LeetCode in Kotlin

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

Medium

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

Solution

class Solution {
    fun minOperations(nums: IntArray): Int {
        var a = 0
        var c = 1
        for (x in nums) {
            if (x != c) {
                a++
                c = c xor 1
            }
        }
        return a
    }
}

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