Maximum score of Array using increasing subsequence and subarray with given conditions

Given an array arr[]. The task is to find the maximum score that can be achieved from arr[] for i=[1, N-2]. The conditions for scoring are given below.

1. If arr[0...j] < arr[i] < arr[i+1...N-1], then score = 2.
2. If arr[i-1] < arr[i] < arr[i+1] and previous condition is not satisfied, then score = 1.
3. If none of the conditions holds, then score = 0.

Examples:

Input: arr[] = {1, 2, 3}
Output: 2
Explanation: The score of arr[1] equals 2, which is maximum possible. 

Input: arr[] = {2, 4, 6, 4}
Output: 1
Explanation: For each index i in the range 1 <= i <= 2:
The score of nums[1] equals 1.
The score of nums[2] equals 0.
Hence 1 is the maximum possible score. 

Approach: This problem can be solved by using Prefix Max and Suffix Min. Follow the steps below to solve the given problem. 

• For an element score to be 2, it should be greater than every element on its left and smaller than every element on its right.
• So Precompute to find prefix max and suffix min for each array element.
• Now check for each array arr[] element at i:
  • If it is greater than prefix max at i-1, and smaller than suffix min at i+1, the score will be 2.
  • else if it is greater than arr[i-1] and smaller than arr[i+1], score will be 1.
  • else score will be 0.
• Sum up all the scores and return that as the final answer.

Below is the implementation of the above approach. 

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find maximum score
int maxScore(vector<int>& nums)
{

    // Size of array
    int n = nums.size(), i;
    int ans = 0;

    // Prefix max
    vector<int> pre(n, 0);

    // Suffix min
    vector<int> suf(n, 0);

    pre[0] = nums[0];

    for (i = 1; i < n; i++)
        pre[i] = max(pre[i - 1], nums[i]);

    suf[n - 1] = nums[n - 1];
    for (i = n - 2; i >= 0; i--)
        suf[i] = min(suf[i + 1], nums[i]);

    for (i = 1; i < n - 1; i++) {
        if (nums[i] > pre[i - 1]
            && nums[i] < suf[i + 1])
            ans += 2;
        else if (nums[i] > nums[i - 1]
                 && nums[i] < nums[i + 1])
            ans += 1;
    }

    return ans;
}

// Driver Code
int main()
{
    int N = 3;

    vector<int> arr = { 1, 2, 3 };

    // Function Call
    cout << maxScore(arr);

    return 0;
}

// Java program for above approach
import java.util.*;
public class GFG
{
  
    // Function to find maximum score
    static int maxScore(ArrayList<Integer> nums)
    {

        // Size of array
        int n = nums.size(), i = 0;

        int ans = 0;

        // Prefix max
        int[] pre = new int[n];

        // Suffix min
        int[] suf = new int[n];

        pre[0] = (int)nums.get(0);

        for (i = 1; i < n; i++)
            pre[i] = Math.max(pre[i - 1], (int)nums.get(i));

        suf[n - 1] = (int)nums.get(n - 1);
        for (i = n - 2; i >= 0; i--)
            suf[i] = Math.min(suf[i + 1], (int)nums.get(i));

        for (i = 1; i < n - 1; i++) {
            if ((int)nums.get(i) > pre[i - 1]
                && (int)nums.get(i) < suf[i + 1])
                ans += 2;
            else if ((int)nums.get(i) > (int)nums.get(i - 1)
                     && (int)nums.get(i) < (int)nums.get(i + 1))
                ans += 1;
        }

        return ans;
    }

    // Driver Code
    public static void main(String args[])
    {

        ArrayList<Integer> arr = new ArrayList<Integer>();
        
        arr.add(1);
        arr.add(2);
        arr.add(3);

        // Function Call
        System.out.println(maxScore(arr));
    }
}

// This code is contributed by Samim Hossain Mondal.

# python program for above approach

# Function to find maximum score
def maxScore(nums):

    # Size of array
    n = len(nums)
    ans = 0

    # Prefix max
    pre = [0 for _ in range(n)]

    # Suffix min
    suf = [0 for _ in range(n)]

    pre[0] = nums[0]

    for i in range(1, n):
        pre[i] = max(pre[i - 1], nums[i])

    suf[n - 1] = nums[n - 1]
    for i in range(n-2, -1, -1):
        suf[i] = min(suf[i + 1], nums[i])

    for i in range(1, n-1):
        if (nums[i] > pre[i - 1] and nums[i] < suf[i + 1]):
            ans += 2
        elif (nums[i] > nums[i - 1] and nums[i] < nums[i + 1]):
            ans += 1

    return ans

# Driver Code
if __name__ == "__main__":
    N = 3
    arr = [1, 2, 3]

    # Function Call
    print(maxScore(arr))

# This code is contributed by rakeshsahni

// C# program for above approach
using System;
using System.Collections.Generic;
class GFG
{
  
    // Function to find maximum score
    static int maxScore(List<int> nums)
    {

        // Size of array
        int n = nums.Count, i = 0;

        int ans = 0;

        // Prefix max
        int[] pre = new int[n];

        // Suffix min
        int[] suf = new int[n];

        pre[0] = nums[0];

        for (i = 1; i < n; i++)
            pre[i] = Math.Max(pre[i - 1], nums[i]);

        suf[n - 1] = nums[n - 1];
        for (i = n - 2; i >= 0; i--)
            suf[i] = Math.Min(suf[i + 1], nums[i]);

        for (i = 1; i < n - 1; i++) {
            if (nums[i] > pre[i - 1]
                && nums[i] < suf[i + 1])
                ans += 2;
            else if (nums[i] > nums[i - 1]
                     && nums[i] < nums[i + 1])
                ans += 1;
        }

        return ans;
    }

    // Driver Code
    public static void Main()
    {

        List<int> arr = new List<int>() { 1, 2, 3 };

        // Function Call
        Console.WriteLine(maxScore(arr));
    }
}

// This code is contributed by ukasp.

  <script>
        // JavaScript code for the above approach

        // Function to find maximum score
        function maxScore(nums) {

            // Size of array
            let n = nums.length, i;
            let ans = 0;

            // Prefix max
            let pre = new Array(n).fill(0)

            // Suffix min
            let suf = new Array(n).fill(0);

            pre[0] = nums[0];

            for (i = 1; i < n; i++)
                pre[i] = Math.max(pre[i - 1], nums[i]);

            suf[n - 1] = nums[n - 1];
            for (i = n - 2; i >= 0; i--)
                suf[i] = Math.min(suf[i + 1], nums[i]);

            for (i = 1; i < n - 1; i++) {
                if (nums[i] > pre[i - 1]
                    && nums[i] < suf[i + 1])
                    ans += 2;
                else if (nums[i] > nums[i - 1]
                    && nums[i] < nums[i + 1])
                    ans += 1;
            }

            return ans;
        }

        // Driver Code

        let N = 3;

        let arr = [1, 2, 3];

        // Function Call
        document.write(maxScore(arr));

  // This code is contributed by Potta Lokesh
    </script>

2

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