Problem Description

You have a binary array nums containing only 0s and 1s. Your goal is to make all elements in the array equal to 1 using the minimum number of operations.

The operation you can perform is:

• Choose any index i in the array
• Flip all elements from index i to the end of the array (changing 0 to 1 and 1 to 0)

For example, if you have nums = [0, 1, 0, 1] and you choose index 1, after flipping from index 1 to the end, the array becomes [0, 0, 1, 0].

The key insight is that each operation affects all elements from the chosen position to the end of the array. When you flip at position i, every element from position i onwards gets flipped - zeros become ones and ones become zeros.

The solution uses a clever bit manipulation approach. It tracks whether the current position has been affected by previous flips using a variable v. As we traverse the array from left to right:

• We XOR each element with v to get its effective current value after all previous flips
• If the effective value is 0, we need to perform a flip operation at this position
• When we flip, all subsequent elements will be affected, so we toggle v to track this change

This greedy approach ensures we only flip when necessary (when we encounter an effective 0), resulting in the minimum number of operations needed to make all elements equal to 1.

Intuition

The key observation is that when we flip at position i, all elements from i to the end get flipped. This means once we decide to flip at a position, it affects everything to the right.

Let's think about this problem from left to right. When we encounter a 0 at position i, we must flip it to make it 1. But this flip affects all elements to the right of i as well.

Here's the crucial insight: instead of actually modifying the array after each flip operation, we can keep track of the "cumulative effect" of all flips we've done so far. If we've done an odd number of flips that affect the current position, then the element's value is flipped from its original; if we've done an even number, it remains the same.

This leads us to use a toggle variable v that tracks whether the current position has been affected by an odd (v = 1) or even (v = 0) number of previous flips.

When we look at element nums[i]:

• The effective value after all previous flips is nums[i] XOR v
• If this effective value is 0, we need to flip at this position to make it 1
• When we flip at position i, all subsequent positions will have one more flip affecting them, so we toggle v

The greedy approach works because:

1. We must fix any 0 we encounter (after accounting for previous flips)
2. Flipping earlier is never worse than flipping later, since earlier flips affect more elements
3. By processing left to right and flipping only when we see an effective 0, we ensure we use the minimum number of operations

This way, we convert the problem from "track all array modifications" to "track whether current position is flipped", reducing both time and space complexity.

Learn more about Greedy and Dynamic Programming patterns.

Solution Approach

The solution uses bit manipulation to efficiently track the state of flips without actually modifying the array.

We initialize two variables:

• ans = 0: counts the number of flip operations performed
• v = 0: tracks whether the current position has been affected by an odd (1) or even (0) number of flips

The algorithm processes each element from left to right:

for x in nums:
    x ^= v
    if x == 0:
        ans += 1
        v ^= 1

For each element x in the array:

1. Calculate effective value: x ^= v

   • This XOR operation gives us the actual current value of the element after all previous flips
   • If v = 0 (even flips), x remains unchanged
   • If v = 1 (odd flips), x is flipped (0→1, 1→0)

2. Check if flip is needed: if x == 0

   • If the effective value is 0, we must perform a flip operation at this position
   • Increment the answer counter: ans += 1
   • Toggle the flip state: v ^= 1 (this changes v from 0→1 or 1→0)

3. Continue to next element

   • The updated v value will affect all subsequent elements

The beauty of this approach is that we never actually modify the array. Instead, we use the XOR operation to compute what the value would be after all previous flips, making the decision to flip or not flip at each position.

This greedy strategy guarantees the minimum number of operations because:

• We only flip when absolutely necessary (when we encounter an effective 0)
• Each flip is optimally placed to convert a 0 to 1 while potentially helping future positions
• The left-to-right traversal ensures we don't revisit positions unnecessarily

Time complexity: O(n) where n is the length of the array Space complexity: O(1) as we only use two variables regardless of input size

Example Walkthrough

Let's walk through the solution with nums = [0, 1, 0, 1].

