Reduces an array to at most one element by the given operation

In this problem, we reduce the array size to 1 or 0 by performing the given operation every round.

We can sort the array in each round to get the maximum element in each iteration. In addition, we can also use the head data structure to improve the performance of the code.

Problem Statement - We are given a nums[] array. We need to reduce the array by doing the following.

• Select the two largest elements in the array.

• If two elements are the same, delete the two elements from the array.

• If the two elements are not the same, delete the two elements from the array and insert abs(first − secondary) into the array.

Print the last element of the array. If the array is empty, print 0.

Example

enter

nums = {5, 9, 8, 3, 2, 5};

Output

0

illustrate

• In the first round, we take 9 and 8 and add their difference to the array. Therefore, the array becomes [5, 3, 2, 5, 1].

• In the second round, we take 5 and 5. Therefore, the array becomes [3, 2, 1].

• Next round, we take 3 and 2. Therefore, the array becomes [1, 1]

• In the last round, we take 1 and 1. Therefore, the array becomes empty and we print 0.

enter

nums = {5, 5, 5, 5, 5};

Output

5

Explanation- We delete a pair of 5 twice and there is one 5 remaining in the array.

enter

nums = {4, 8, 7, 6};

Output

1

Explanation - First, we select 8 and 7. Therefore, the array becomes [4, 1, 6]. After that, we select 4 and 6. Therefore, the array becomes [1, 2]. On the last operation, the array becomes [1].

method 1

In this method, we will iterate through the array until the size of the array becomes 1 or 0. In each iteration, we will sort the array and perform the given operation on the first 2 elements of the sorted array. Finally, we will print the output based on the array size.

algorithm

Step 1- Store the size of the array in the "len" variable.

Step 2- Start traversing the array using a while loop.

Step 3- Use the sort() method in a loop to sort the array in reverse.

Step 4- Get the first and second elements of the array. Also, calculate the difference between the first and second elements of the array.

Step 5− If the difference is 0, delete the first two elements of the array and reduce 'len' by 2. If the difference is not 0, delete the first 2 elements and decrement 'len' minus 1.

Step 6- Finally, if the size of the array is 0, return 0. Otherwise, the first element of the array is returned.

Example

#include <bits/stdc++.h>
using namespace std;

int findLast(vector<int> &nums) {
    int len = nums.size();
    int p = 0;
    while (len > 1) {
        // Sort array in reverse order
        sort(nums.begin(), nums.end(), greater<int>());
        // Take the first and second elements of the array
        int a = nums[0];
        int b = nums[1];
        // Take the difference between the first and second element
        int diff = a - b;
        if (diff == 0) {
            nums.erase(nums.begin());
            nums.erase(nums.begin());
            len -= 2;
        } else {
            nums.erase(nums.begin());
            nums.erase(nums.begin());
            nums.push_back(diff);
            len -= 1;
        }
    }
    // When the size of the array is 0
    if (nums.size() == 0)
        return 0;
    return nums[0];
}
int main() {
    vector<int> nums = {5, 9, 8, 3, 2, 5};
    cout << "The last remaining element after performing the given operations is " << findLast(nums) << "\n";
    return 0;
}

Output

The last remaining element after performing the given operations is 0

Time complexity - O(N*NlogN), where O(N) is used to iterate over the array and O(NlogN) is used to sort the array in each iteration.

Space complexity - O(N) to sort the array.

Method 2

In this approach, we will use a priority queue, which implements the heap data structure. It always stores elements in sorted order. So we can easily remove the first 2 largest elements.

algorithm

Step 1- Define "p_queue" named priority queue.

Step 2- Insert all array elements into the priority queue.

Step 3- Iterate until the size of the priority queue is greater than 1.

Step 4- Delete the first 2 elements of the priority queue one by one.

Step 5- Find the difference between two elements.

Step 6- If the difference is not 0, push it into the priority queue.

Step 7− Finally, if the queue size is 0, return 0.

Step 8 - Otherwise, return the element at the top of the queue.

Example

#include <bits/stdc++.h>
using namespace std;

int findLast(vector<int> &nums) {
    // Defining a priority queue
    priority_queue<int> p_queue;
    // Inserting array elements in priority queue
    for (int p = 0; p < nums.size(); ++p)
        p_queue.push(nums[p]);
    // Make iterations
    while (p_queue.size() > 1) {
        // Take the first element from queue
        int first = p_queue.top();
        p_queue.pop();
        // Get the second element from queue
        int second = p_queue.top();
        p_queue.pop();
        // Take the difference of first and second elements
        int diff = first - second;
        if (diff != 0)
            p_queue.push(diff);
    }
    // When queue is empty
    if (p_queue.size() == 0)
        return 0;
    // Return the last remaining element
    return p_queue.top();
}
int main() {
    vector<int> nums = {5, 9, 8, 3, 2, 5};
    cout << "The last remaining element after performing the given operations is " << findLast(nums) << "\n";
    return 0;
}

Output

The last remaining element after performing the given operations is 0

Time complexity - The time complexity of inserting and deleting elements in the priority queue is O(NlogN).

Space Complexity - O(N) to store elements in the priority queue.

Priority queue data structure is always useful when we need to arrange the array data in a specific order after inserting or deleting any element. It implements the heap data structure, thereby enabling insertion and deletion.

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