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3495. Minimum Operations to Make Array Elements Zero

Description

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

Solutions

Solution 1

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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;
      }
  }
  
  

• class Solution {
  public:
      long long minOperations(vector<vector<int>>& queries) {
          auto f = [&](long long x) {
              long long res = 0;
              long long p = 1;
              int i = 1;
              while (p <= x) {
                  long long cnt = min(p * 4 - 1, x) - p + 1;
                  res += cnt * i;
                  i++;
                  p *= 4;
              }
              return res;
          };
  
          long long ans = 0;
          for (auto& q : queries) {
              int l = q[0], r = q[1];
              long long s = f(r) - f(l - 1);
              long long mx = f(r) - f(r - 1);
              ans += max((s + 1) / 2, mx);
          }
          return ans;
      }
  };
  
  

• 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
  
  

• 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
  }
  
  

• 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;
  }
  
  

• 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
      }
  }
  
  

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