Longest Subarrays having each Array element as the maximum

Given an array arr[] of length N, the task is to find the longest subarray for each array element arr[i], which contains arr[i] as the maximum.

Examples:

Input: arr[] = {1, 2, 3, 0, 1} 
Output: 1 2 5 1 2 
Explanation:
The longest subarray having arr[0] as the largest is {1} 
The longest subarray having arr[1] as the largest is {1, 2} 
The longest subarray having arr[2] as the largest is {1, 2, 3, 0, 1} 
The longest subarray having arr[3] as the largest is {0} 
The longest subarray having arr[4 as the largest is {0, 1}

Input: arr[] = {3, 3, 3, 1, 6, 2} 
Output: 4 4 4 1 6 1

Approach: The idea is to use Two Pointer technique to solve the problem: 

• Initialize two pointers, left and right. such that for every element arr[i], the left points to indices [i – 1, 0] to find elements smaller than or equal to arr[i] in a contiguous manner. The moment an element larger than arr[i] is found, the pointer stops.
• Similarly, right points to indices [i + 1, n – 1] to find elements smaller than or equal to arr[i] in a contiguous manner, and stops on finding any element larger than arr[i].
• Therefore, the largest contiguous subarray where arr[i] is largest is of length 1 + right – left.
• Repeat the above steps for every array element.

Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the maximum length of
// Subarrays for each element of the array
// having it as the maximum
void solve(int n, int arr[])
{
    int i, ans = 0;
    for (i = 0; i < n; i++) {

        // Initialize the bounds
        int left = max(i - 1, 0);
        int right = min(n - 1, i + 1);

        // Iterate to find greater
        // element on the left
        while (left >= 0) {

            // If greater element is found
            if (arr[left] > arr[i]) {
                left++;
                break;
            }

            // Decrement left pointer
            left--;
        }

        // If boundary is exceeded
        if (left < 0)
            left++;

        // Iterate to find greater
        // element on the right
        while (right < n) {

            // If greater element is found
            if (arr[right] > arr[i]) {
                right--;
                break;
            }
            // Increment right pointer
            right++;
        }

        // If boundary is exceeded
        if (right >= n)
            right--;

        // Length of longest subarray where
        // arr[i] is the largest
        ans = 1 + right - left;

        // Print the answer
        cout << ans << " ";
    }
}

// Driver Code
int main()
{
    int arr[] = { 4, 2, 1 };
    int n = sizeof arr / sizeof arr[0];

    solve(n, arr);

    return 0;
}

Java

// Java Program to implement
// the above approach
import java.util.*;
class GFG{

// Function to find the maximum length of
// Subarrays for each element of the array
// having it as the maximum
static void solve(int n, int arr[])
{
    int i, ans = 0;
    for (i = 0; i < n; i++)
    {

        // Initialize the bounds
        int left = Math.max(i - 1, 0);
        int right = Math.min(n - 1, i + 1);

        // Iterate to find greater
        // element on the left
        while (left >= 0)
        {

            // If greater element is found
            if (arr[left] > arr[i])
            {
                left++;
                break;
            }

            // Decrement left pointer
            left--;
        }

        // If boundary is exceeded
        if (left < 0)
            left++;

        // Iterate to find greater
        // element on the right
        while (right < n)
        {

            // If greater element is found
            if (arr[right] > arr[i])
            {
                right--;
                break;
            }

            // Increment right pointer
            right++;
        }

        // If boundary is exceeded
        if (right >= n)
            right--;

        // Length of longest subarray where
        // arr[i] is the largest
        ans = 1 + right - left;

        // Print the answer
        System.out.print(ans + " ");
    }
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 4, 2, 1 };
    int n = arr.length;

    solve(n, arr);
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 program to implement
# the above approach

# Function to find the maximum length of
# Subarrays for each element of the array
# having it as the maximum
def solve(n, arr):

    ans = 0

    for i in range(n):

        # Initialise the bounds
        left = max(i - 1, 0)
        right = min(n - 1, i + 1)

        # Iterate to find greater
        # element on the left
        while left >= 0:

            # If greater element is found
            if arr[left] > arr[i]:
                left += 1
                break

            # Decrement left pointer
            left -= 1

        # If boundary is exceeded
        if left < 0:
            left += 1

        # Iterate to find greater
        # element on the right
        while right < n:

            # If greater element is found
            if arr[right] > arr[i]:
                right -= 1
                break

            # Increment right pointer
            right += 1

        # if boundary is exceeded
        if right >= n:
            right -= 1

        # Length of longest subarray where
        # arr[i] is the largest
        ans = 1 + right - left

        # Print the answer
        print(ans, end = " ")

# Driver code
arr = [ 4, 2, 1 ]
n = len(arr)

solve(n, arr)

# This code is contributed by Stuti Pathak

C#

// C# Program to implement
// the above approach
using System;
class GFG{

// Function to find the maximum length of
// Subarrays for each element of the array
// having it as the maximum
static void solve(int n, int []arr)
{
    int i, ans = 0;
    for (i = 0; i < n; i++)
    {

        // Initialize the bounds
        int left = Math.Max(i - 1, 0);
        int right = Math.Min(n - 1, i + 1);

        // Iterate to find greater
        // element on the left
        while (left >= 0)
        {

            // If greater element is found
            if (arr[left] > arr[i])
            {
                left++;
                break;
            }

            // Decrement left pointer
            left--;
        }

        // If boundary is exceeded
        if (left < 0)
            left++;

        // Iterate to find greater
        // element on the right
        while (right < n)
        {

            // If greater element is found
            if (arr[right] > arr[i])
            {
                right--;
                break;
            }

            // Increment right pointer
            right++;
        }

        // If boundary is exceeded
        if (right >= n)
            right--;

        // Length of longest subarray where
        // arr[i] is the largest
        ans = 1 + right - left;

        // Print the answer
        Console.Write(ans + " ");
    }
}

// Driver Code
public static void Main(String[] args)
{
    int []arr = { 4, 2, 1 };
    int n = arr.Length;

    solve(n, arr);
}
}

// This code is contributed by Princi Singh

Javascript

<script>

// JavaScript program to implement
// the above approach

// Function to find the maximum length of
// Subarrays for each element of the array
// having it as the maximum
function solve(n, arr)
{
    let i, ans = 0;
    for (i = 0; i < n; i++)
    {

        // Initialize the bounds
        let left = Math.max(i - 1, 0);
        let right = Math.min(n - 1, i + 1);

        // Iterate to find greater
        // element on the left
        while (left >= 0)
        {

            // If greater element is found
            if (arr[left] > arr[i])
            {
                left++;
                break;
            }

            // Decrement left pointer
            left--;
        }

        // If boundary is exceeded
        if (left < 0)
            left++;

        // Iterate to find greater
        // element on the right
        while (right < n)
        {

            // If greater element is found
            if (arr[right] > arr[i])
            {
                right--;
                break;
            }

            // Increment right pointer
            right++;
        }

        // If boundary is exceeded
        if (right >= n)
            right--;

        // Length of longest subarray where
        // arr[i] is the largest
        ans = 1 + right - left;

        // Print the answer
        document.write(ans + " ");
    }
}

// Driver Code

    let arr = [ 4, 2, 1 ];
    let n = arr.length;

    solve(n, arr);

</script>

Output:

3 2 1

Time Complexity: O(N²)
Auxiliary Space: O(1)