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

MediumGreedyArrayDynamic Programming

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Problem Description

You are given a binary array nums.

You can perform the following operation on the array any number of times (including zero times):

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.

Intuition

To solve this problem, notice that flipping an array portion involves toggling each binary value starting from a chosen index i to the end. Because of this, we can use a variable v to track the current flip state of the array from any index onward. Initially, this flip state v is set to 0.

As we iterate through the array nums, for each element, we perform an XOR operation with v to effectively simulate the flipping effect up to that point. If the result is 0, it means this position and all to its right are currently 0 (or effectively need to be flipped to reach 1), hence, we need to perform a flip operation from this position. We increment the flip count and toggle v to indicate the entire remaining portion has been flipped.

The solution, therefore, leverages bit manipulation with XOR operations to determine the minimal flips necessary, using v to efficiently track the flip status of the array beyond each position we consider.

Learn more about Greedy and Dynamic Programming patterns.

Solution Approach

The solution uses a simple algorithm with a bit manipulation mindset, revolving around the use of XOR operations and a tracking variable. Here’s the breakdown of the approach:

Initialize Variables:
- ans to count the number of flip operations required. Initially set to 0.
- v to track the current flip state (0 or 1). Initially set to 0.
Iterate Through the Array:
- Loop through each element x in the array nums.
- Apply XOR between x and v (x ^= v). This simulates the effect of flips that have been applied to the current position.
- If the result of x is 0, it means that all current position values need an additional flip to become 1. Thus:
  - Increment ans by 1 because a new flip operation is needed.
  - Toggle v using v ^= 1 to indicate that from this point onward (to the end of the array) the array has undergone one more flip operation.
Return the Answer:
- After processing all elements, return ans which contains the minimum number of flips needed to turn all elements into 1s.

This approach efficiently traverses the list while using logical operations to minimize the flip count with an overall time complexity of O(n), where n is the number of elements in the binary array.

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Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

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

Step-by-Step Execution:

Initialization:
- Set ans = 0 to keep track of flip operations.
- Set v = 0 to track the current flip state.
Iteration through the array:
- Index 0 (Value 0):
  - Apply XOR: nums[0] ^= v -> 0 ^= 0 = 0
  - Since the result is 0, perform a flip to make it 1.
  - Increment ans to 1.
  - Toggle v: v ^= 1 -> 0 ^= 1 = 1.
- Index 1 (Value 1):
  - Apply XOR: nums[1] ^= v -> 1 ^= 1 = 0
  - Since the result is 0, perform a flip to make it 1.
  - Increment ans to 2.
  - Toggle v: v ^= 1 -> 1 ^= 1 = 0.
- Index 2 (Value 0):
  - Apply XOR: nums[2] ^= v -> 0 ^= 0 = 0
  - The result is 0, perform a flip to make it 1.
  - Increment ans to 3.
  - Toggle v: v ^= 1 -> 0 ^= 1 = 1.
- Index 3 (Value 0):
  - Apply XOR: nums[3] ^= v -> 0 ^= 1 = 1
  - The result is 1. No flip needed, continue.
- Index 4 (Value 1):
  - Apply XOR: nums[4] ^= v -> 1 ^= 1 = 0
  - The result is 0. Perform a flip to make it 1.
  - Increment ans to 4.
  - Toggle v: v ^= 1 -> 1 ^= 1 = 0.
Final Result:
- Return ans = 4 as the minimum number of operations required to make all elements in nums equal to 1.

This example clearly demonstrates how using XOR and a tracking variable v allows us to efficiently reach the optimal flipping strategy.

