Queries to find the maximum array element after removing elements from a given range

Given an array arr[] and an array Q[][] consisting of queries of the form of {L, R}, the task for each query is to find the maximum array element after removing array elements from the range of indices [L, R]. If the array becomes empty after removing the elements from given range of indices, then print 0.

Examples:

Input: arr[] = {5, 6, 8, 10, 15}, Q = {{0, 1}, {0, 2}, {1, 4}}
Output:
15
15
5
Explanation:
For the first query {0 1}: Remove array elements in range [0, 1], array modifies to {8, 10, 15}. Hence, the maximum element is 15.
For the second query {0, 2}: Remove array elements in range [0, 2], array modifies to {10, 15}. Hence, the maximum element is 15.
For the third query {1 4}: Remove array elements in range [1, 4], array modifies to {5}. Hence, the maximum element is 5.

Input: arr[] = {8, 12, 14, 10, 13}, Q = {{0, 3}, {0, 4}, {4, 4}}
Output:
13
-1
14

Naive Approach: The simplest approach is to traverse the array to find the maximum element after removing the array elements in the given range in each query.

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

Efficient Approach: To optimize the above approach, the idea is to use two auxiliary arrays: one will store the prefix maximum (maximum element in the range [0, i]) and the other array will store the suffix maximum (maximum element in the range [i, N – 1]). Then for every query [L, R], the answer will be a maximum of prefixMax[l – 1] and suffixMax[r + 1]. Follow the steps below to solve the problem:

• Initialize two auxiliary arrays: prefixMax and suffixMax of size N + 1 and initialize prefixMax[0] as arr[0] and suffixMax[N – 1] as arr[N – 1].
• Iterate over the range [1, N – 1] using the variable i and set prefixMax[i] to the maximum of prefixMax[i – 1] and arr[i].
• Iterate over the range [N – 2, 0] using the variable i and set suffixMax[i] to the maximum of suffixMax[i + 1] and arr[i].
• Traverse the array Q[][] and for each query, check the following conditions:
  • If L = 0 and R = N – 1, print 0 as the result.
  • Otherwise, if L = 0, print suffixMax[R + 1] as the result.
  • Otherwise, if R = N – 1, print prefixMax[L – 1] as the result.
  • For all other cases, print the maximum of prefixMax[L – 1] and suffixMax[R + 1].

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 the maximum element
// after removing elements in range [l, r]
int findMaximum(int arr[], int N, int Q,
                int queries[][2])
{
    // Store prefix maximum element
    int prefix_max[N + 1] = { 0 };

    // Store suffix maximum element
    int suffix_max[N + 1] = { 0 };

    // Prefix max till first index
    // is first index itself
    prefix_max[0] = arr[0];

    // Traverse the array to fill
    // prefix max array
    for (int i = 1; i < N; i++) {

        // Store maximum till index i
        prefix_max[i]
            = max(prefix_max[i - 1],
                  arr[i]);
    }

    // Suffix max till last index
    // is last index itself
    suffix_max[N - 1] = arr[N - 1];

    // Traverse the array to fill
    // suffix max array
    for (int i = N - 2; i >= 0; i--) {

        // Store maximum till index i
        suffix_max[i]
            = max(suffix_max[i + 1],
                  arr[i]);
    }

    // Traverse all queries
    for (int i = 0; i < Q; i++) {

        // Store the starting and the
        // ending index of the query
        int l = queries[i][0];
        int r = queries[i][1];

        // Edge Cases
        if (l == 0 && r == (N - 1))
            cout << "0\n";

        else if (l == 0)
            cout << suffix_max[r + 1]
                 << "\n";

        else if (r == (N - 1))
            cout << prefix_max[l - 1]
                 << "\n";

        // Otherwise
        else
            cout << max(prefix_max[l - 1],
                        suffix_max[r + 1])
                 << "\n";
    }
}

