870. Advantage Shuffle

Problem Description

In the given problem, we are provided with two integer arrays nums1 and nums2 of the same length. The term advantage refers to the number of positions i at which the element in nums1 is greater than the corresponding element in nums2, meaning nums1[i] > nums2[i]. The goal is to rearrange nums1 in such a way that it maximizes the advantage with respect to nums2. This means we need to reorder the elements in nums1 so that as many elements as possible are greater than the corresponding elements in nums2 when they are compared index-wise.

Intuition

The intuition behind solving this problem involves sorting and greedy strategy. First, we sort nums1 because we want to arrange its elements in an increasing order to match them with the elements from nums2 efficiently. We also sort nums2, but since we need to create a resulting array that corresponds to the original indices of nums2, we track the original indices by creating tuples (value, index).

For the solution approach, we employ a two-pointer technique. We consider the smallest element in nums1 and try to match it with the smallest element in nums2. If the current element in nums1 does not exceed the smallest remaining element of nums2, it cannot contribute to the advantage. In such a case, we assign it to the position of the largest element of nums2 where it's less likely to affect the advantage negatively.

On the other hand, if the current element in nums1 can surpass the smallest element in nums2, we place it in the result array at the corresponding index and move to the next element in both nums1 and nums2. This process is repeated until all elements in nums1 are placed into the result. This greedy approach ensures that we maximize the advantage by matching each 'nums1' element with the best possible counterpart in 'nums2'.

Learn more about Greedy, Two Pointers and Sorting patterns.

Solution Approach

The implementation of the solution approach is based on a sorted array, greedy algorithm, and two-pointer technique. Here's a step-by-step breakdown:

Sorting nums1: We start by sorting nums1 in non-decreasing order, which allows us to consider the smallest elements first and match them against nums2.
Create and Sort tuple array for nums2: We create tuples containing the value and the original index from nums2 and sort this array. This sorting helps us to consider the elements of nums2 from the smallest to the largest, while remembering their original positions.
Initialize the result array: We initialize an empty result array ans with the same length as nums1 and nums2, which will store the final permutation of nums1 maximizing the advantage.
Two-pointer approach: We set up two pointers, i to point at the start and j to point at the end of the sorted nums2 tuple array. These pointers will be used to traverse the elements in the tuple array.
Iterating over nums1 and placing elements into ans:
- We iterate over each element v in nums1. For each element v, we look at the smallest yet-to-be-assigned element in nums2 (pointed to by i).
- If v is less than or equal to t[i][0] (the smallest element in nums2), the element v cannot contribute to the advantage. We then assign v to ans at the index of the largest yet-to-be-assigned element in nums2 (pointed to by j), and decrement j.
- If v is greater than t[i][0], v can contribute to the advantage. We assign v to ans at the index of the smallest yet-to-be-assigned element in nums2 (pointed to by i), and increment i.
Returning the result: After iterating through all elements in nums1, the ans array now represents a permutation of nums1 that has been greedily arranged to maximize the advantage over nums2. We return ans as the final output.

The algorithm uses sorting and greedy matching to ensure each element from nums1 is used optimally against an element in nums2, thus achieving the maximum advantage. Data structures used include an array of tuples for tracking nums2 elements with their original indices, and an additional array for constructing the result.

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Example Walkthrough

Let's consider two example arrays nums1 = [12, 24, 8, 32] and nums2 = [13, 25, 32, 11]. We aim to find a permutation of nums1 that maximizes the number of indices at which nums1[i] is greater than nums2[i].

