3542. Minimum Operations to Convert All Elements to Zero

Description

You are given an array nums of size n, consisting of non-negative integers. Your task is to apply some (possibly zero) operations on the array so that all elements become 0.

In one operation, you can select a [i, j] (where 0 <= i <= j < n) and set all occurrences of the minimum non-negative integer in that subarray to 0.

Return the minimum number of operations required to make all elements in the array 0.

 

Example 1:

Input: nums = [0,2]

Output: 1

Explanation:

• Select the subarray [1,1] (which is [2]), where the minimum non-negative integer is 2. Setting all occurrences of 2 to 0 results in [0,0].
• Thus, the minimum number of operations required is 1.

Example 2:

Input: nums = [3,1,2,1]

Output: 3

Explanation:

• Select subarray [1,3] (which is [1,2,1]), where the minimum non-negative integer is 1. Setting all occurrences of 1 to 0 results in [3,0,2,0].
• Select subarray [2,2] (which is [2]), where the minimum non-negative integer is 2. Setting all occurrences of 2 to 0 results in [3,0,0,0].
• Select subarray [0,0] (which is [3]), where the minimum non-negative integer is 3. Setting all occurrences of 3 to 0 results in [0,0,0,0].
• Thus, the minimum number of operations required is 3.

Example 3:

Input: nums = [1,2,1,2,1,2]

Output: 4

Explanation:

• Select subarray [0,5] (which is [1,2,1,2,1,2]), where the minimum non-negative integer is 1. Setting all occurrences of 1 to 0 results in [0,2,0,2,0,2].
• Select subarray [1,1] (which is [2]), where the minimum non-negative integer is 2. Setting all occurrences of 2 to 0 results in [0,0,0,2,0,2].
• Select subarray [3,3] (which is [2]), where the minimum non-negative integer is 2. Setting all occurrences of 2 to 0 results in [0,0,0,0,0,2].
• Select subarray [5,5] (which is [2]), where the minimum non-negative integer is 2. Setting all occurrences of 2 to 0 results in [0,0,0,0,0,0].
• Thus, the minimum number of operations required is 4.

 

Constraints:

• 1 <= n == nums.length <= 105
• 0 <= nums[i] <= 105

 

Solutions

Solution: Greedy

• Time complexity: O(n)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
const minOperations = function (nums) {
  const stack = [];
  let result = 0;

  for (const num of nums) {
    while (stack.length && stack.at(-1) > num) {
      stack.pop();
      result += 1;
    }

    if (num && stack.at(-1) !== num) {
      stack.push(num);
    }
  }

  return result + stack.length;
};