Longest unique subarray of an Array with maximum sum in another Array

Given two arrays X[] and Y[] of size N, the task is to find the longest subarray in X[] containing only unique values such that a subarray with similar indices in Y[] should have a maximum sum. The value of array elements is in the range [0, 1000].

Examples:

Input: N = 5,  
X[] = {0, 1, 2, 0, 2},  
Y[] = {5, 6, 7, 8, 22}
Output: 21
Explanation: The largest unique subarray in X[] with maximum sum in Y[] is {1, 2, 0}. 
So, the subarray with same indices in Y[] is {6, 7, 8}.
Therefore maximum sum is 21.

Input: N = 3,  
X[] = {1, 1, 1},  
Y[] = {2, 6, 7}
Output: 7

Naive Approach: The task can be solved by generating all the subarrays of the array X[], checking for each subarray if it is valid, and then calculating the sum in the array for corresponding indices in Y.

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

Efficient Approach: The task can be solved using the concept of the sliding window. Follow the below steps to solve the problem:

• Create an array m of size 1001 and initialize all elements as -1. For index i, m[i] stores the index at which i is present in the subarray. If m[i] is -1, it means the element doesn’t exist in the subarray.
• Initialize low = 0, high = 0, these two pointers will define the indices of the current subarray.
• currSum and maxSum, define the sum of the current subarray and the maximum sum in the array.
• Iterate over a loop and check if the current element at index high exists in the subarray already, if it does find the sum of elements in the subarray, update maxSum (if needed) and update low. Now, finally, move to the next element by incrementing high.
• Note that you will be encountering a corner case which can result in wrong answer, for instance, let’s consider our first Input, then the subarray {8, 2} is the right choice and 30 is our right answer. So handle that corner case separately by summing all elements from the previous low to high.

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 max sum
int findMaxSumSubarray(int X[], int Y[],
                       int N)
{
    // Array to store the elements
    // and their indices
    int m[1001];

    // Initialize all elements as -1
    for (int i = 0; i < 1001; i++)
        m[i] = -1;

    // low and high represent
    // beginning and end of subarray
    int low = 0, high = 0;
    int currSum = 0, maxSum = 0;

    // Iterate through the array
    // Note that the array is traversed till high <= N
    // so that the corner case can be handled
    while (high <= N) {
      if(high==N){
        //Calculate the currSum for the subarray
        //after the last updated low to high-1
        currSum=0;
        for (int i = low; i <= high - 1;
                 i++)
                currSum += Y[i];

            // Find the maximum sum
            maxSum = max(maxSum, currSum);
        break;
      }
        // If the current element already
        // exists in the current subarray
        if (m[X[high]] != -1
            && m[X[high]] >= low) {
            currSum = 0;

            // Calculate the sum
            // of current subarray
            for (int i = low; i <= high - 1;
                 i++)
                currSum += Y[i];

            // Find the maximum sum
            maxSum = max(maxSum, currSum);

            // Starting index of new subarray
            low = m[X[high]] + 1;
        }

        // Keep expanding the subarray
        // and mark the index
        m[X[high]] = high;
        high++;
    }

    // Return the maxSum
    return maxSum;
}

// Driver code
int main()
{
    int X[] = { 0, 1, 2, 0, 2 };
    int Y[] = { 5, 6, 7, 8, 22 };
    int N = sizeof(X) / sizeof(X[0]);

    // Function call to find the sum
    int maxSum = findMaxSumSubarray(X, Y, N);

    // Print the result
    cout << maxSum << endl;
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class GFG
{

  // Function to find the max sum
  static int findMaxSumSubarray(int X[], int Y[],
                                int N)
  {

    // Array to store the elements
    // and their indices
    int m[] = new int[1001];

    // Initialize all elements as -1
    for (int i = 0; i < 1001; i++)
      m[i] = -1;

    // low and high represent
    // beginning and end of subarray
    int low = 0, high = 0;
    int currSum = 0, maxSum = 0;

    // Iterate through the array
    // Note that the array is traversed till high <= N
    // so that the corner case can be handled
    while (high <= N) {
      if(high==N){
           // Calculate the currSum for the subarray
        // after the last updated low to high-1
        currSum = 0;
        for (int i = low; i <= high - 1;i++)
             currSum += Y[i];
        // Find the maximum sum
        maxSum = Math.max(maxSum, currSum);
        break;
      }
      // If the current element already
      // exists in the current subarray
      if (m[X[high]] != -1
          && m[X[high]] >= low) {
        currSum = 0;

        // Calculate the sum
        // of current subarray
        for (int i = low; i <= high - 1;
             i++)
          currSum += Y[i];

        // Find the maximum sum
        maxSum = Math.max(maxSum, currSum);

        // Starting index of new subarray
        low = m[X[high]] + 1;
      }

      // Keep expanding the subarray
      // and mark the index
      m[X[high]] = high;
      high++;
    }

    // Return the maxSum
    return maxSum;
  }

  // Driver code
  public static void main(String args[])
  {
    int X[] = { 0, 1, 2, 0, 2 };
    int Y[] = { 5, 6, 7, 8, 22 };
    int N = X.length;

    // Function call to find the sum
    int maxSum = findMaxSumSubarray(X, Y, N);

    // Print the result
    System.out.println(maxSum);
  }
}

// This code is contributed by Samim Hossain Mondal.

