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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 ``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);``    ``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

 ``
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 ``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);``    ``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

 ``
Output:
`4`

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

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