n - 1; i >= 0; i--) {
 
            // If the value 2*arr[i] is
            // greater than mx
            if (arr[i] * 2 > mx) {
 
                // Update the value of R
                // and break out of loop
                R = i;
                break;
            }
        }
 
        // Print the final answer
        System.out.println((L + 1) * (n - R));
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] arr = { 1, 6, 10, 9, 7, 3 };
        int N = arr.length;
        countSubarray(arr, N);
    }
}
 
// This code is contributed by rishavmahato348.

Python3




# Python3 program for the above approach
 
# Function to find count of subarrays
# which have max element greater than
# twice maximum of all other elements
def countSubarray(arr, n):
     
    count = 0
    L = 0
    R = 0
 
    # Stores the maximum element of
    # the array
    mx = max(arr)
 
    # Traverse the given array
    for i in range(n):
         
        # If the value of 2*arr[i] is
        # greater than mx
        if (arr[i] * 2 > mx):
 
            # Update the value of L
            # and break out of loop
            L = i
            break
     
    i = n - 1
     
    while (i >= 0):
         
        # If the value 2*arr[i] is
        # greater than mx
        if (arr[i] * 2 > mx):
 
            # Update the value of R
            # and break out of loop
            R = i
            break
         
        i -= 1
 
    # Print the final answer
    print((L + 1) * (n - R))
 
# Driver Code
if __name__ == '__main__':
     
    arr = [ 1, 6, 10, 9, 7, 3 ]
    N = len(arr)
     
    countSubarray(arr, N)
         
# This code is contributed by SURENDRA_GANGWAR

C#




// C# program for the above approach
using System;
class GFG {
 
    // Function to find count of subarrays
    // which have max element greater than
    // twice maximum of all other elements
    static void countSubarray(int[] arr, int n)
    {
        int L = 0, R = 0;
 
        // Stores the maximum element of
        // the array
 
        int mx = Int32.MinValue;
        for (int i = 0; i < n; i++)
            mx = Math.Max(mx, arr[i]);
 
        // Traverse the given array
        for (int i = 0; i < n; i++) {
 
            // If the value of 2*arr[i] is
            // greater than mx
            if (arr[i] * 2 > mx) {
 
                // Update the value of L
                // and break out of loop
                L = i;
                break;
            }
        }
 
        for (int i = n - 1; i >= 0; i--) {
 
            // If the value 2*arr[i] is
            // greater than mx
            if (arr[i] * 2 > mx) {
 
                // Update the value of R
                // and break out of loop
                R = i;
                break;
            }
        }
 
        // Print the final answer
        Console.WriteLine((L + 1) * (n - R));
    }
 
    // Driver Code
    public static void Main()
    {
        int[] arr = { 1, 6, 10, 9, 7, 3 };
        int N = arr.Length;
        countSubarray(arr, N);
    }
}
 
// This code is contributed by ukasp.

Javascript




<script>
 
// javascript program for the above approach
 
// Function to find count of subarrays
// which have max element greater than
// twice maximum of all other elements
function countSubarray(arr, n)
{
    var count = 0, L = 0, R = 0;
 
    // Stores the maximum element of
    // the array
    var mx = Math.max.apply(null, arr);
     
    var i;
    // Traverse the given array
    for (i = 0; i < n; i++) {
 
        // If the value of 2*arr[i] is
        // greater than mx
        if (arr[i] * 2 > mx) {
 
            // Update the value of L
            // and break out of loop
            L = i;
            break;
        }
    }
 
    for (i = n - 1; i >= 0; i--) {
 
        // If the value 2*arr[i] is
        // greater than mx
        if (arr[i] * 2 > mx) {
 
            // Update the value of R
            // and break out of loop
            R = i;
            break;
        }
    }
 
    // Print the final answer
    document.write((L + 1) * (n - R));
}
 
// Driver Code
 
    var arr = [1, 6, 10, 9, 7, 3]
    var N = arr.length;
    countSubarray(arr, N);
 
// This code is contributed by ipg2016107.
</script>
Output: 
4

 

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

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