LeetCode 3066 - Minimum Operations to Exceed Threshold Value II

LeetCode Problem 3066

Difficulty: 🟡 Medium
Topics: Array, Heap (Priority Queue), Simulation

Solution

Problem Understanding

This problem asks us to repeatedly combine the two smallest elements of an array nums until all elements are greater than or equal to a threshold value k. The combination operation is not a simple sum, but a formula: min(x, y) * 2 + max(x, y). Essentially, this operation increases the smaller numbers in the array while gradually reducing the total number of elements.

The input is a 0-indexed array of integers and a threshold k. The expected output is the minimum number of operations needed to make every element in nums at least k. We are guaranteed that the input is valid, meaning it is always possible to reach this threshold with some sequence of operations.

Constraints are significant. The array length can go up to 200,000 and the integers themselves can reach 10^9. This rules out naive approaches that repeatedly scan the entire array for the smallest two elements, since that would be too slow. Key edge cases include arrays that already satisfy the threshold (0 operations needed), arrays with very small elements relative to k (requiring many operations), or arrays with repeated elements that could affect the order of operations.

Approaches

Brute Force

The brute-force approach would repeatedly find the two smallest numbers in nums, compute the new number using the operation formula, and then insert it back into the array. We would repeat this until all elements are greater than or equal to k. While this is correct, it is extremely inefficient. Finding the two smallest elements in an unsorted array takes O(n) each time, and inserting back into an array may take another O(n) per operation. With up to 2 * 10^5 elements, this becomes impractical.

Optimal Approach

The optimal solution leverages a min-heap (priority queue). By maintaining a heap of the elements, we can efficiently access and remove the two smallest elements in O(log n) time and insert the new combined element in O(log n) time. This ensures that each operation is efficient, and we always combine the smallest numbers, which is the key to minimizing the number of operations.

Using a min-heap preserves the invariant that the smallest element is always accessible in O(1) time. This is crucial because the operation must always select the two smallest numbers. Without a heap, a naive search for the minimum each time would be too slow.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n^2)	O(1)	Repeated linear scans to find smallest elements; slow for large n
Optimal	O(n log n + m log n)	O(n)	Use a min-heap to efficiently find and replace smallest elements, where m is the number of operations

Algorithm Walkthrough

1. Initialize a min-heap with all elements of nums. This allows fast access to the smallest elements.

2. Initialize a counter operations to zero. This will track the number of operations performed.

3. While the smallest element in the heap is less than k and the heap contains at least two elements:

4. Pop the two smallest elements from the heap.

5. Compute the new element using the formula: new_val = min(x, y) * 2 + max(x, y).

6. Push new_val back into the heap.

7. Increment the operations counter.

8. After the loop, check if the heap's smallest element is still less than k. If so, additional operations may be needed, but the problem guarantees a solution exists, so the loop should suffice.

9. Return the total number of operations.

Why it works: Each operation strictly increases the smaller numbers, reducing the number of elements below k. By always combining the two smallest numbers, we guarantee the fastest growth of the minimal elements, which minimizes the number of operations.

Python Solution

import heapq
from typing import List

class Solution:
    def minOperations(self, nums: List[int], k: int) -> int:
        heapq.heapify(nums)
        operations = 0
        
        while len(nums) > 1 and nums[0] < k:
            x = heapq.heappop(nums)
            y = heapq.heappop(nums)
            new_val = 2 * min(x, y) + max(x, y)
            heapq.heappush(nums, new_val)
            operations += 1
        
        return operations

Implementation Walkthrough: We first convert nums into a min-heap to efficiently retrieve the smallest elements. Inside the loop, we remove the two smallest numbers, compute the new number according to the given formula, and push it back into the heap. We count each operation. The loop stops when all elements are at least k or when fewer than two elements remain, ensuring correctness due to the problem guarantees.

Go Solution

package main

import (
	"container/heap"
)

type IntHeap []int

func (h IntHeap) Len() int           { return len(h) }
func (h IntHeap) Less(i, j int) bool { return h[i] < h[j] }
func (h IntHeap) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }

func (h *IntHeap) Push(x interface{}) {
	*h = append(*h, x.(int))
}

func (h *IntHeap) Pop() interface{} {
	old := *h
	n := len(old)
	x := old[n-1]
	*h = old[0 : n-1]
	return x
}

func minOperations(nums []int, k int) int {
	h := IntHeap(nums)
	heap.Init(&h)
	operations := 0

	for h.Len() > 1 && h[0] < k {
		x := heap.Pop(&h).(int)
		y := heap.Pop(&h).(int)
		newVal := 2*min(x, y) + max(x, y)
		heap.Push(&h, newVal)
		operations++
	}
	return operations
}

func min(a, b int) int {
	if a < b { return a }
	return b
}

func max(a, b int) int {
	if a > b { return a }
	return b
}

Go-specific Notes: We implement a custom IntHeap type for container/heap. All operations use interface conversions, and we explicitly define min and max helper functions. Slice management is automatic, so no additional memory management is required.

Worked Examples

Example 1: nums = [2,11,10,1,3], k = 10

Operation	Heap State	Notes
Initial	[1,2,10,11,3]	Min-heap built
1	[3,4,10,11]	Pop 1,2 → new 4
2	[10,11,10]	Pop 3,4 → new 10
Done	[10,11,10]	All elements ≥ 10

Example 2: nums = [1,1,2,4,9], k = 20

Operation	Heap State	Notes
Initial	[1,1,2,4,9]	Min-heap built
1	[2,3,4,9]	Pop 1,1 → new 3
2	[4,7,9]	Pop 2,3 → new 7
3	[9,15]	Pop 4,7 → new 15
4	[33]	Pop 9,15 → new 33
Done	[33]	All elements ≥ 20

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n + m log n)	Building the heap is O(n); each operation is O(log n) and there are m operations
Space	O(n)	The heap stores all elements of nums

The complexity is dominated by heap operations, which are logarithmic in the number of elements. This is efficient enough for n up to 2 * 10^5.

Test Cases

# Provided examples
assert Solution().minOperations([2,11,10,1,3], 10) == 2  # Example 1
assert Solution().minOperations([1,1,2,4,9], 20) == 4    # Example 2

# Edge cases
assert Solution().minOperations([10, 12], 5) == 0         # Already above threshold
assert Solution().minOperations([1, 1], 10) == 1         # Small array needing one operation
assert Solution().minOperations([1]*10, 50) > 0          # Multiple repeated elements
assert Solution().minOperations([5, 5, 5, 5, 5], 25) > 0 # All same elements

Test	Why
[2,11,10,1,3], 10	Basic example
[1,1,2,4,9], 20	Larger threshold requiring multiple operations
[10,12], 5	Already satisfies threshold
[1,1], 10	Minimum-length array
[1]*10,