3191. Minimum Operations to Make Binary Array Elements Equal to One I

MediumBit ManipulationQueueArrayPrefix SumSliding Window

Problem Description

You are given a binary array nums.

You can perform 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 involves changing its value from 0 to 1, and from 1 to 0.

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

Intuition

To solve this problem, we can take advantage of the operation which flips three consecutive elements in the array. The key observation here is that any segment of the array which starts with 0 will need to be flipped if we want the entire array to become all 1s.

The strategy involves traversing the array sequentially and whenever we encounter a 0, we flip it along with the next two elements. This ensures that the current 0 is turned into a 1.

By accumulating the number of flips (operations), we can determine the minimum number of operations needed. If at any point, we encounter a 0 and there aren't two more elements after it to flip, it's impossible to convert the entire array to 1s, and we return -1.

This sequential traversal and simulation of the flipping process help arrive at the solution efficiently.

Learn more about Queue, Prefix Sum and Sliding Window patterns.

Solution Approach

The solution follows a sequential traversal and simulation approach. Let's break down the steps involved in the implementation:

Initialize a Counter: Start by initializing a counter ans to 0. This will keep track of the number of flip operations performed.
Traverse the Array: Iterate over each element x in the array nums using its index i.
Check for Flipping: For each element, check if it is 0. If it is, this means a flip operation is necessary to convert it to 1.
Edge Condition: Before performing a flip, ensure that there are at least two elements following the current element (i.e., i + 2 < len(nums)). If not, it is impossible to make all the elements 1, and hence return -1.
Perform the Flip: If the conditions are met, perform the flip by using the XOR operation ^= 1 on the next two consecutive elements (nums[i + 1] ^= 1 and nums[i + 2] ^= 1). This effectively flips the value of these elements.
Increment the Counter: After flipping, increment the operation counter ans by 1.
Return the Result: After the entire array is processed, return the value of ans, which represents the minimum number of flip operations needed.

This approach efficiently traverses the array and ensures each group of consecutive 0s is flipped appropriately to achieve a complete transformation to all 1s.

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Example Walkthrough

Let's use a small example to illustrate the solution approach step-by-step.

Consider the binary array: nums = [0, 1, 0, 0, 1, 0].

Initialize a Counter: Start with ans = 0.
Traverse the Array: Begin iterating through the array starting from index 0.
Index 0:
- Current element is 0. It needs to be flipped.
- Check if there are at least two elements available to flip after it. (Indexes 1 and 2)
- Perform the Flip: Flip elements at indices 0, 1, 2.
  - Resulting array: [1, 0, 1, 0, 1, 0]
- Increment the Counter: ans = 1
Index 1:
- Current element is 0. It needs to be flipped.
- Check if there are at least two elements available to flip after it. (Indexes 2 and 3)
- Perform the Flip: Flip elements at indices 1, 2, 3.
  - Resulting array: [1, 1, 0, 1, 1, 0]
- Increment the Counter: ans = 2
Index 2:
- Current element is 0. It needs to be flipped.
- Check if there are at least two elements available to flip after it. (Indexes 3 and 4)
- Perform the Flip: Flip elements at indices 2, 3, 4.
  - Resulting array: [1, 1, 1, 0, 0, 0]
- Increment the Counter: ans = 3
Index 3:
- Current element is 0. It needs to be flipped.
- Check if there are at least two elements available to flip after it. (Indexes 4 and 5)
- Perform the Flip: Flip elements at indices 3, 4, 5.
  - Resulting array: [1, 1, 1, 1, 1, 1]
- Increment the Counter: ans = 4
Indexes 4 and 5:
- Both elements are 1, no flips needed.

After processing the full array, all elements are 1, and the minimum number of operations needed is 4. Therefore, the answer is 4.

Solution Implementation

1from typing import List
2
3class Solution:
4    def minOperations(self, nums: List[int]) -> int:
5        # Initialize the number of operations required to 0
6        operations = 0
7      
8        # Iterate over each element in the list with its index
9        for index, value in enumerate(nums):
10            # If the current element is 0, we need to flip the next two elements
11            if value == 0:
12                # Check if there are enough elements to perform the flip
13                if index + 2 >= len(nums):
14                    # It's not possible to fulfill the requirements, return -1
15                    return -1
16                # Flip the next two elements using XOR operation
17                nums[index + 1] ^= 1
18                nums[index + 2] ^= 1
19                # Increment the count of operations performed
20                operations += 1
21      
22        # Return the total number of operations performed
23        return operations
24

1class Solution {
2    public int minOperations(int[] nums) {
3        int operations = 0; // Initialize the count of operations
4        int length = nums.length; // Get the length of the array
5
6        // Traverse through the array
7        for (int i = 0; i < length; ++i) {
8            // If we encounter a 0 in the array
9            if (nums[i] == 0) {
10                // Check if there's room to flip the next two bits
11                if (i + 2 >= length) {
12                    return -1; // Return -1 if we can't perform the flip operation
13                }
14                // Flip the bits at positions i + 1 and i + 2
15                nums[i + 1] ^= 1;
16                nums[i + 2] ^= 1;
17                ++operations; // Increment the operation count
18            }
19        }
20        return operations; // Return the total count of operations needed
21    }
22}
23

1#include <vector>
2
3class Solution {
4public:
5    int minOperations(std::vector<int>& nums) {
6        int operations = 0;
7        int length = nums.size();
8      
9        // Iterate over each element of the array
10        for (int i = 0; i < length; ++i) {
11            // If the current element is 0, move to the next operation
12            if (nums[i] == 0) {
13                // Check if it is possible to flip at least two more elements
14                if (i + 2 >= length) {
15                    return -1; // Return -1 if flipping is not possible
16                }
17                // Flip the next two elements to make current zero into one
18                nums[i + 1] ^= 1;
19                nums[i + 2] ^= 1;
20                ++operations; // Increment the operations count
21            }
22        }
23      
24        // Return the total number of operations needed
25        return operations;
26    }
27};
28

1/**
2 * Function to count the minimum number of operations required
3 * to change an array where each '0' needs to be eliminated by flipping
4 * the current and the next two consecutive elements.
5 * 
6 * If it is not possible to perform the operations, return -1.
7 * 
8 * @param nums - Array of numbers consisting of 0s and 1s.
9 * @returns The minimum number of operations or -1 if not possible.
10 */
11function minOperations(nums: number[]): number {
12    const n: number = nums.length; // Length of the input array
13    let ans: number = 0; // Variable to count the number of operations
14
15    // Iterate through the array to find and handle '0's
16    for (let i = 0; i < n; ++i) {
17        if (nums[i] === 0) { // If we encounter a '0'
18            // Check if we can flip the next two elements
19            if (i + 2 >= n) {
20                return -1; // Return -1 if the flip cannot be performed
21            }
22            // Flip the next two elements using XOR operation
23            nums[i + 1] ^= 1;
24            nums[i + 2] ^= 1;
25            ++ans; // Increment the operation counter
26        }
27    }
28
29    return ans; // Return the total count of operations needed
30}
31

Time and Space Complexity

The given code iterates through each element of the list nums exactly once, performing constant-time operations for each element. Therefore, the time complexity is O(n), where n is the length of the array nums.

For space complexity, the code operates in-place by modifying the input list nums directly and uses a constant amount of additional space for variables like ans, i, and x. Consequently, the space complexity is O(1).

Learn more about how to find time and space complexity quickly.

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