Python3

# Python code for the above approach

# Function to find the max sum
def findMaxSumSubarray(X, Y, N):

    # Array to store the elements
    # and their indices
    m = [0] * (1001);

    # Initialize all elements as -1
    for i in range(1001):
        m[i] = -1;

    # low and high represent
    # beginning and end of subarray
    low = 0
    high = 0
    currSum = 0
    maxSum = 0;

    # Iterate through the array
    # Note that the array is traversed till high <= N
    # so that the corner case can be handled
    while (high <= N):
        if(high == N):
            currSum=0;
            # Calculate the currSum for the subarray
            # after the last updated low to high-1
            for i in range(low, high):
                currSum += Y[i];
            maxSum = max(maxSum, currSum);
            break;

        # If the current element already
        # exists in the current subarray
        if (m[X[high]] != -1 and m[X[high]] >= low):
            currSum = 0;

            # Calculate the sum
            # of current subarray
            for i in range(low, high):
                currSum += Y[i];

            # Find the maximum sum
            maxSum = max(maxSum, currSum);

            # Starting index of new subarray
            low = m[X[high]] + 1;


        # Keep expanding the subarray
        # and mark the index
        m[X[high]] = high;
        high += 1


    # Return the maxSum
    return maxSum;


# Driver code
X = [0, 1, 2, 0, 2];
Y = [5, 6, 7, 8, 22];
N = len(X)

# Function call to find the sum
maxSum = findMaxSumSubarray(X, Y, N);

# Print the result
print(maxSum, end="")

# This code is contributed by Saurabh Jaiswal

C#

// C# program for above approach
using System;
using System.Collections.Generic;

public class GFG{

  // Function to find the max sum
  static int findMaxSumSubarray(int[] X, int[] Y, int N)
  {

    // Array to store the elements
    // and their indices
    int[] m = new int[1001];

    // Initialize all elements as -1
    for (int i = 0; i < 1001; i++)
      m[i] = -1;

    // low and high represent
    // beginning and end of subarray
    int low = 0, high = 0;
    int currSum = 0, maxSum = 0;

    // Iterate through the array
    // Note that the array is traversed till high <= N
    // so that the corner case can be handled
    while (high <= N)
    {
      if(high==N){
        // Calculate the currSum for the subarray
        // after the last updated low to high-1
        currSum=0;
        for (int i = low; i <= high - 1;
             i++)
          currSum += Y[i];

        // Find the maximum sum
        maxSum = Math.Max(maxSum, currSum);
        break;
      }

      // If the current element already
      // exists in the current subarray
      if (m[X[high]] != -1
          && m[X[high]] >= low) {
        currSum = 0;

        // Calculate the sum
        // of current subarray
        for (int i = low; i <= high - 1;
             i++)
          currSum += Y[i];

        // Find the maximum sum
        maxSum = Math.Max(maxSum, currSum);

        // Starting index of new subarray
        low = m[X[high]] + 1;
      }

      // Keep expanding the subarray
      // and mark the index
      m[X[high]] = high;
      high++;
    }

    // Return the maxSum
    return maxSum;
  }

  // Driver code
  static public void Main ()
  {

    int[] X = { 0, 1, 2, 0, 2 };
    int[] Y = { 5, 6, 7, 8, 22 };
    int N = X.Length;

    // Function call to find the sum
    int maxSum = findMaxSumSubarray(X, Y, N);

    // Print the result
    Console.WriteLine(maxSum);
  }
}

// This code is contributed by hrithikgarg03188.

Javascript

<script>
       // JavaScript code for the above approach

       // Function to find the max sum
       function findMaxSumSubarray(X, Y, N)
       {

           // Array to store the elements
           // and their indices
           let m = new Array(1001);

           // Initialize all elements as -1
           for (let i = 0; i < 1001; i++)
               m[i] = -1;

           // low and high represent
           // beginning and end of subarray
           let low = 0, high = 0;
           let currSum = 0, maxSum = 0;

           // Iterate through the array
           // Note that the array is traversed till high <= N
           // so that the corner case can be handled
           while (high <= N)
           {
               if(high==N){
               // Calculate the currSum for the subarray
               // after the last updated low to high-1
               currSum=0;
               for (let i = low; i <= high - 1;
                       i++)
                       currSum += Y[i];

                   // Find the maximum sum
                   maxSum = Math.max(maxSum, currSum);
                   break;
               }
               // If the current element already
               // exists in the current subarray
               if (m[X[high]] != -1
                   && m[X[high]] >= low) {
                   currSum = 0;

                   // Calculate the sum
                   // of current subarray
                   for (let i = low; i <= high - 1;
                       i++)
                       currSum += Y[i];

                   // Find the maximum sum
                   maxSum = Math.max(maxSum, currSum);

                   // Starting index of new subarray
                   low = m[X[high]] + 1;
               }

               // Keep expanding the subarray
               // and mark the index
               m[X[high]] = high;
               high++;
           }

           // Return the maxSum
           return maxSum;
       }

       // Driver code
       let X = [0, 1, 2, 0, 2];
       let Y = [5, 6, 7, 8, 22];
       let N = X.length

       // Function call to find the sum
       let maxSum = findMaxSumSubarray(X, Y, N);

       // Print the result
       document.write(maxSum + '<br>')

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

Output

30

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