Following the solution approach:

Sorting nums1: We sort nums1 to [8, 12, 24, 32].
Create and Sort tuple array for nums2: We create tuple pairs of nums2 with their indices: [(13, 0), (25, 1), (32, 2), (11, 3)]. After sorting, we get [(11, 3), (13, 0), (25, 1), (32, 2)].
Initialize the result array (ans): We set ans as an empty array of the same length: [0, 0, 0, 0].
Two-pointer approach: We initialize two pointers: i starts at 0 and j starts at 3 (pointing at the first and last index of the sorted nums2 tuple array).
Iterating over nums1:
- We compare 8 from nums1 with 11 (the smallest element in nums2 tuple array). Since 8 is less than 11, it can't contribute to the advantage. Place 8 at ans[2] (index of largest nums2 which is 32) and decrement j to 2.
- Now, compare 12 from nums1 with 11 from nums2 (current smallest). 12 is greater, so it can contribute to the advantage. Place 12 at ans[3] (index of current smallest in nums2) and increment i to 1.
- Next, compare 24 from nums1 with 13 (new smallest in nums2). 24 is greater, so it can also contribute to the advantage. Place 24 at ans[0] and increment i to 2.
- Lastly, 32 from nums1 is compared with 25 (new smallest in nums2). 32 is greater and contributes to the advantage. Place 32 at ans[1] and increment i to 3.
Returning the result: The final ans array is [24, 32, 8, 12], representing the permutation of nums1 that gives us the maximum advantage over nums2.

By applying this solution approach to the given arrays, we successfully arranged nums1 ([24, 32, 8, 12]) to maximize the advantage against nums2 ([13, 25, 32, 11]), resulting in an advantage at 3 positions: at indices 0, 1, and 3.

Solution Implementation

1from typing import List
2
3class Solution:
4    def advantageCount(self, A: List[int], B: List[int]) -> List[int]:
5        # Sort the first list to efficiently assign elements
6        A.sort()
7
8        # Create tuples of value and index from list B, then sort these tuples.
9        # This will allow us to compare elements in A with the sorted elements of B.
10        sorted_B_with_indices = sorted((value, index) for index, value in enumerate(B))
11
12        # Initialize the length variable for convenience and readability.
13        length = len(B)
14
15        # Initialize the answer list with placeholder zeros.
16        answer = [0] * length
17
18        # Initialize two pointers for the sorted B list.
19        left_pointer, right_pointer = 0, length - 1
20
21        # Iterate over the sorted list A and try to assign an advantage
22        for value in A:
23            # If the current value in A is greater than the smallest unassigned value in B,
24            # we can assign it as an advantage over B.
25            if value > sorted_B_with_indices[left_pointer][0]:
26                answer[sorted_B_with_indices[left_pointer][1]] = value
27                left_pointer += 1
28            # Otherwise, there is no advantage, so assign the value to the largest 
29            # remaining element in B to discard it as efficiently as possible.
30            else:
31                answer[sorted_B_with_indices[right_pointer][1]] = value
32                right_pointer -= 1
33
34        return answer
35
36# Example of using the class and method above:
37# solution = Solution()
38# print(solution.advantageCount([12,24,8,32], [13,25,32,11]))
39

1import java.util.Arrays;
2
3class Solution {
4    public int[] advantageCount(int[] A, int[] B) {
5        int n = A.length; // Length of input arrays
6        int[][] sortedBWithIndex = new int[n][2]; // Array to keep track of B's elements and their indices
7      
8        // Fill the array with pairs {value, index} for B
9        for (int i = 0; i < n; ++i) {
10            sortedBWithIndex[i] = new int[] {B[i], i};
11        }
12      
13        // Sort the array of pairs by their values (the values from B)
14        Arrays.sort(sortedBWithIndex, (a, b) -> a[0] - b[0]);
15        // Sort array A to efficiently find the advantage count
16        Arrays.sort(A);
17      
18        int[] result = new int[n]; // Result array to store the advantage count
19        int left = 0; // Pointer for the smallest value in A
20        int right = n - 1; // Pointer for the largest value in A
21      
22        // Iterate through A to determine the advantage
23        for (int value : A) {
24            // If the current value is less than or equal to the smallest in sortedBWithIndex,
25            // put the value at the end of result (because it doesn't have an advantage)
26            // and decrease the right pointer
27            if (value <= sortedBWithIndex[left][0]) {
28                result[sortedBWithIndex[right--][1]] = value;
29            } else {
30                // If the current value has an advantage (is larger),
31                // assign it to the corresponding index in the result array
32                // and increase the left pointer
33                result[sortedBWithIndex[left++][1]] = value;
34            }
35        }
36      
37        return result; // Return the advantage array
38    }
39}
40

