LeetCode #3495 — HARD

Minimum Operations to Make Array Elements Zero

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given a 2D array queries, where queries[i] is of the form [l, r]. Each queries[i] defines an array of integers nums consisting of elements ranging from l to r, both inclusive.

In one operation, you can:

Select two integers a and b from the array.
Replace them with floor(a / 4) and floor(b / 4).

Your task is to determine the minimum number of operations required to reduce all elements of the array to zero for each query. Return the sum of the results for all queries.

Example 1:

Input: queries = [[1,2],[2,4]]

Output: 3

Explanation:

For queries[0]:

The initial array is nums = [1, 2].
In the first operation, select nums[0] and nums[1]. The array becomes [0, 0].
The minimum number of operations required is 1.

For queries[1]:

The initial array is nums = [2, 3, 4].
In the first operation, select nums[0] and nums[2]. The array becomes [0, 3, 1].
In the second operation, select nums[1] and nums[2]. The array becomes [0, 0, 0].
The minimum number of operations required is 2.

The output is 1 + 2 = 3.

Example 2:

Input: queries = [[2,6]]

Output: 4

Explanation:

For queries[0]:

The initial array is nums = [2, 3, 4, 5, 6].
In the first operation, select nums[0] and nums[3]. The array becomes [0, 3, 4, 1, 6].
In the second operation, select nums[2] and nums[4]. The array becomes [0, 3, 1, 1, 1].
In the third operation, select nums[1] and nums[2]. The array becomes [0, 0, 0, 1, 1].
In the fourth operation, select nums[3] and nums[4]. The array becomes [0, 0, 0, 0, 0].
The minimum number of operations required is 4.

The output is 4.

Constraints:

1 <= queries.length <= 10⁵
queries[i].length == 2
queries[i] == [l, r]
1 <= l < r <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given a 2D array queries, where queries[i] is of the form [l, r]. Each queries[i] defines an array of integers nums consisting of elements ranging from l to r, both inclusive. In one operation, you can: Select two integers a and b from the array. Replace them with floor(a / 4) and floor(b / 4). Your task is to determine the minimum number of operations required to reduce all elements of the array to zero for each query. Return the sum of the results for all queries.

Baseline thinking

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

Pattern signal: Array · Math · Bit Manipulation

Example 1

[[1,2],[2,4]]

Example 2

[[2,6]]

Step 02

Core Insight

What unlocks the optimal approach

For a number <code>x</code>, the number of <code>"/4"</code> operations to change it to 0 is <code>floor(log4(x)) + 1</code>.
Always pair the 2 numbers with the maximum <code>"/4"</code> operations needed.

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

Largest constraint values

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 #3495: Minimum Operations to Make Array Elements Zero
class Solution {
    public long minOperations(int[][] queries) {
        long ans = 0;
        for (int[] q : queries) {
            int l = q[0], r = q[1];
            long s = f(r) - f(l - 1);
            long mx = f(r) - f(r - 1);
            ans += Math.max((s + 1) / 2, mx);
        }
        return ans;
    }

    private long f(long x) {
        long res = 0;
        long p = 1;
        int i = 1;
        while (p <= x) {
            long cnt = Math.min(p * 4 - 1, x) - p + 1;
            res += cnt * i;
            i++;
            p *= 4;
        }
        return res;
    }
}

// Accepted solution for LeetCode #3495: Minimum Operations to Make Array Elements Zero
func minOperations(queries [][]int) (ans int64) {
	f := func(x int64) (res int64) {
		var p int64 = 1
		i := int64(1)
		for p <= x {
			cnt := min(p*4-1, x) - p + 1
			res += cnt * i
			i++
			p *= 4
		}
		return
	}
	for _, q := range queries {
		l, r := int64(q[0]), int64(q[1])
		s := f(r) - f(l-1)
		mx := f(r) - f(r-1)
		ans += max((s+1)/2, mx)
	}
	return
}

# Accepted solution for LeetCode #3495: Minimum Operations to Make Array Elements Zero
class Solution:
    def minOperations(self, queries: List[List[int]]) -> int:
        def f(x: int) -> int:
            res = 0
            p = i = 1
            while p <= x:
                cnt = min(p * 4 - 1, x) - p + 1
                res += cnt * i
                i += 1
                p *= 4
            return res

        ans = 0
        for l, r in queries:
            s = f(r) - f(l - 1)
            mx = f(r) - f(r - 1)
            ans += max((s + 1) // 2, mx)
        return ans

// Accepted solution for LeetCode #3495: Minimum Operations to Make Array Elements Zero
impl Solution {
    pub fn min_operations(queries: Vec<Vec<i32>>) -> i64 {
        let f = |x: i64| -> i64 {
            let mut res: i64 = 0;
            let mut p: i64 = 1;
            let mut i: i64 = 1;
            while p <= x {
                let cnt = std::cmp::min(p * 4 - 1, x) - p + 1;
                res += cnt * i;
                i += 1;
                p *= 4;
            }
            res
        };

        let mut ans: i64 = 0;
        for q in queries {
            let l = q[0] as i64;
            let r = q[1] as i64;
            let s = f(r) - f(l - 1);
            let mx = f(r) - f(r - 1);
            ans += std::cmp::max((s + 1) / 2, mx);
        }
        ans
    }
}

// Accepted solution for LeetCode #3495: Minimum Operations to Make Array Elements Zero
function minOperations(queries: number[][]): number {
    const f = (x: number): number => {
        let res = 0;
        let p = 1;
        let i = 1;
        while (p <= x) {
            const cnt = Math.min(p * 4 - 1, x) - p + 1;
            res += cnt * i;
            i++;
            p *= 4;
        }
        return res;
    };

    let ans = 0;
    for (const [l, r] of queries) {
        const s = f(r) - f(l - 1);
        const mx = f(r) - f(r - 1);
        ans += Math.max(Math.ceil(s / 2), mx);
    }
    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)

Space

O(1)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.