Minimum Operations to Make Binary Array Elements Equal to One II

Problem Statement

You are given a binary array nums.

You can perform the following operation any number of times: choose two indices L and R such that 0 <= L, R < nums.length and flip every 0 to 1 and every 1 to 0 in the subarray nums[L...R] (inclusive).

Return the minimum number of operations needed to make nums contain only 1s.

Example 1:

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

Explanation: Flip the element at index 2 (0-indexed) to get [1,1,1,1].

Example 2:

Input: nums = [0,1,1,0] Output: 2

Explanation: Flip the elements at index 0 and 3 (0-indexed) to get [1,1,1,1].

Constraints:

• 1 <= nums.length <= 10^5
• nums[i] is either 0 or 1.

Solutions

Approach

The goal is to minimize the number of operations required to convert all elements of the array nums to 1. The operations allowed involve flipping subarrays of nums.

Intuition

To achieve this efficiently:

1. Prefix Sum Calculation: Compute the prefix sums of nums such that prefix[i] represents the number of 0s in nums[0...i-1].

2. Two Pointers Technique: Use two pointers (left and right) to determine the range [L, R] where flipping would be most effective. This involves:

   • Calculating the total number of 0s in nums initially.
   • Iteratively adjusting the range [L, R] to minimize the number of 0s.

Steps

1. Initialization: Initialize left and right pointers at the start of the array. Compute the initial count of 0s in nums.

2. Iterative Adjustment: Iterate through possible ranges [L, R]:

   • Update the count of 0s by including nums[right] and excluding nums[left].
   • Update the minimum operations required based on the count of 0s.

3. Edge Cases: Handle arrays where all elements are initially 1.

Complexity

• Time Complexity: O(n) where n is the length of nums. This is because we iterate through the array with two pointers.
• Space Complexity: O(1) extra space for variables.

Implementation (Java)

class Solution {
    public int minOperations(int[] nums) {
        int n = nums.length;
        int left = 0, right = 0;
        int countZeros = 0;
        int minOps = Integer.MAX_VALUE;
        
        for (int num : nums) {
            if (num == 0) {
                countZeros++;
            }
        }
        
        int currentOps = countZeros;
        
        while (right < n) {
            if (nums[right] == 0) {
                countZeros--;
            }
            right++;
            
            while (countZeros * 2 <= right - left) {
                if (nums[left] == 0) {
                    countZeros++;
                }
                left++;
            }
            
            minOps = Math.min(minOps, currentOps);
        }
        
        return minOps == Integer.MAX_VALUE ? 0 : minOps;
    }
}

Implementation (Python)

class Solution:
    def minOperations(self, nums):
        n = len(nums)
        left = 0
        right = 0
        countZeros = sum(1 for num in nums if num == 0)
        minOps = float('inf')
        currentOps = countZeros
        
        while right < n:
            if nums[right] == 0:
                countZeros -= 1
            right += 1
            
            while countZeros * 2 <= right - left:
                if nums[left] == 0:
                    countZeros += 1
                left += 1
            
            minOps = min(minOps, currentOps)
        
        return minOps if minOps != float('inf') else 0

Conclusion

This implementation efficiently calculates the minimum operations required to convert the binary array nums into an array containing only 1s using a two-pointer technique. Adjustments can be made according to specific platform requirements or further customization needs.