LeetCode #3066 — MEDIUM

Minimum Operations to Exceed Threshold Value II

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

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 1:

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.

Example 2:

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 * 10⁵
1 <= nums[i] <= 10⁹
1 <= k <= 10⁹
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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[2,11,10,1,3]
10

Example 2

[1,1,2,4,9]
20

Core Insight

What unlocks the optimal approach

Use priority queue to keep track of minimum elements.
Remove the minimum two elements, perform the operation, and insert the resulting number into the priority queue.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3066: Minimum Operations to Exceed Threshold Value II
class Solution {
    public int minOperations(int[] nums, int k) {
        PriorityQueue<Long> pq = new PriorityQueue<>();
        for (int x : nums) {
            pq.offer((long) x);
        }
        int ans = 0;
        for (; pq.size() > 1 && pq.peek() < k; ++ans) {
            long x = pq.poll(), y = pq.poll();
            pq.offer(x * 2 + y);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3066: Minimum Operations to Exceed Threshold Value II
func minOperations(nums []int, k int) (ans int) {
	pq := &hp{nums}
	heap.Init(pq)
	for ; pq.Len() > 1 && pq.IntSlice[0] < k; ans++ {
		x, y := heap.Pop(pq).(int), heap.Pop(pq).(int)
		heap.Push(pq, x*2+y)
	}
	return
}

type hp struct{ sort.IntSlice }

func (h *hp) Less(i, j int) bool { return h.IntSlice[i] < h.IntSlice[j] }
func (h *hp) Pop() interface{} {
	old := h.IntSlice
	n := len(old)
	x := old[n-1]
	h.IntSlice = old[0 : n-1]
	return x
}
func (h *hp) Push(x interface{}) {
	h.IntSlice = append(h.IntSlice, x.(int))
}

# Accepted solution for LeetCode #3066: Minimum Operations to Exceed Threshold Value II
class Solution:
    def minOperations(self, nums: List[int], k: int) -> int:
        heapify(nums)
        ans = 0
        while len(nums) > 1 and nums[0] < k:
            x, y = heappop(nums), heappop(nums)
            heappush(nums, x * 2 + y)
            ans += 1
        return ans

// Accepted solution for LeetCode #3066: Minimum Operations to Exceed Threshold Value II
use std::collections::BinaryHeap;

impl Solution {
    pub fn min_operations(nums: Vec<i32>, k: i32) -> i32 {
        let mut pq = BinaryHeap::new();

        for &x in &nums {
            pq.push(-(x as i64));
        }

        let mut ans = 0;

        while pq.len() > 1 && -pq.peek().unwrap() < k as i64 {
            let x = -pq.pop().unwrap();
            let y = -pq.pop().unwrap();
            pq.push(-(x * 2 + y));
            ans += 1;
        }

        ans
    }
}

// Accepted solution for LeetCode #3066: Minimum Operations to Exceed Threshold Value II
function minOperations(nums: number[], k: number): number {
    const pq = new MinPriorityQueue<number>();
    for (const x of nums) {
        pq.enqueue(x);
    }
    let ans = 0;
    for (; pq.size() > 1 && pq.front() < k; ++ans) {
        const x = pq.dequeue();
        const y = pq.dequeue();
        pq.enqueue(x * 2 + y);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.