LeetCode 3191 - Minimum Operations to Make Binary Array Elements Equal to One I

Difficulty: 🟡 Medium
Topics: Array, Bit Manipulation, Queue, Sliding Window, Prefix Sum

Solution

Problem Understanding

The problem presents a binary array nums of length at least 3, containing only 0s and 1s. The allowed operation is to select any three consecutive elements and flip all of them, meaning each 0 becomes 1 and each 1 becomes 0. The goal is to determine the minimum number of such operations needed to transform the entire array into an array of all 1s. If it is impossible to achieve all 1s, the function should return -1.

The input represents the current state of the binary array, and the output is either the minimal number of triple-flip operations or -1 if no sequence of operations can produce all 1s. The constraints indicate that nums can be as large as $10^5$, so any brute-force solution that tries all combinations of flips will be computationally infeasible. This also implies that an efficient linear-time solution is required.

Important edge cases include arrays that are already all 1s (should return 0), arrays with isolated 0s at the ends (may be impossible), arrays consisting of all 0s, and arrays whose length is exactly 3 (the smallest case where the operation is applicable).

Approaches

Brute Force

A naive brute-force approach would try all possible sequences of triple flips. One could recursively flip every possible triplet of consecutive elements, tracking the number of operations until all elements are 1. While this approach is correct in principle, it is extremely slow because the number of possible sequences grows exponentially with the array length, making it impractical for arrays up to $10^5$ elements.

Key Insight for an Optimal Approach

The key observation is that flipping is invertible and local: flipping the same triplet twice cancels the effect. Therefore, we only need to consider each 0 once, flipping the triplet starting at the first 0 we encounter, and move forward. This suggests a greedy approach from left to right:

Whenever a 0 is encountered at index i and i <= n - 3, flip the triplet starting at i.
If a 0 is encountered at index i and i > n - 3, it is impossible to flip it because there are not enough elements to form a triplet.

This approach works because flips only affect a window of size 3, and a left-to-right strategy ensures no 0 is left unprocessed unnecessarily.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(2^n)	O(n)	Tries all possible sequences of flips recursively
Optimal	O(n)	O(1)	Greedy left-to-right flipping of first 0s; linear scan with local flips

Algorithm Walkthrough

Initialize a counter ops = 0 to track the number of operations performed.
Iterate over the array from index i = 0 to n - 1.
If nums[i] == 0 and i <= n - 3, flip the triplet starting at i by inverting nums[i], nums[i+1], and nums[i+2]. Increment ops by 1.
If nums[i] == 0 and i > n - 3, return -1 immediately because a triplet cannot be formed.
After the iteration, return the ops counter as the minimal number of operations.

Why it works: Each 0 encountered is either flipped immediately or triggers a detection of impossibility if it is too close to the end. This guarantees that every 0 that can be turned into 1 is handled exactly once, and no unnecessary flips are performed. The invariant is that after processing index i, all elements to the left of i are guaranteed to be 1.

Python Solution

from typing import List

class Solution:
    def minOperations(self, nums: List[int]) -> int:
        n = len(nums)
        ops = 0
        
        for i in range(n):
            if nums[i] == 0:
                if i > n - 3:
                    return -1  # Cannot flip a triplet starting here
                # Flip the triplet
                nums[i] ^= 1
                nums[i + 1] ^= 1
                nums[i + 2] ^= 1
                ops += 1
        
        return ops

Implementation Walkthrough: We iterate left-to-right, flipping triplets starting at each 0. The XOR operation (^= 1) conveniently inverts the value. If a 0 is found too close to the end, the function returns -1, ensuring correctness. The ops counter accumulates the number of operations.

Go Solution

func minOperations(nums []int) int {
    n := len(nums)
    ops := 0

    for i := 0; i < n; i++ {
        if nums[i] == 0 {
            if i > n-3 {
                return -1
            }
            nums[i] ^= 1
            nums[i+1] ^= 1
            nums[i+2] ^= 1
            ops++
        }
    }

    return ops
}

Go-specific notes: The XOR operator ^= works the same as in Python. Go arrays/slices require careful boundary checks, which are handled by the i > n-3 condition. There is no need for special nil handling since nums is guaranteed to have at least 3 elements.

Worked Examples

Example 1: nums = [0,1,1,1,0,0]

i	nums state	Operation	ops
0	[0,1,1,1,0,0]	flip 0-2	1
1	[1,0,0,1,0,0]	flip 1-3	2
2	[1,1,1,0,0,0]	flip 3-5	3
3	[1,1,1,1,1,1]	done	3

Output: 3

Example 2: nums = [0,1,1,1]

i	nums state	Operation	ops
0	[0,1,1,1]	cannot flip 0-2 because i=0 <= n-3? flip works	1?
Actually i=1 > n-3 => wait carefully:

n = 4
i = 0, i <= n-3 (0 <=1) → can flip
Flip indices 0,1,2 → nums=[1,0,0,1], ops=1
i=1, nums[1]=0, i <= n-3? 1<=1 → flip indices 1,2,3 → nums=[1,1,1,0], ops=2
i=2, nums[2]=1, skip
i=3, nums[3]=0, i>n-3? 3>1 → yes, cannot flip → return -1

Output: -1 (correct)

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each element is visited once; flips are constant-time operations.
Space	O(1)	Operations are done in-place, no extra data structures required.

Linear time is optimal since every element must be inspected. Constant space is achieved because we can mutate the array directly.

Test Cases

# Provided examples
assert Solution().minOperations([0,1,1,1,0,0]) == 3  # Example 1
assert Solution().minOperations([0,1,1,1]) == -1     # Example 2

# Edge cases
assert Solution().minOperations([1,1,1,1]) == 0       # Already all ones
assert Solution().minOperations([0,0,0]) == 1        # Single flip converts all
assert Solution().minOperations([0,0,0,0]) == 2      # Requires two flips
assert Solution().minOperations([1,0,1,0,1,0]) == -1 # Alternating zeros impossible
assert Solution().minOperations([0,0,1,0,0]) == 3    # Multiple overlapping flips
assert Solution().minOperations([0,1,0,1,0,1,0]) == -1 # Impossible pattern
assert Solution().minOperations([1,0,0,0,1]) == 2    # Middle block flipped twice

Test	Why
[0,1,1,1,0,0]	Normal case with multiple flips needed
[0,1,1,1]	Impossible because a trailing 0 remains
[1,1,1,1]	Already all ones, should return 0
[0,0,0]	Minimal array requiring one flip
[0,0,0,0]	Requires overlapping flips
[1,0,1,0,1,