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.
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; }}Video walkthrough
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