2025-03-19 Daily Challenge

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In: DailyChallenge

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

March LeetCoding Challenge 19

Description

Minimum Operations to Make Binary Array Elements Equal to One I

You are given a binary array nums.

You can do the following operation on the array any number of times (possibly zero):

  • Choose any 3 consecutive elements from the array and flip all of them.

Flipping an element means changing its value from 0 to 1, and from 1 to 0.

Return the minimum number of operations required to make all elements in nums equal to 1. If it is impossible, return -1.

 

Example 1:

Input: nums = [0,1,1,1,0,0]

Output: 3

Explanation:
We can do the following operations:

  • Choose the elements at indices 0, 1 and 2. The resulting array is nums = [1,0,0,1,0,0].
  • Choose the elements at indices 1, 2 and 3. The resulting array is nums = [1,1,1,0,0,0].
  • Choose the elements at indices 3, 4 and 5. The resulting array is nums = [1,1,1,1,1,1].

Example 2:

Input: nums = [0,1,1,1]

Output: -1

Explanation:
It is impossible to make all elements equal to 1.

 

Constraints:

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

Solution

class Solution {
public:
  int minOperations(vector<int>& nums) {
    int flip = 0;
    int len = nums.size();
    queue<int> flipPos;
    for(int i = 0; i < len; ++i) {
      if(flipPos.front() == i) flipPos.pop();
      if(nums[i] ^ (flipPos.size() & 1)) continue;
      if(i + 2 >= len) return -1;
      flip += 1;
      flipPos.push(i + 3);
    }
    return flip;
  }
};

// Accepted
// 689/689 cases passed (43 ms)
// Your runtime beats 5.04 % of cpp submissions
// Your memory usage beats 5.33 % of cpp submissions (144.2 MB)
Algorithm