1#include <vector>
2#include <algorithm>
3using namespace std;
4
5class Solution {
6public:
7    // The function is intended to find an "advantage" permutation of nums1
8    // such that for each element in nums2, there is a corresponding element in
9    // nums1 that is greater. The output is a permutation of nums1 that maximizes
10    // the number of elements in nums1 that are greater than elements in nums2 at
11    // the same index.
12    vector<int> advantageCount(vector<int>& nums1, vector<int>& nums2) {
13        // Get the size of the input vectors
14        int n = nums1.size();
15      
16        // Create a vector of pairs to hold elements from nums2 and their indices
17        vector<pair<int, int>> nums2WithIndices;
18        for (int i = 0; i < n; ++i) {
19            nums2WithIndices.push_back({nums2[i], i});
20        }
21      
22        // Sort the nums2WithIndices based on the values of nums2
23        sort(nums2WithIndices.begin(), nums2WithIndices.end());
24      
25        // Sort nums1 in ascending order
26        sort(nums1.begin(), nums1.end());
27      
28        // Use a two-pointer approach to assign elements from nums1 to nums2
29        // Start i from the beginning and j from the end
30        int i = 0, j = n - 1;
31      
32        // The ans vector will store the "advantaged" permutation of nums1
33        vector<int> ans(n);
34      
35        for (int num : nums1) {
36            // If the current num in nums1 is less than or equal to the smallest
37            // unprocessed num in nums2, then this num in nums1 cannot have advantage
38            // over any unprocessed nums in nums2. So, assign it to the largest remaining
39            // num in nums2 (by decreasing index "j").
40            //
41            // If num is greater, assign it to the current smallest unprocessed num in
42            // nums2 (by increasing index "i") for the advantage.
43            if (num <= nums2WithIndices[i].first) {
44                ans[nums2WithIndices[j--].second] = num;
45            } else {
46                ans[nums2WithIndices[i++].second] = num;
47            }
48        }
49      
50        // Return the final "advantaged" permutation
51        return ans;
52    }
53};
54

1function advantageCount(nums1: number[], nums2: number[]): number[] {
2    // Determine the length of the arrays.
3    const length = nums1.length;
4  
5    // Create an index array for sorting the indices based on nums2 values.
6    const indexArray = Array.from({ length }, (_, i) => i);
7    indexArray.sort((i, j) => nums2[i] - nums2[j]);
8
9    // Sort nums1 in ascending order to easily find the next greater element.
10    nums1.sort((a, b) => a - b);
11
12    // Initialize an answer array with zeroes to store results.
13    const answerArray = new Array(length).fill(0);
14    let leftPointer = 0;
15    let rightPointer = length - 1;
16
17    for (let i = 0; i < length; i++) {
18        // If current element is greater than the smallest element in nums2,
19        // assign it to the index where nums2 is smallest.
20        if (nums1[i] > nums2[indexArray[leftPointer]]) {
21            answerArray[indexArray[leftPointer]] = nums1[i];
22            leftPointer++;
23        } else {
24            // Otherwise, assign it to the index where nums2 is the largest.
25            answerArray[indexArray[rightPointer]] = nums1[i];
26            rightPointer--;
27        }
28    }
29
30    // Return the answer array where each element in nums1 has been placed
31    // to maximise the number of elements in nums1 that are greater than 
32    // the element at the same index in nums2.
33    return answerArray;
34}
35

Time and Space Complexity

The given Python code sorts both arrays and then iterates through them to match elements in nums1 with elements in nums2 in a way that optimizes the advantage condition. Here's the computational complexity analysis of the code provided:

Time Complexity

Sorting nums1: Sorting an array of size n using a comparison-based sort like Timsort (the default in Python) usually takes O(n log n) time.
Creating and sorting t: The list comprehension first iterates over nums2 to create a list of tuples where each tuple contains an element and its index. This operation takes O(n). The list t is then sorted, which again takes O(n log n) time.
Iterating and building the ans array: The main loop iterates over every element in nums1, which is n operations. Within each operation, it performs constant time checks and assignments, so this step is O(n).

