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3192. Minimum Operations to Make Binary Array Elements Equal to One II

Description

You are given a binary array nums.

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

  • Choose any index i from the array and flip all the elements from index i to the end of the array.

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.

 

Example 1:

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

Output: 4

Explanation:
We can do the following operations:

  • Choose the index i = 1. The resulting array will be nums = [0,0,0,1,0].
  • Choose the index i = 0. The resulting array will be nums = [1,1,1,0,1].
  • Choose the index i = 4. The resulting array will be nums = [1,1,1,0,0].
  • Choose the index i = 3. The resulting array will be nums = [1,1,1,1,1].

Example 2:

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

Output: 1

Explanation:
We can do the following operation:

  • Choose the index i = 1. The resulting array will be nums = [1,1,1,1].

 

Constraints:

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

Solutions

Solution 1: Bit Manipulation

We notice that whenever we change an element at a certain position to 1, all the elements to its right are flipped. Therefore, we can use a variable \(v\) to record whether the current position and all elements to its right have been flipped. If flipped, the value of \(v\) is 1, otherwise, it is 0.

We iterate through the array \(\textit{nums}\). For each element \(x\), we perform an XOR operation between \(x\) and \(v\). If \(x\) is 0, then we need to change \(x\) to 1, which requires a flip operation. We increment the answer by one and flip the value of \(v\).

After the iteration, we can obtain the minimum number of operations.

The time complexity is \(O(n)\), where \(n\) is the length of the array \(\textit{nums}\). The space complexity is \(O(1)\).

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class Solution:
    def minOperations(self, nums: List[int]) -> int:
        ans = v = 0
        for x in nums:
            x ^= v
            if x == 0:
                ans += 1
                v ^= 1
        return ans

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class Solution {
    public int minOperations(int[] nums) {
        int ans = 0, v = 0;
        for (int x : nums) {
            x ^= v;
            if (x == 0) {
                v ^= 1;
                ++ans;
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int minOperations(vector<int>& nums) {
        int ans = 0, v = 0;
        for (int x : nums) {
            x ^= v;
            if (x == 0) {
                v ^= 1;
                ++ans;
            }
        }
        return ans;
    }
};

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func minOperations(nums []int) (ans int) {
    v := 0
    for _, x := range nums {
        x ^= v
        if x == 0 {
            v ^= 1
            ans++
        }
    }
    return
}

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function minOperations(nums: number[]): number {
    let [ans, v] = [0, 0];
    for (let x of nums) {
        x ^= v;
        if (x === 0) {
            v ^= 1;
            ++ans;
        }
    }
    return ans;
}

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