LeetCode #3192 — MEDIUM

Minimum Operations to Make Binary Array Elements Equal to One II

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 index i from the array and flip all the elements from index i to 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 be nums = [0,0,0,1,0].
Choose the index i = 0. The resulting array will be nums = [1,1,1,0,1].
Choose the index i = 4. The resulting array will be nums = [1,1,1,0,0].
Choose the index i = 3. The resulting array will be nums = [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 be nums = [1,1,1,1].

Constraints:

1 <= 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 index i from the array and flip all the elements from index i to 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.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming · Greedy

Example 1

[0,1,1,0,1]

Example 2

[1,0,0,0]

Core Insight

What unlocks the optimal approach

The only way to change <code>nums[0]</code> to 1 is by performing an operation with index <code>i = 0</code>.
Iterate from left to right and perform an operation at each index i where nums[i] is 0, and keep track of how many operations are currently performed on the suffix.

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 #3192: Minimum Operations to Make Binary Array Elements Equal to One II
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;
    }
}

// Accepted solution for LeetCode #3192: Minimum Operations to Make Binary Array Elements Equal to One II
func minOperations(nums []int) (ans int) {
	v := 0
	for _, x := range nums {
		x ^= v
		if x == 0 {
			v ^= 1
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #3192: Minimum Operations to Make Binary Array Elements Equal to One II
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 ans

// Accepted solution for LeetCode #3192: Minimum Operations to Make Binary Array Elements Equal to One II
/**
 * [3192] Minimum Operations to Make Binary Array Elements Equal to One II
 */
pub struct Solution {}

// submission codes start here

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

        for i in 0..nums.len() {
            let num = if change_time % 2 == 1 {
                if nums[i] == 1 {
                    0
                } else {
                    1
                }
            } else {
                nums[i]
            };

            if num == 1 {
                continue;
            }

            change_time += 1;
            result += 1;
        }

        result
    }
}

// submission codes end

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

    #[test]
    fn test_3192() {
        assert_eq!(4, Solution::min_operations(vec![0, 1, 1, 0, 1]));
        assert_eq!(1, Solution::min_operations(vec![1, 0, 0, 0]));
    }
}

// Accepted solution for LeetCode #3192: Minimum Operations to Make Binary Array Elements Equal to One II
function minOperations(nums: number[]): number {
    let [ans, v] = [0, 0];
    for (let x of nums) {
        x ^= v;
        if (x === 0) {
            v ^= 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 × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.