When combining these steps, the sorting operations dominate the complexity, so the total time complexity is O(n log n).

Space Complexity

The sorted list of tuples t: This requires O(n) space to store the elements of nums2 along with their indices.
The answer list ans: This also requires O(n) space to store the final output.
The temporary variables used for indexing (i, j, etc.): These are a constant number of extra space, O(1).

Adding these up, the total space complexity of the algorithm is O(n) (since the O(n) space for the ans array and for the list t is the significant factor, and the constant space is negligible).

Learn more about how to find time and space complexity quickly using problem constraints.

Discover Your Strengths and Weaknesses: Take Our 2-Minute Quiz to Tailor Your Study Plan:

Question 1 out of 10

What's the output of running the following function using input [30, 20, 10, 100, 33, 12]?

1def fun(arr: List[int]) -> List[int]:
2    import heapq
3    heapq.heapify(arr)
4    res = []
5    for i in range(3):
6        res.append(heapq.heappop(arr))
7    return res
8

1public static int[] fun(int[] arr) {
2    int[] res = new int[3];
3    PriorityQueue<Integer> heap = new PriorityQueue<>();
4    for (int i = 0; i < arr.length; i++) {
5        heap.add(arr[i]);
6    }
7    for (int i = 0; i < 3; i++) {
8        res[i] = heap.poll();
9    }
10    return res;
11}
12

1class HeapItem {
2    constructor(item, priority = item) {
3        this.item = item;
4        this.priority = priority;
5    }
6}
7
8class MinHeap {
9    constructor() {
10        this.heap = [];
11    }
12
13    push(node) {
14        // insert the new node at the end of the heap array
15        this.heap.push(node);
16        // find the correct position for the new node
17        this.bubble_up();
18    }
19
20    bubble_up() {
21        let index = this.heap.length - 1;
22
23        while (index > 0) {
24            const element = this.heap[index];
25            const parentIndex = Math.floor((index - 1) / 2);
26            const parent = this.heap[parentIndex];
27
28            if (parent.priority <= element.priority) break;
29            // if the parent is bigger than the child then swap the parent and child
30            this.heap[index] = parent;
31            this.heap[parentIndex] = element;
32            index = parentIndex;
33        }
34    }
35
36    pop() {
37        const min = this.heap[0];
38        this.heap[0] = this.heap[this.size() - 1];
39        this.heap.pop();
40        this.bubble_down();
41        return min;
42    }
43
44    bubble_down() {
45        let index = 0;
46        let min = index;
47        const n = this.heap.length;
48
49        while (index < n) {
50            const left = 2 * index + 1;
51            const right = left + 1;
52
53            if (left < n && this.heap[left].priority < this.heap[min].priority) {
54                min = left;
55            }
56            if (right < n && this.heap[right].priority < this.heap[min].priority) {
57                min = right;
58            }
59            if (min === index) break;
60            [this.heap[min], this.heap[index]] = [this.heap[index], this.heap[min]];
61            index = min;
62        }
63    }
64
65    peek() {
66        return this.heap[0];
67    }
68
69    size() {
70        return this.heap.length;
71    }
72}
73
74function fun(arr) {
75    const heap = new MinHeap();
76    for (const x of arr) {
77        heap.push(new HeapItem(x));
78    }
79    const res = [];
80    for (let i = 0; i < 3; i++) {
81        res.push(heap.pop().item);
82    }
83    return res;
84}
85

[30, 20, 10]

[100, 33, 30]

[10, 12, 20]

I don’t know