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

• Difficulty Level : Medium
• Last Updated : 20 Jan, 2022

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, 2}
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.

Below is the implementation of the above approach:

C++

 // C++ program for the above approach #include 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;       // 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 throught the array     while (high < N) {         // If the current elemetn 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, 2 };     int N = sizeof(X) / sizeof(X);       // 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;       // 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 throught the array     while (high < N) {       // If the current elemetn 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, 2 };     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.

Javascript



Output

21

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

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