// Driver Code
int main()
{
    int arr[] = { 5, 6, 8, 10, 15 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int queries[][2] = { { 0, 1 }, { 0, 2 }, { 1, 4 } };
    int Q = sizeof(queries) / sizeof(queries[0]);

    findMaximum(arr, N, Q, queries);

    return 0;
}

Java

// Java program for the above approach
class GFG
{

// Function to find the maximum element
// after removing elements in range [l, r]
static void findMaximum(int arr[], int N, int Q,
                int queries[][])
{

    // Store prefix maximum element
    int prefix_max[] = new int[N + 1];

    // Store suffix maximum element
    int suffix_max[] = new int[N + 1];

    // Prefix max till first index
    // is first index itself
    prefix_max[0] = arr[0];

    // Traverse the array to fill
    // prefix max array
    for (int i = 1; i < N; i++) {

        // Store maximum till index i
        prefix_max[i]
            = Math.max(prefix_max[i - 1],
                  arr[i]);
    }

    // Suffix max till last index
    // is last index itself
    suffix_max[N - 1] = arr[N - 1];

    // Traverse the array to fill
    // suffix max array
    for (int i = N - 2; i >= 0; i--) {

        // Store maximum till index i
        suffix_max[i]
            = Math.max(suffix_max[i + 1],
                  arr[i]);
    }

    // Traverse all queries
    for (int i = 0; i < Q; i++) {

        // Store the starting and the
        // ending index of the query
        int l = queries[i][0];
        int r = queries[i][1];

        // Edge Cases
        if (l == 0 && r == (N - 1))
            System.out.print("0\n");
        else if (l == 0)
            System.out.print(suffix_max[r + 1]
                + "\n");
        else if (r == (N - 1))
            System.out.print(prefix_max[l - 1]
                + "\n");

        // Otherwise
        else
            System.out.print(Math.max(prefix_max[l - 1],
                        suffix_max[r + 1])
                + "\n");
    }
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 5, 6, 8, 10, 15 };
    int N = arr.length;
    int queries[][] = { { 0, 1 }, { 0, 2 }, { 1, 4 } };
    int Q = queries.length;
    findMaximum(arr, N, Q, queries);
}
}

// This code is contributed by shikhasingrajput

Python3

# Python3 program for the above approach

# Function to find the maximum element
# after removing elements in range [l, r]
def findMaximum(arr, N, Q, queries):

    # Store prefix maximum element
    prefix_max = [0]*(N + 1)

    # Store suffix maximum element
    suffix_max = [0]*(N + 1)

    # Prefix max till first index
    # is first index itself
    prefix_max[0] = arr[0]

    # Traverse the array to fill
    # prefix max array
    for i in range(1, N):

        # Store maximum till index i
        prefix_max[i]= max(prefix_max[i - 1], arr[i])

    # Suffix max till last index
    # is last index itself
    suffix_max[N - 1] = arr[N - 1]

    # Traverse the array to fill
    # suffix max array
    for i in range(N - 2, -1, -1):

        # Store maximum till index i
        suffix_max[i] = max(suffix_max[i + 1], arr[i])

    # Traverse all queries
    for i in range(Q):

        # Store the starting and the
        # ending index of the query
        l = queries[i][0]
        r = queries[i][1]

        # Edge Cases
        if (l == 0 and r == (N - 1)):
            print("0")
        elif (l == 0):
            print(suffix_max[r + 1])
        elif (r == (N - 1)):
            print(prefix_max[l - 1])

        # Otherwise
        else:
            print(max(prefix_max[l - 1], suffix_max[r + 1]))

# Driver Code
if __name__ == '__main__':
    arr = [5, 6, 8, 10, 15]
    N = len(arr)
    queries = [ [ 0, 1 ], [ 0, 2 ], [ 1, 4 ] ]
    Q = len(queries)
    findMaximum(arr, N, Q, queries)

    # This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
public class GFG
{

// Function to find the maximum element
// after removing elements in range [l, r]
static void findMaximum(int[] arr, int N, int Q,
                int[,] queries)
{

