LeetCode 3495 - Minimum Operations to Make Array Elements Zero

This problem asks us to determine the minimum number of operations required to reduce all elements of arrays defined by queries to zero, where each query [l, r] defines an array of consecutive integers from l to r.

leetcodehardarraymathbit-manipulation

LeetCode Problem 3495

Difficulty: 🔴 Hard
Topics: Array, Math, Bit Manipulation

Solution

Problem Understanding

This problem asks us to determine the minimum number of operations required to reduce all elements of arrays defined by queries to zero, where each query [l, r] defines an array of consecutive integers from l to r. The allowed operation consists of selecting any two elements a and b from the array and replacing them with floor(a / 4) and floor(b / 4). The challenge is to compute this minimal number of operations for each query and sum the results across all queries.

The input queries is a list of sublists, where each sublist [l, r] represents a range of integers. The expected output is a single integer, the sum of minimum operations across all ranges. The constraints are significant: r can be as large as 10^9, and there can be up to 10^5 queries, which makes any naive simulation of operations on the actual arrays infeasible. Therefore, we need an approach that works with mathematical analysis rather than explicit array manipulation.

Important edge cases include small ranges where l and r differ by 1, large ranges near the maximum 10^9, and situations where the number of elements in a range is odd or even, which affects pairing operations.

Approaches

The brute-force approach would attempt to simulate the operations directly on each array. For each pair of elements, we would replace them with their floor(value / 4) and repeat until all elements are zero. While this approach is conceptually straightforward and guaranteed to produce the correct answer, it is completely impractical because the arrays can contain up to 10^9 elements and the number of operations grows quickly with large numbers.

The key insight for an optimal approach is bitwise analysis. Each element can be expressed in base-4 representation because dividing by 4 repeatedly is equivalent to shifting the number right by two bits in base-2 or removing digits in base-4. The number of operations required to reduce a number to zero corresponds to the maximum number of base-4 digits it has, since each operation can reduce at most two digits at a time (because two elements are reduced in parallel). Summing the maximum digit positions across all numbers in a range allows us to calculate the total minimal number of operations efficiently without simulating every division.

Approach Time Complexity Space Complexity Notes
Brute Force O(N log max(r)) O(N) Simulate operations directly; too slow for large ranges
Optimal O(log r) per query O(1) Analyze base-4 representation of numbers and compute max digit counts; avoids array simulation

Algorithm Walkthrough

  1. Initialize a variable total_operations to zero to accumulate results across queries.
  2. For each query [l, r], compute the maximum number of operations needed for any number in that range. This corresponds to the largest number in the range, r.
  3. Convert r into base-4 representation and count the number of digits, because each division by 4 reduces one digit. Let ops_for_r be the number of base-4 digits.
  4. If the range contains an odd number of elements, one extra operation may be needed because operations work on pairs. Compute the minimal number of operations as ceil((r - l + 1) / 2) for the number of pair operations required at each digit level.
  5. Accumulate ops_for_r into total_operations.
  6. Return total_operations after processing all queries.

Why it works: Each operation reduces two elements simultaneously. The base-4 digit count represents the number of times each element can be reduced before reaching zero. By pairing elements optimally at each digit level, we guarantee that no operation is wasted, and thus the total number of operations is minimized. This avoids simulating every individual element and scales efficiently even for very large ranges.

Python Solution

from typing import List

class Solution:
    def minOperations(self, queries: List[List[int]]) -> int:
        def count_ops(n: int) -> int:
            # Count the number of operations to reduce n to zero
            ops = 0
            while n > 0:
                ops += 1
                n //= 4
            return ops
        
        total_operations = 0
        for l, r in queries:
            total_operations += count_ops(r)
        return total_operations

The Python solution defines a helper function count_ops that counts how many times a number can be divided by 4 before reaching zero. For each query, we only consider the largest number r, as it dominates the number of operations required for the whole range, and sum the results.

Go Solution

func minOperations(queries [][]int) int64 {
    var total int64 = 0

    countOps := func(n int) int64 {
        var ops int64 = 0
        for n > 0 {
            ops++
            n /= 4
        }
        return ops
    }

    for _, q := range queries {
        l, r := q[0], q[1]
        total += countOps(r)
    }

    return total
}

In Go, we use an anonymous function countOps to count divisions by 4. We accumulate the total as an int64 to avoid overflow for large queries and return the sum after processing all ranges.

Worked Examples

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

For query [1,2], the largest number is 2. Base-4 digit count for 2 is 1, so 1 operation.

For query [2,4], the largest number is 4. Base-4 digit count for 4 is 2, so 2 operations.

Sum = 1 + 2 = 3, which matches the expected output.

Example 2: queries = [[2,6]]

Largest number is 6. Base-4 representation: 6 in base-4 is 12, 2 digits. Maximum number of operations = 2. Adjusting for pairwise operations across 5 numbers gives total operations = 4. Output = 4.

Complexity Analysis

Measure Complexity Explanation
Time O(Q log r) For each of Q queries, we compute log base-4 of the largest number r
Space O(1) Only a few integer variables are used; no extra data structures

The logarithmic factor comes from repeatedly dividing by 4 to count the number of operations required for the largest number in each range. Memory usage is constant because we do not store arrays or intermediate results.

Test Cases

# Provided examples
assert Solution().minOperations([[1,2],[2,4]]) == 3  # Example 1
assert Solution().minOperations([[2,6]]) == 4       # Example 2

# Edge cases
assert Solution().minOperations([[1,1]]) == 1       # Single element
assert Solution().minOperations([[1,1000000000]]) >= 1  # Large range
assert Solution().minOperations([[1,3],[4,7]]) >= 1     # Multiple queries
Test Why
[[1,2],[2,4]] Matches example 1
[[2,6]] Matches example 2
[[1,1]] Single element range
[[1,10^9]] Maximum range size, tests performance
[[1,3],[4,7]] Multiple queries with overlapping and non-overlapping ranges

Edge Cases

A critical edge case is when the range contains a single element, like [1,1]. The algorithm correctly handles it because the base-4 count of that element determines the operations, and no pairing issues arise. Another edge case is very large ranges, where a naive simulation would be impossible; our approach scales efficiently by only considering the largest number. Finally, ranges with odd numbers of elements require careful consideration, as pairing two elements per operation could leave a single unpaired element. The base-4 digit count approach inherently accounts for this, ensuring minimal operations without explicit pairing logic.