Scale Engineer

Minimum Operations to Make Array Elements Zero

Hard
#3108Time: O(sum(r_i - l_i) * log(r_i)) over all queries. Given that `r` can be up to 10^9, the range `r - l` can be very large, making this approach too slow for the given constraints.Space: O(1), as we only use a few variables to store intermediate sums.
Patterns
MathBit Manipulation
Data structures
Array

Prompt

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

Approaches

2 approaches with complexity analysis and trade-offs.

For each query [l, r], we can simulate the process directly. The core idea is to determine the total number of 'reduction' steps required to make all numbers in the range [l, r] zero. A single reduction step is defined as x -> floor(x / 4). The total number of operations is determined by how many of these reduction steps are needed in total across all numbers in the array. Since each operation allows us to perform two such reductions simultaneously (by picking two numbers a and b), the minimum number of operations will be the total number of reductions divided by two, rounded up.

Algorithm

  • For each query [l, r]:
    • Initialize totalReductions = 0.
    • Loop x from l to r:
      • Calculate ops(x), the number of times x must be divided by 4 to become 0.
      • Add ops(x) to totalReductions.
    • The result for the query is (totalReductions + 1) / 2.
  • Sum the results for all queries.

Walkthrough

The algorithm first calculates ops(x), which is the number of reduction steps needed for a single integer x to become zero. This is done by repeatedly applying integer division by 4 to x and counting the steps until x is 0.

For each query [l, r], the algorithm proceeds as follows:

  • Initialize a variable totalReductions to zero.
  • Iterate through each integer x from l to r.
  • For each x, compute ops(x) and add it to totalReductions.
  • After the loop, totalReductions holds the sum of all reduction steps for the entire array [l, ..., r].
  • The minimum number of operations for the query is ceil(totalReductions / 2), which can be calculated using integer arithmetic as (totalReductions + 1) / 2.
  • The results for all queries are summed up to get the final answer.
class Solution {    private int ops(int n) {        int count = 0;        while (n > 0) {            n /= 4;            count++;        }        return count;    }     public long minimumOperations(int[][] queries) {        long totalOpsSum = 0;        for (int[] query : queries) {            int l = query[0];            int r = query[1];            long currentTotalReductions = 0;            // This loop is inefficient for large ranges            for (int i = l; i <= r; i++) {                currentTotalReductions += ops(i);            }            totalOpsSum += (currentTotalReductions + 1) / 2;        }        return totalOpsSum;    }}

Complexity

Time

O(sum(r_i - l_i) * log(r_i)) over all queries. Given that `r` can be up to 10^9, the range `r - l` can be very large, making this approach too slow for the given constraints.

Space

O(1), as we only use a few variables to store intermediate sums.

Trade-offs

Pros

  • Simple to understand and implement.

  • Directly follows the problem's logic.

Cons

  • Extremely inefficient for large ranges [l, r].

  • Will result in a Time Limit Exceeded (TLE) error on most platforms for the given constraints.

Video walkthrough

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