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Heaps

Minimum Operations to Exceed Threshold Value II Solution In C++/Java/Python/Javascript

Problem Description

You are given a 0-indexed integer array nums, and an integer k.
You are allowed to perform some operations on nums, where in a single operation, you can:
Select the two smallest integers x and y from nums.
Remove x and y from nums.
Insert (min(x, y) * 2 + max(x, y)) at any position in the array.
Note that you can only apply the described operation if nums contains at least two elements.
Return the minimum number of operations needed so that all elements of the array are greater than or equal to k.

Example

Input: nums = [2,11,10,1,3], k = 10
Output: 2
Explanation:
In the first operation, we remove elements 1 and 2, then add 1 * 2 + 2 to nums. nums becomes equal to [4, 11, 10, 3].
In the second operation, we remove elements 3 and 4, then add 3 * 2 + 4 to nums. nums becomes equal to [10, 11, 10].
At this stage, all the elements of nums are greater than or equal to 10 so we can stop. 
It can be shown that 2 is the minimum number of operations needed so that all elements of the array are greater than or equal to 10.

Input: nums = [1,1,2,4,9], k = 20
Output: 4
Explanation:
After one operation, nums becomes equal to [2, 4, 9, 3]. 
After two operations, nums becomes equal to [7, 4, 9]. 
After three operations, nums becomes equal to [15, 9]. 
After four operations, nums becomes equal to [33].
At this stage, all the elements of nums are greater than 20 so we can stop. 
It can be shown that 4 is the minimum number of operations needed so that all elements of the array are greater than or equal to 20.

Constraints:

  • 2 <= nums.length <= 2 * 10^5
  • 1 <= nums[i] <= 10^9
  • 1 <= k <= 10^9
  • The input is generated such that an answer always exists. That is, after performing some number of operations, all elements of the array are greater than or equal to k.
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Try it Yourself
The problem asks for the fewest operations needed so that every number in the array is at least k. In each operation, you pick the two smallest numbers (x and y), remove them, and then insert a new number calculated as (min(x, y) * 2 + max(x, y)). The initial thought is to always find the top two smallest elements so that we can combine them and form a larger element.

Optimal Approach

Intuition

Imagine you're managing a team where each member has a specific skill level, and your goal is to ensure every team member reaches a minimum threshold, k. In this scenario, each team member's skill is represented by a number in a list. The challenge is that not every team member is up to par, and you need a strategy to boost their skills efficiently.

To address this, consider a strategy where you focus on the two team members with the lowest skills. Instead of training everyone separately, you pair the two weakest links together, effectively concentrating your resources on the most critical areas. This pairing process combines their skills using a specific formula—doubling the skill of the weakest member and then adding the skill of the second weakest. This method represents an intense mentorship or collaborative effort where the weaker member receives extra support to catch up.

By always selecting the two lowest performers, you ensure that every operation directly targets the largest gap in your team's abilities. The process is repeated until even the weakest member meets or exceeds the minimum required skill level, k. This approach not only raises the overall performance of the team but also ensures that no one is left behind, similar to a strategy in real life where addressing the most significant weaknesses first can lead to a more uniformly strong team.

Now, the question arises: how can we always find the two players with the lowest skill set? One might think of scanning through the entire team each time to identify the least skilled players, but this approach is feasible only when the team is small. As the number of players grows, repeatedly searching through every member becomes inefficient and slow.

To address this challenge, we need a data structure that can efficiently retrieve the smallest elements at any moment. This is where a heap comes into play. Specifically, a min-heap (or priority queue) is designed to always provide access to the smallest element in constant time, and it allows insertion and deletion operations in logarithmic time. By using a min-heap, we can quickly extract the two players with the lowest skills, combine their skills, and then reinsert the new, improved skill level back into the heap—all without having to search through the entire list each time.

What is Heap?

A heap is a specialized tree-based data structure often used to implement priority queues. Heaps are usually implemented as complete binary trees (every level is fully filled except possibly the last, and all nodes are as far left as possible), which allows them to be efficiently stored in an array. With a heap, you can quickly access the minimum or maximum element in constant time, and insertion or deletion operations take logarithmic time. This makes heaps extremely useful in algorithms that require repeated retrieval of the "best" element, like scheduling tasks or sorting data with heapsort.

What is min-heap?
A min-heap is a specialized binary tree-based data structure where the smallest element is always at the root. This means that for any given node in the tree, its value is less than or equal to the values of its children. Because of this property, you can quickly access the minimum element at any time. Additionally, a min-heap is usually implemented as a complete binary tree, which means all levels of the tree are fully filled except possibly the last level, and all nodes in the last level are as far left as possible.