We'll track:

• Current index and value
• The flip state v (0 = even flips affecting this position, 1 = odd flips)
• Effective value after XOR with v
• Whether we need to flip
• Running operation count ans

Initial state: v = 0, ans = 0

Index 0: nums[0] = 0

• Effective value: 0 XOR 0 = 0
• Since effective value is 0, we need to flip
• Perform flip: ans = 1, toggle v to 1
• After this flip, all elements from index 0 to end are flipped
• Array would effectively be: [1, 0, 1, 0] (but we don't actually modify it)

Index 1: nums[1] = 1

• Effective value: 1 XOR 1 = 0 (original was 1, but one flip affects it)
• Since effective value is 0, we need to flip
• Perform flip: ans = 2, toggle v to 0
• After this flip, all elements from index 1 to end are flipped again
• Array would effectively be: [1, 1, 0, 1]

Index 2: nums[2] = 0

• Effective value: 0 XOR 0 = 0 (original was 0, two flips = even, so stays 0)
• Since effective value is 0, we need to flip
• Perform flip: ans = 3, toggle v to 1
• After this flip, all elements from index 2 to end are flipped
• Array would effectively be: [1, 1, 1, 0]

Index 3: nums[3] = 1

• Effective value: 1 XOR 1 = 0 (original was 1, but three flips affect it)
• Since effective value is 0, we need to flip
• Perform flip: ans = 4, toggle v to 0
• After this flip, element at index 3 is flipped
• Array would effectively be: [1, 1, 1, 1]

Result: Minimum operations needed = 4

The key insight: We never actually modify the array. The XOR with v tells us what the current value would be after all previous flips, allowing us to make optimal decisions in a single pass.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array nums. This is because the algorithm iterates through the array exactly once with a single for loop, and each operation inside the loop (XOR operation, comparison, increment, and XOR assignment) takes constant time O(1).

The space complexity is O(1). The algorithm only uses a fixed amount of extra space regardless of the input size - it maintains only two variables: ans to count the operations and v to track the current XOR state. These variables do not grow with the size of the input array.

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

Common Pitfalls

1. Misunderstanding the XOR Logic

Pitfall: Many developers struggle to understand why x ^= v gives us the "effective value" of the current element. They might think we need to actually modify the array or track all previous flips explicitly.

Why it happens: The XOR operation seems counterintuitive - why does XORing with 0 or 1 tell us the current state after flips?

Solution: Remember that:

• v = 0 means an even number of flips have affected this position (including zero flips)
• v = 1 means an odd number of flips have affected this position
• XOR with 0 keeps the value unchanged: 0^0=0, 1^0=1
• XOR with 1 flips the value: 0^1=1, 1^1=0

2. Incorrectly Checking for Flip Condition

Pitfall: Writing the condition as if nums[i] == 0: instead of if x == 0: (where x is the XORed value).

# WRONG
for num in nums:
    if num == 0:  # This checks the original value, not the effective value!
        operation_count += 1
        current_state ^= 1

# CORRECT
for num in nums:
    x = num ^ current_state
    if x == 0:  # Check the effective value after previous flips
        operation_count += 1
        current_state ^= 1

3. Trying to Simulate Actual Array Flips

Pitfall: Attempting to actually modify the array elements from position i to the end during each operation.

# INEFFICIENT - O(n²) time complexity
def minOperations(nums):
    operations = 0
    n = len(nums)
    for i in range(n):
        if nums[i] == 0:
            operations += 1
            # Actually flipping all elements from i to end
            for j in range(i, n):
                nums[j] = 1 - nums[j]
    return operations

Solution: Use the state tracking approach with XOR to avoid actual modifications, achieving O(n) time complexity.

4. Over-complicating the State Toggle

Pitfall: Using complex logic to toggle the state variable instead of simple XOR.

# OVERCOMPLICATED
if current_state == 0:
    current_state = 1
else:
    current_state = 0

# SIMPLE AND ELEGANT
current_state ^= 1

5. Not Understanding Why Greedy Works

Pitfall: Doubting whether the greedy approach (always flipping when we see an effective 0) truly gives the minimum operations.

Solution: Understand that:

• We process left to right, so each position is visited exactly once
• When we encounter an effective 0 at position i, we MUST flip at some point from position i or earlier
• Flipping at position i is optimal because it affects all remaining elements, potentially helping future positions
• Delaying the flip would not reduce the total number of operations needed