LeetCode #2208 — MEDIUM

Minimum Operations to Halve Array Sum

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

The Problem

Problem Statement

You are given an array nums of positive integers. In one operation, you can choose any number from nums and reduce it to exactly half the number. (Note that you may choose this reduced number in future operations.)

Return the minimum number of operations to reduce the sum of nums by at least half.

Example 1:

Input: nums = [5,19,8,1]
Output: 3
Explanation: The initial sum of nums is equal to 5 + 19 + 8 + 1 = 33.
The following is one of the ways to reduce the sum by at least half:
Pick the number 19 and reduce it to 9.5.
Pick the number 9.5 and reduce it to 4.75.
Pick the number 8 and reduce it to 4.
The final array is [5, 4.75, 4, 1] with a total sum of 5 + 4.75 + 4 + 1 = 14.75. 
The sum of nums has been reduced by 33 - 14.75 = 18.25, which is at least half of the initial sum, 18.25 >= 33/2 = 16.5.
Overall, 3 operations were used so we return 3.
It can be shown that we cannot reduce the sum by at least half in less than 3 operations.

Example 2:

Input: nums = [3,8,20]
Output: 3
Explanation: The initial sum of nums is equal to 3 + 8 + 20 = 31.
The following is one of the ways to reduce the sum by at least half:
Pick the number 20 and reduce it to 10.
Pick the number 10 and reduce it to 5.
Pick the number 3 and reduce it to 1.5.
The final array is [1.5, 8, 5] with a total sum of 1.5 + 8 + 5 = 14.5. 
The sum of nums has been reduced by 31 - 14.5 = 16.5, which is at least half of the initial sum, 16.5 >= 31/2 = 15.5.
Overall, 3 operations were used so we return 3.
It can be shown that we cannot reduce the sum by at least half in less than 3 operations.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁷

Step 01

Brute Force Baseline

Problem summary: You are given an array nums of positive integers. In one operation, you can choose any number from nums and reduce it to exactly half the number. (Note that you may choose this reduced number in future operations.) Return the minimum number of operations to reduce the sum of nums by at least half.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

[5,19,8,1]

Example 2

[3,8,20]

Core Insight

What unlocks the optimal approach

It is always optimal to halve the largest element.
What data structure allows for an efficient query of the maximum element?
Use a heap or priority queue to maintain the current elements.

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 #2208: Minimum Operations to Halve Array Sum
class Solution {
    public int halveArray(int[] nums) {
        PriorityQueue<Double> pq = new PriorityQueue<>(Collections.reverseOrder());
        double s = 0;
        for (int x : nums) {
            s += x;
            pq.offer((double) x);
        }
        s /= 2.0;
        int ans = 0;
        while (s > 0) {
            double t = pq.poll() / 2.0;
            s -= t;
            pq.offer(t);
            ++ans;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2208: Minimum Operations to Halve Array Sum
func halveArray(nums []int) (ans int) {
	var s float64
	pq := &hp{}
	for _, x := range nums {
		s += float64(x)
		heap.Push(pq, float64(x))
	}
	s /= 2
	for s > 0 {
		t := heap.Pop(pq).(float64) / 2
		s -= t
		ans++
		heap.Push(pq, t)
	}
	return
}

type hp struct{ sort.Float64Slice }

func (h hp) Less(i, j int) bool { return h.Float64Slice[i] > h.Float64Slice[j] }
func (h *hp) Push(v any)        { h.Float64Slice = append(h.Float64Slice, v.(float64)) }
func (h *hp) Pop() any {
	a := h.Float64Slice
	v := a[len(a)-1]
	h.Float64Slice = a[:len(a)-1]
	return v
}

# Accepted solution for LeetCode #2208: Minimum Operations to Halve Array Sum
class Solution:
    def halveArray(self, nums: List[int]) -> int:
        s = sum(nums) / 2
        pq = []
        for x in nums:
            heappush(pq, -x)
        ans = 0
        while s > 0:
            t = -heappop(pq) / 2
            s -= t
            heappush(pq, -t)
            ans += 1
        return ans

// Accepted solution for LeetCode #2208: Minimum Operations to Halve Array Sum
use std::collections::BinaryHeap;

impl Solution {
    pub fn halve_array(nums: Vec<i32>) -> i32 {
        let mut pq: BinaryHeap<i64> = BinaryHeap::new();
        let mut s: i64 = 0;

        for x in nums {
            let v = (x as i64) << 20;
            s += v;
            pq.push(v);
        }

        s /= 2;
        let mut ans = 0;

        while s > 0 {
            let t = pq.pop().unwrap() / 2;
            s -= t;
            pq.push(t);
            ans += 1;
        }

        ans
    }
}

// Accepted solution for LeetCode #2208: Minimum Operations to Halve Array Sum
function halveArray(nums: number[]): number {
    let s: number = nums.reduce((a, b) => a + b) / 2;
    const pq = new MaxPriorityQueue<number>();
    for (const x of nums) {
        pq.enqueue(x);
    }
    let ans = 0;
    while (s > 0) {
        const t = pq.dequeue() / 2;
        s -= t;
        ++ans;
        pq.enqueue(t);
    }
    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

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.