What is max-heap?
A max-heap is a complete binary tree data structure where every parent node has a value greater than or equal to its children, ensuring that the largest element is always at the root. This property makes it easy to quickly access the maximum element. Like a min-heap, a max-heap is typically implemented using an array, which efficiently represents the complete binary tree.

General Operations on Heap:

  • Insertion:
    Inserting an element takes O(log n) time because after placing the new element at the end of the array, you "bubble up" (or "sift up") to restore the heap property.
  • Removal (Extract-Min/Extract-Max):
    Removing the root element (the minimum in a min-heap or maximum in a max-heap) also takes O(log n) time since after removal, you "bubble down" (or "sift down") the last element placed at the root to reestablish the heap property.
  • Peek (Find-Min/Find-Max):
    Retrieving the root element is an O(1) operation because it is always at the front of the array.
  • Building a Heap:
    Building a heap from an unsorted array can be done in O(n) time using a bottom-up heapify process.
  • Space Complexity:
    The heap uses O(n) space to store n elements.

What happens if all numbers in the list are already greater than or equal to k?

If every number in the list is already at least k, the condition for the while loop fails immediately. The algorithm then returns 0 operations because no further work is needed—the team already meets the desired threshold.

What if there’s only one element left in the heap when an operation is needed?

The problem assumes that it is always possible to combine two elements until all numbers reach or exceed kkk. However, in a more robust solution, you would include checks to handle cases where fewer than two numbers remain. This might involve handling edge cases or verifying that operations can be performed before attempting to extract two elements.

Approach

  • Step 1: Initialize Data Structures
  • Create a min-heap and insert every element from the array into it.

Step 2: Set Up the Operation Counter

  • Initialize a counter (e.g., ops = 0) to keep track of the number of operations.

Step 3: Check Initial Condition

  • Verify if the smallest element in the heap is already greater than or equal to k.
  • If the condition is met, return the counter (which would be 0).

Step 4: Process the Heap Until Requirement is Met

  • While the smallest element in the heap is less than k, perform the following:
  • Ensure the heap contains at least two elements; if not, return an indicator (such as -1) for failure.
  • Remove the two smallest elements from the heap.
  • Calculate the new value using the formula: new_value = (2 * smaller_value) + larger_value.
  • Insert the new value back into the heap.
  • Increment the operation counter by one.

Step 5: Final Check and Return

  • Once every element in the heap is greater than or equal to k, return the operation counter as the minimum number of operations needed.

Dry Run

Step 1: Define the Example and Initialize

  • Let nums = [1, 2, 3, 9, 10, 12] and k = 7.
  • Create a min-heap from nums: [1, 2, 3, 9, 10, 12].
  • Set the operation counter to 0.

Step 2: First Check

  • The smallest element is 1, which is less than 7.
  • Proceed with an operation.

Step 3: Operation 1

  • Remove the two smallest elements: 1 and 2.
  • Calculate the new value: (2 * 1) + 2 = 4.
  • Insert 4 into the heap.
  • The new heap becomes: [3, 4, 9, 10, 12].
  • Increment the operation counter to 1.

Step 4: Second Check

  • The smallest element is now 3, which is still less than 7.
  • Proceed with another operation.

Step 5: Operation 2

  • Remove the two smallest elements: 3 and 4.
  • Calculate the new value: (2 * 3) + 4 = 10.
  • Insert 10 into the heap.
  • The new heap becomes: [9, 10, 10, 12].
  • Increment the operation counter to 2.

Step 6: Final Check

  • The smallest element in the heap is now 9, which is greater than or equal to 7.
  • All elements meet the requirement.

Step 7: Return the Result

  • The minimum number of operations required is 2.
Code for All Languages
Minimum Operations to Exceed Threshold Value II solution in C++ :Optimal 
class Solution {
public:
    int minOperations(vector<int>& nums, int k) {
        // Initialize a min-heap (priority queue) with the given numbers.
        // This ensures that the smallest element is always at the top.
        priority_queue<ll, vector<ll>, greater<ll>> pq(nums.begin(), nums.end());
        
        ll ans = 0; // Counter for the number of operations
        
        // Continue processing until the smallest number in the heap is at least k.
        while (pq.top() < k) {
            // Extract the two smallest numbers from the min-heap.
            ll num1 = pq.top(); 
            pq.pop();
            ll num2 = pq.top(); 
            pq.pop();
            
            // Combine them by doubling the first and adding the second.
            // Push the result back into the min-heap.
            pq.push(num1 * 2 + num2);
            ans++; // Increment the operation count.
        }
        return ans;
    }
};

