Minimum Operations to Make Binary Array Elements Equal to One II
MedPrompt
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
ifrom the array and flip all the elements from indexito 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 benums = [0,0,0,1,0]. - Choose the index
i = 0. The resulting array will benums = [1,1,1,0,1]. - Choose the index
i = 4. The resulting array will benums = [1,1,1,0,0]. - Choose the index
i = 3. The resulting array will benums = [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 benums = [1,1,1,1].
Constraints:
1 <= nums.length <= 1050 <= nums[i] <= 1
Approaches
2 approaches with complexity analysis and trade-offs.
This approach directly simulates the process described in the problem. We iterate through the array from left to right. Whenever we encounter a 0, we perform a flip operation starting from that index to make it a 1. This is a greedy strategy because we fix each element from left to right, and a flip at index i does not affect elements before it.
Algorithm
- Initialize a variable
operationsto 0. - Iterate through the array
numsfrom indexi = 0ton-1. - At each index
i, ifnums[i]is 0, it means we must perform a flip operation to make it 1. - Increment
operations. - To simulate the flip, start another loop from
j = iton-1and changenums[j]to1 - nums[j]. - If
nums[i]is already 1, do nothing and continue to the next element. - After the loop finishes, return
operations.
Walkthrough
The core idea is to iterate through the array and fix each element one by one. When we are at index i, we want to ensure nums[i] becomes 1. If it's already 1, we move on. If it's 0, the only way to change it without altering the already-fixed elements nums[0...i-1] is to perform an operation starting at index i. This operation flips all elements from nums[i] to the end of the array. We count this as one operation and perform the actual flip on the array. We repeat this process for all elements.
class Solution { public int minOperations(int[] nums) { int operations = 0; int n = nums.length; for (int i = 0; i < n; i++) { // If the current element is 0, we need to flip it. if (nums[i] == 0) { operations++; // Simulate the flip for all elements from i to the end. for (int j = i; j < n; j++) { nums[j] = 1 - nums[j]; } } } return operations; }}Complexity
Time
O(N^2), where N is the length of `nums`. The outer loop runs N times. In the worst case (e.g., an array of all zeros), the inner loop for flipping also runs up to N times for each outer loop iteration, resulting in a quadratic runtime.
Space
O(1), as we modify the input array in-place. If modifying the input is not allowed, the space complexity would be O(N) to maintain a copy of the array.
Trade-offs
Pros
The logic is straightforward and directly follows the problem statement, making it easy to understand and implement.
It correctly solves the problem by making locally optimal choices that lead to a global optimum.
Cons
The time complexity of O(N^2) makes it too slow for large inputs, likely causing a 'Time Limit Exceeded' error on competitive programming platforms.
This approach improves upon the naive simulation by avoiding the costly step of explicitly flipping the subarray. Instead of modifying the array, we track the cumulative effect of flips. The state of any element nums[i] depends on its original value and the total number of flips performed at indices j <= i. By maintaining a state variable that represents the parity of flips, we can determine the effective value of nums[i] in constant time.
Algorithm
- Initialize
operations = 0. - Initialize a state variable,
flipState = 0. This tracks the parity of flips affecting the current element (0 for even, 1 for odd). - Iterate through the array
numsfromi = 0ton-1. - Determine the effective value of
nums[i]. Ifnums[i]is the same asflipState, the effective value is 0. - If the effective value is 0 (i.e.,
nums[i] == flipState), we must perform an operation.- Increment
operations. - Toggle
flipState(e.g.,flipState = 1 - flipState) to reflect the new flip affecting subsequent elements.
- Increment
- Return
operations.
Walkthrough
The key insight is that we don't need to perform the flips. We only need to know their effect. A flip at index j inverts the state of all elements from j onwards. So, for an element at index i, its final state depends on its initial value and how many times it has been flipped by operations at indices 0, 1, ..., i. We can track the parity of the number of flips with a single variable, let's call it flipState.
We iterate through the array. At each index i, we check the effective value of nums[i]. The effective value is nums[i] if an even number of flips have occurred (flipState = 0), and 1 - nums[i] if an odd number of flips have occurred (flipState = 1). We need this effective value to be 1. If it's 0, we must perform a flip at index i. This increments our operation count and toggles the flipState for all subsequent elements.
class Solution { public int minOperations(int[] nums) { int operations = 0; // Tracks the current flip state. 0 means not flipped (even flips), 1 means flipped (odd flips). int flipState = 0; for (int num : nums) { // If the current element's value matches the flip state, its effective value is 0. // For example: // - num=0, flipState=0 (no flip) -> effective value is 0. // - num=1, flipState=1 (flipped) -> effective value is 1-1=0. // In both cases, we need to flip. if (num == flipState) { operations++; // This new flip toggles the state for all subsequent elements. flipState = 1 - flipState; } } return operations; }}Complexity
Time
O(N), where N is the length of `nums`. We iterate through the array a single time, performing constant time operations at each step.
Space
O(1), as we only use a few variables (`operations`, `flipState`) to keep track of the state, regardless of the input size.
Trade-offs
Pros
Extremely efficient with a linear time complexity, making it suitable for large inputs.
Optimal space complexity, using only a constant amount of extra memory.
Provides a clean and concise implementation.
Cons
The logic, while simple, is an abstraction of the physical process and might be slightly less intuitive to derive compared to the direct simulation.
Solutions
Solution
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; }}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 ansclass 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; }};Video walkthrough
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