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Minimum Operations to Make Array Elements Zero

Number: 3744

Difficulty: Hard

Paid? No

Companies: N/A

Problem Description

Given a query [l, r], consider the array nums containing every integer from l to r (inclusive). In one operation you are allowed to pick any two nonzero elements a and b from nums and replace them with floor(a/4) and floor(b/4) respectively. The task is to find the minimum number of operations required to reduce every element in nums to zero for each query, and then output the sum of the minimum operations over all queries.

Key Insights

  • Every number x needs to be repeatedly replaced by floor(x/4) until it becomes zero.
  • For a given number, the number of times (steps) we need to perform floor(x/4) is determined by which power-of-4 interval x falls into.
  • If x is in the interval [4^(k-1), 4^k - 1] then it requires exactly k operations.
  • Since an operation can process two numbers simultaneously, the minimum number of operations for a set of required steps is the ceiling of (total individual steps / 2).
  • Instead of iterating through every integer from l to r, a mathematical summation based on intervals can be used to efficiently compute the total steps required for a range.

Space and Time Complexity

Time Complexity: O(Q * log(max(r))) where Q is the number of queries and log(max(r)) is the number of interval blocks (~40 iterations max for r up to 10^9).
Space Complexity: O(1) additional space apart from input storage.

Solution

We start by noting that each number x requires a number of operations equal to:   operations(x) = k if x ∈ [4^(k-1), 4^k - 1].

For any range [l, r], if we define a function F(n) as:   F(n) = sum { operations(x) for x = 1 to n }, then the total steps for [l, r] is:   total_steps = F(r) - F(l-1).

Since we can reduce two numbers in one operation, the minimum operations required for that range is:   min_ops = ceil(total_steps / 2).

To efficiently compute F(n), we partition the numbers into intervals where the operation count is constant:  For k = 1 to K (such that 4^(k-1) ≤ n),   the count of numbers in the interval [4^(k-1), min(n, 4^k - 1)] is:    count = min(n, 4^k - 1) - 4^(k-1) + 1,   and the contribution is k * count. Finally, sum the contributions for all valid k.

Repeat the above for each query and add the results.

Key data structures:  • Simple integer arithmetic and loop iterations over logarithmic intervals.

Algorithmic approach:  1. Implement sumF(n) that computes F(n) as described.  2. For each query, compute total_steps = sumF(r) - sumF(l-1) and then result = ceil(total_steps / 2).  3. Output the sum of results across all queries.

Code Solutions

# Python solution
import math

def sumF(n):
    # Computes F(n) = sum_{x=1}^{n} f(x), where f(x) is the number of operations required for x
    if n <= 0:
        return 0
    total = 0
    k = 1
    lower = 1
    while lower <= n:
        # upper bound for numbers with k operations is 4^k - 1
        upper = 4**k - 1
        # bound the interval to n
        interval_end = min(n, upper)
        count = max(0, interval_end - lower + 1)
        total += k * count
        # move to next interval: lower becomes 4^k
        k += 1
        lower = 4**(k-1)
    return total

def minimumOperationsSum(queries):
    total_operations = 0
    for l, r in queries:
        # Compute total steps for range [l, r]
        steps = sumF(r) - sumF(l - 1)
        # Each operation reduces steps of two numbers simultaneously
        # Use math.ceil to get the minimum operations required.
        ops = (steps + 1) // 2  # equivalent to math.ceil(steps / 2)
        total_operations += ops
    return total_operations

# Example usage:
queries1 = [[1,2],[2,4]]
print(minimumOperationsSum(queries1))  # Output: 3

queries2 = [[2,6]]
print(minimumOperationsSum(queries2))  # Output: 4
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