Solution Implementation

1from typing import List
2
3class Solution:
4    def minOperations(self, nums: List[int]) -> int:
5        operations_count = 0  # This will store the total number of operations needed
6        toggle_value = 0  # This is used to toggle between 0 and 1
7      
8        # Iterate through each element in the list 'nums'
9        for num in nums:
10            # Apply XOR operation with the current toggle_value
11            num ^= toggle_value
12          
13            # If the result after XOR is 0, it means an operation is needed
14            if num == 0:
15                operations_count += 1  # Increment the operations count
16                toggle_value ^= 1  # Toggle the value for future iterations
17      
18        return operations_count  # Return the total operations needed
19

1class Solution {
2    public int minOperations(int[] nums) {
3        int operationsCount = 0; // Initialize the count of operations
4        int flipState = 0;       // Variable to keep track of the XOR state
5
6        // Iterate through each element in the nums array
7        for (int num : nums) {
8            // Apply XOR with the current flip state
9            num ^= flipState;
10          
11            // If the result is zero, a flip is needed
12            if (num == 0) {
13                flipState ^= 1;  // Toggle the flip state
14                ++operationsCount;  // Increment the number of operations
15            }
16        }
17
18        // Return the total number of operations performed
19        return operationsCount;
20    }
21}
22

1#include <vector>
2
3class Solution {
4public:
5    // Method to calculate the minimum number of operations to make the XOR of all elements non-zero.
6    int minOperations(std::vector<int>& nums) {
7        int operations = 0; // Count the number of operations needed.
8        int toggle = 0;     // Toggle state to keep track of XOR changes.
9
10        // Loop through each number in the list
11        for (int num : nums) {
12            // Apply XOR with the toggle to determine necessary changes
13            num ^= toggle;
14          
15            // If the number becomes zero, toggle state to ensure non-zero XOR
16            if (num == 0) {
17                toggle ^= 1;  // Change the toggle state
18                ++operations; // Increment operations as a non-zero is required
19            }
20        }
21        return operations; // Return the total number of operations
22    }
23};
24

1// This function finds the minimum number of operations needed to 
2// make all elements in the array non-zero by using XOR operations.
3function minOperations(nums: number[]): number {
4    // Initialize the count of operations (ans) and the variable v used for XOR operation.
5    let ans = 0;
6    let v = 0;
7
8    // Iterate over each number in the array.
9    for (let x of nums) {
10        // Perform an XOR operation between the current element and v.
11        x ^= v;
12      
13        // If the result of the XOR is zero, increment operation count and toggle v.
14        if (x === 0) {
15            // Toggle v using XOR with 1.
16            v ^= 1;
17            // Increment the number of operations.
18            ++ans;
19        }
20    }
21  
22    // Return the total number of operations performed.
23    return ans;
24}
25

Time and Space Complexity

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

The space complexity of the code is O(1), as the algorithm uses a fixed amount of extra space regardless of the input size, storing only a few integer variables (ans and v).

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

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Question 1 out of 10

What's the output of running the following function using the following tree as input?

1def serialize(root):
2    res = []
3    def dfs(root):
4        if not root:
5            res.append('x')
6            return
7        res.append(root.val)
8        dfs(root.left)
9        dfs(root.right)
10    dfs(root)
11    return ' '.join(res)
12

1import java.util.StringJoiner;
2
3public static String serialize(Node root) {
4    StringJoiner res = new StringJoiner(" ");
5    serializeDFS(root, res);
6    return res.toString();
7}
8
9private static void serializeDFS(Node root, StringJoiner result) {
10    if (root == null) {
11        result.add("x");
12        return;
13    }
14    result.add(Integer.toString(root.val));
15    serializeDFS(root.left, result);
16    serializeDFS(root.right, result);
17}
18

1function serialize(root) {
2    let res = [];
3    serialize_dfs(root, res);
4    return res.join(" ");
5}
6
7function serialize_dfs(root, res) {
8    if (!root) {
9        res.push("x");
10        return;
11    }
12    res.push(root.val);
13    serialize_dfs(root.left, res);
14    serialize_dfs(root.right, res);
15}
16

1 2 3 6 x x

1 2 3 x x x 6 x x

1 2 6 x x x 3 x x

I don’t know