    // Store prefix maximum element
    int[] prefix_max = new int[N + 1];

    // Store suffix maximum element
    int[] suffix_max = new int[N + 1];

    // Prefix max till first index
    // is first index itself
    prefix_max[0] = arr[0];

    // Traverse the array to fill
    // prefix max array
    for (int i = 1; i < N; i++)
    {

        // Store maximum till index i
        prefix_max[i]
            = Math.Max(prefix_max[i - 1],
                  arr[i]);
    }

    // Suffix max till last index
    // is last index itself
    suffix_max[N - 1] = arr[N - 1];

    // Traverse the array to fill
    // suffix max array
    for (int i = N - 2; i >= 0; i--)
    {

        // Store maximum till index i
        suffix_max[i]
            = Math.Max(suffix_max[i + 1],
                  arr[i]);
    }

    // Traverse all queries
    for (int i = 0; i < Q; i++)
    {

        // Store the starting and the
        // ending index of the query
        int l = queries[i, 0];
        int r = queries[i, 1];

        // Edge Cases
        if (l == 0 && r == (N - 1))
            Console.Write("0\n");
        else if (l == 0)
            Console.Write(suffix_max[r + 1]
                + "\n");
        else if (r == (N - 1))
            Console.Write(prefix_max[l - 1]
                + "\n");

        // Otherwise
        else
            Console.Write(Math.Max(prefix_max[l - 1],
                        suffix_max[r + 1])
                + "\n");
    }
}


// Driver Code
static public void Main ()
{
    int[] arr = { 5, 6, 8, 10, 15 };
    int N = arr.Length;
    int[,] queries = { { 0, 1 }, { 0, 2 }, { 1, 4 } };
    int Q = queries.GetLength(0);
    findMaximum(arr, N, Q, queries);
}
}

// This code is contributed by sanjoy_62.

Javascript

<script>

// JavaScript program of the above approach

// Function to find the maximum element
// after removing elements in range [l, r]
function findMaximum(arr, N, Q,
                queries)
{

    // Store prefix maximum element
    let prefix_max = [];

    // Store suffix maximum element
    let suffix_max = [];

    // Prefix max till first index
    // is first index itself
    prefix_max[0] = arr[0];

    // Traverse the array to fill
    // prefix max array
    for (let i = 1; i < N; i++) {

        // Store maximum till index i
        prefix_max[i]
            = Math.max(prefix_max[i - 1],
                  arr[i]);
    }

    // Suffix max till last index
    // is last index itself
    suffix_max[N - 1] = arr[N - 1];

    // Traverse the array to fill
    // suffix max array
    for (let i = N - 2; i >= 0; i--) {

        // Store maximum till index i
        suffix_max[i]
            = Math.max(suffix_max[i + 1],
                  arr[i]);
    }

    // Traverse all queries
    for (let i = 0; i < Q; i++)
    {

        // Store the starting and the
        // ending index of the query
        let l = queries[i][0];
        let r = queries[i][1];

        // Edge Cases
        if (l == 0 && r == (N - 1))
            document.write("0");
        else if (l == 0)
            document.write(suffix_max[r + 1]
                + "<br/>");
        else if (r == (N - 1))
            document.write(prefix_max[l - 1]
                + "<br/>");

        // Otherwise
        else
            document.write(Math.max(prefix_max[l - 1],
                        suffix_max[r + 1])
                + "<br/>");
    }
}

    // Driver Code
    let arr = [ 5, 6, 8, 10, 15 ];
    let N = arr.length;
    let queries = [[ 0, 1 ], [ 0, 2 ], [ 1, 4 ]];
    let Q = queries.length;
    findMaximum(arr, N, Q, queries);

// This code is contributed by avijitmondal1998.
</script>

Output:

15
15
5

Time Complexity: O(N + Q)
Auxiliary Space: O(N)