LeetCode #3191 — MEDIUM

Minimum Operations to Make Binary Array Elements Equal to One I

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given a binary array nums.

You can do the following operation on the array any number of times (possibly zero):

Choose any 3 consecutive elements from the array and flip all of them.

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. If it is impossible, return -1.

Example 1:

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

Output: 3

Explanation:
We can do the following operations:

Choose the elements at indices 0, 1 and 2. The resulting array is nums = [1,0,0,1,0,0].
Choose the elements at indices 1, 2 and 3. The resulting array is nums = [1,1,1,0,0,0].
Choose the elements at indices 3, 4 and 5. The resulting array is nums = [1,1,1,1,1,1].

Example 2:

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

Output: -1

Explanation:
It is impossible to make all elements equal to 1.

Constraints:

3 <= nums.length <= 10⁵
0 <= nums[i] <= 1

Step 01

Brute Force Baseline

Problem summary: You are given a binary array nums. You can do the following operation on the array any number of times (possibly zero): Choose any 3 consecutive elements from the array and flip all of them. 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. If it is impossible, return -1.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Bit Manipulation · Sliding Window

Example 1

[0,1,1,1,0,0]

Example 2

[0,1,1,1]

Core Insight

What unlocks the optimal approach

If <code>nums[0]</code> is 0, then the only way to change it to 1 is by doing an operation on the first 3 elements of the array.
After Changing <code>nums[0]</code> to 1, use the same logic on the remaining array.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3191: Minimum Operations to Make Binary Array Elements Equal to One I
class Solution {
    public int minOperations(int[] nums) {
        int ans = 0;
        int n = nums.length;
        for (int i = 0; i < n; ++i) {
            if (nums[i] == 0) {
                if (i + 2 >= n) {
                    return -1;
                }
                nums[i + 1] ^= 1;
                nums[i + 2] ^= 1;
                ++ans;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3191: Minimum Operations to Make Binary Array Elements Equal to One I
func minOperations(nums []int) (ans int) {
	for i, x := range nums {
		if x == 0 {
			if i+2 >= len(nums) {
				return -1
			}
			nums[i+1] ^= 1
			nums[i+2] ^= 1
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #3191: Minimum Operations to Make Binary Array Elements Equal to One I
class Solution:
    def minOperations(self, nums: List[int]) -> int:
        ans = 0
        for i, x in enumerate(nums):
            if x == 0:
                if i + 2 >= len(nums):
                    return -1
                nums[i + 1] ^= 1
                nums[i + 2] ^= 1
                ans += 1
        return ans

// Accepted solution for LeetCode #3191: Minimum Operations to Make Binary Array Elements Equal to One I
/**
 * [3191] Minimum Operations to Make Binary Array Elements Equal to One I
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn min_operations(nums: Vec<i32>) -> i32 {
        let mut result = 0;
        let mut nums = nums;

        let mut pos = 0;

        while pos < nums.len() - 3 {
            if nums[pos] == 1 {
                pos += 1;
                continue;
            }

            nums[pos] = 1;
            Self::reverse(pos + 1, &mut nums);
            Self::reverse(pos + 2, &mut nums);

            result += 1;
            pos += 1;
        }

        if nums[pos] ^ nums[pos + 1] == 0 && nums[pos + 1] ^ nums[pos + 2] == 0 {
            if nums[pos] == 1 {
                result
            } else {
                result + 1
            }
        } else {
            -1
        }
    }

    fn reverse(pos: usize, nums: &mut Vec<i32>) {
        nums[pos] = if nums[pos] == 1 { 0 } else { 1 }
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3191() {
        assert_eq!(-1, Solution::min_operations(vec![0, 1, 1, 0, 1, 0, 0]));
        assert_eq!(3, Solution::min_operations(vec![0, 1, 1, 1, 0, 0]));
        assert_eq!(-1, Solution::min_operations(vec![0, 1, 1, 1]));
    }
}

// Accepted solution for LeetCode #3191: Minimum Operations to Make Binary Array Elements Equal to One I
function minOperations(nums: number[]): number {
    const n = nums.length;
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        if (nums[i] === 0) {
            if (i + 2 >= n) {
                return -1;
            }
            nums[i + 1] ^= 1;
            nums[i + 2] ^= 1;
            ++ans;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.