Leetcode 2233: Maximum Product After K Increments
You are given an array of non-negative integers and an integer k. In one operation, you can choose any element from the array and increment it by 1. Your task is to determine the maximum product of the array elements after at most k operations. The result should be returned modulo 10^9 + 7.
Problem
Approach
Steps
Complexity
Input: The input consists of an array nums of integers and an integer k. The array nums contains non-negative integers.
Example: nums = [1, 2, 3], k = 4
Constraints:
• 1 <= nums.length <= 10^5
• 0 <= nums[i] <= 10^6
• 1 <= k <= 10^5
Output: Return the maximum product of the array after performing at most k operations, modulo 10^9 + 7.
Example: Output: 72
Constraints:
• The result should fit within the 32-bit signed integer range after taking the modulo.
Goal: The goal is to maximize the product of the array elements by distributing k increments across the elements. To achieve this, we want to increment the smallest elements first, as this maximizes the overall product.
Steps:
• 1. Use a priority queue (min-heap) to store the elements in the array, ensuring that the smallest element can be incremented efficiently.
• 2. Increment the smallest element by 1 at each operation, and reinsert it back into the priority queue.
• 3. After performing all k operations, compute the product of all the elements in the array.
• 4. Return the product modulo 10^9 + 7.
Goal: The solution must handle large inputs efficiently and ensure the product calculation is within the bounds of a 32-bit integer after applying the modulo.
Steps:
• The array can have up to 10^5 elements.
• Each element in the array can be as large as 10^6.
Assumptions:
• The array is non-empty and contains at least one element.
• The value of k is guaranteed to be a positive integer.
• Input: Input: nums = [1, 2, 3], k = 4
• Explanation: To maximize the product, we first increment 1 to 2, then 2 to 3, and finally increment 3 to 4. The array becomes [2, 3, 4], and the product is 2 * 3 * 4 = 24.
• Input: Input: nums = [5, 1, 2], k = 3
• Explanation: To maximize the product, we first increment 1 to 2, then 2 to 3, and finally 3 to 4. The array becomes [5, 3, 4], and the product is 5 * 3 * 4 = 60.
• Input: Input: nums = [10, 20], k = 10
• Explanation: Here, we increment the smaller number, 10, 10 times. The array becomes [20, 20], and the product is 20 * 20 = 400.
Approach: We approach the problem using a greedy strategy where we increment the smallest elements first to maximize the product. This can be efficiently achieved using a priority queue.
Observations:
• The product will be maximized by focusing on increasing the smallest numbers in the array.
• Using a priority queue (min-heap) allows us to efficiently increment the smallest element and track changes as we proceed with the operations.
Steps:
• 1. Initialize a priority queue with the elements from the input array.
• 2. Perform k operations by incrementing the smallest element in the priority queue and reinserting it.
• 3. After k operations, calculate the product of all the elements, applying modulo 10^9 + 7 at each step to prevent overflow.
• 4. Return the product modulo 10^9 + 7.
Empty Inputs:
• The array is guaranteed to have at least one element.
Large Inputs:
• The algorithm must handle up to 100,000 elements and efficiently perform up to 100,000 operations.
Special Values:
• The array elements could be zero, which would require handling increment operations to avoid multiplication by zero.
Constraints:
• The priority queue approach is efficient enough to handle the constraints within the time limit.
int maximumProduct(vector<int>& nums, int k) {
priority_queue<int, vector<int>, greater<int>> pq(nums.begin(), nums.end());
while(k--) {
int mini = pq.top();
pq.pop();
pq.push(mini + 1);
}
long long ans = 1, mod = (long) 1e9 + 7;
while(!pq.empty()) {
ans = ((ans % mod) * (pq.top() % mod)) % mod;
pq.pop();
}
return ans;
}
1 : Function Definition
int maximumProduct(vector<int>& nums, int k) {
Defines the function 'maximumProduct', which takes a vector of integers 'nums' and an integer 'k', and returns the maximum product after performing 'k' increments on the smallest elements in 'nums'.
2 : Priority Queue Initialization
priority_queue<int, vector<int>, greater<int>> pq(nums.begin(), nums.end());
Initializes a priority queue 'pq' to store the elements of 'nums' in ascending order, allowing easy access to the smallest element.
3 : Loop Initialization
The next step involves performing a loop to increment the smallest elements in the priority queue.
4 : Loop Control
while(k--) {
This loop will run 'k' times, decrementing 'k' each time to perform the increment operation on the smallest element.
5 : Priority Queue Operations
int mini = pq.top();
Extracts the smallest element from the priority queue and stores it in 'mini'.
6 : Priority Queue Operations
pq.pop();
Removes the smallest element (stored in 'mini') from the priority queue.
7 : Priority Queue Operations
pq.push(mini + 1);
Increments the smallest element and pushes it back into the priority queue.
8 : Variable Initialization
long long ans = 1, mod = (long) 1e9 + 7;
Initializes 'ans' to 1 for storing the product result, and 'mod' to 1e9+7 to perform modulo operations and avoid overflow.
9 : Loop Control
while(!pq.empty()) {
This loop will continue until the priority queue is empty, multiplying the elements to compute the product.
10 : Product Calculation
ans = ((ans % mod) * (pq.top() % mod)) % mod;
Calculates the product of the current 'ans' and the top element of the priority queue, applying modulo to prevent overflow.
11 : Priority Queue Operations
pq.pop();
Removes the top element from the priority queue after it has been used in the product calculation.
12 : Return Statement
return ans;
Returns the final result of the product of all elements in the priority queue after performing 'k' increments, modulo 1e9+7.
Best Case: O(n log n)
Average Case: O(n log n)
Worst Case: O(k log n)
Description: Each of the k operations involves extracting and inserting an element into the priority queue, which takes O(log n) time. Hence, the total time complexity is O(k log n).
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n) due to the storage of the array in the priority queue.
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