Minimum Operations to Exceed Threshold Value II solution in Java :Optimal 
class Solution {
    // The use of 'long' ensures that we can handle large integer values,
    // which is especially important for the calculations in this algorithm.
    public int minOperations(int[] nums, int k) {
        // Initialize a min-heap (priority queue) with the given numbers.
        // The PriorityQueue in Java naturally orders elements in ascending order.
        PriorityQueue<Long> pq = new PriorityQueue<>();
        for (int num : nums) {
            pq.add((long) num);
        }
        
        long ans = 0; // Counter for the number of operations
        
        // Continue processing until the smallest number in the heap is at least k.
        while (pq.peek() < k) {
            // Extract the two smallest numbers from the min-heap.
            long num1 = pq.poll();
            long num2 = pq.poll();
            
            // Combine them by doubling the first and adding the second.
            // Push the result back into the min-heap.
            pq.add(num1 * 2 + num2);
            ans++; // Increment the operation count.
        }
        
        return (int) ans;
    }
}

Minimum Operations to Exceed Threshold Value II solution in Python :Optimal 
import heapq
import sys

class Solution:
    # Python's int type automatically handles large values,
    # so we don't need a separate long type.
    def minOperations(self, nums, k):
        # Initialize a min-heap with the given numbers using heapq
        heapq.heapify(nums)
        
        ans = 0  # Counter for the number of operations
        
        # Continue processing until the smallest element in the heap is at least k.
        while nums[0] < k:
            # Extract the two smallest numbers from the min-heap.
            num1 = heapq.heappop(nums)
            num2 = heapq.heappop(nums)
            
            # Combine them by doubling the first and adding the second.
            # Push the result back into the min-heap.
            heapq.heappush(nums, num1 * 2 + num2)
            ans += 1  # Increment the operation count
        
        return ans

Minimum Operations to Exceed Threshold Value II solution in Javascript :Optimal 
// Import the 'fs' module to read input from standard input.
const fs = require('fs');

// A class representing a simple Min-Heap (priority queue).
class MinHeap {
    constructor() {
        this.heap = [];
    }
    
    // Push a value into the heap and sort it to maintain the min-heap property.
    // Note: This is not the most efficient way to implement a heap, but it's simple.
    push(val) {
        this.heap.push(val);
        this.heap.sort((a, b) => a - b);
    }
    
    // Remove and return the smallest element from the heap.
    pop() {
        return this.heap.shift();
    }
    
    // Return the smallest element without removing it.
    peek() {
        return this.heap[0];
    }
}

// The Solution class encapsulates the minOperations method.
class Solution {
    // This method computes the minimum number of operations required
    // so that every number in the heap is at least k.
    // It uses a min-heap to always process the smallest elements first.
    minOperations(nums, k) {
        let pq = new MinHeap();
        
        // Add all numbers into the min-heap.
        for (let num of nums) {
            pq.push(num);
        }
        let ans = 0;  // Counter for the number of operations
        // Continue until the smallest element in the heap is at least k.
        while (pq.peek() < k) {
            // Extract the two smallest numbers from the heap.
            let num1 = pq.pop();
            let num2 = pq.pop();
            
            // Combine them by doubling the first and adding the second,
            // then push the result back into the heap.
            pq.push(num1 * 2 + num2);
            ans++;  // Increment the operation count.
        }
        return ans;
    }
}

Minimum Operations to Exceed Threshold Value II complexity analysis: Time Complexity O(n log n).

Step 1: Building the Heap : O(n)

  • Construct the min-heap from the array. This process takes linear time, or O(n).

Step 2: Removing Elements : O(log n)

  • In each operation, the two smallest elements are removed from the heap. Each removal takes O(log n) time, so removing two elements takes O(2 log n), which simplifies to O(log n).

Step 3: Inserting the New Element : O(log n)

  • After removal, a new element is inserted into the heap. The insertion operation requires O(log n) time.

Step 4: Combining the Operations : O(log n)

  • Each full operation (removing two elements and inserting one) takes O(log n) time in total.

Step 5: Number of Operations: O(n)

  • In the worst-case scenario, up to O(n) operations might be required if almost every element is combined before the condition is met.

Step 6: Overall Time Complexity: O(n log n)

  • The total time complexity is O(n) for building the heap plus O(n log n) for the operations, resulting in an overall worst-case time complexity of O(n log n).

Minimum Operations to Exceed Threshold Value II complexity analysis: Space Complexity O(n log n).

Step 1: Input Space

  • The input array itself consists of n elements, requiring O(n) space.

Step 2: Auxiliary Space

  • A min-heap is built to manage the elements. In the worst case, the heap holds up to n elements, resulting in an auxiliary space complexity of O(n).

Step 3: Total Space Complexity

  • Combining the input storage and the auxiliary space, the overall space complexity remains O(n).

Learning Tip

Now we have successfully tackled this problem, let's try these similar problems.

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