2025-09-06 Daily Challenge

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In: DailyChallenge

Today I have done leetcode's September LeetCoding Challenge with cpp.

September LeetCoding Challenge 6

Description

Minimum Operations to Make Array Elements Zero

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

Solution

class Solution {
  long long countOps(long long target) {
    long long base = 4;
    long long ops = 2;
    long long result = 0;
    while(target >= base) {
      base *= 4;
      ops += 1;
    }
    base /= 4;
    ops -= 1;
    while(base) {
      result += ops * (target - base + 1);
      target = base - 1;
      ops -= 1;
      base /= 4;
    }
    result += target;
    return result;
  }
public:
  long long minOperations(vector<vector<int>>& queries) {
    long long answer = 0;
    for(auto const &q : queries) {
      // cout << q[1] << ' ' << countOps(q[1]) << endl;
      // cout << q[0] - 1 << ' ' << countOps(q[0] - 1) << endl;
      answer += (countOps(q[1]) - countOps(q[0] - 1) + 1) / 2;
    }
    return answer;
  }
};

// Accepted
// 646/646 cases passed (21 ms)
// Your runtime beats 61.27 % of cpp submissions
// Your memory usage beats 63.38 % of cpp submissions (183.4 MB)
Algorithm