LeetCode 3192 - Minimum Operations to Make Binary Array Elements Equal to One II

LeetCode Problem 3192

Difficulty: 🟡 Medium
Topics: Array, Dynamic Programming, Greedy

Solution

Problem Understanding

This problem asks us to take a binary array nums, which contains only 0s and 1s, and transform it so that all elements become 1 using the minimum number of allowed operations. The operation is defined as choosing any index i and flipping all elements from that index to the end of the array. Flipping a 0 turns it into 1, and flipping a 1 turns it into 0.

The input nums represents a sequence of binary values, and the output is a single integer indicating the minimum number of flips required. Constraints indicate that the array can be large, up to 10^5 elements, so any naive O(n²) approach that tries all possible flips sequentially will be too slow. We are guaranteed that nums is non-empty and only contains 0s and 1s.

Important edge cases include arrays that are already all 1s (requiring 0 operations), arrays that are all 0s, or alternating sequences of 0s and 1s which could trick a naive implementation into unnecessary flips.

Approaches

Brute Force

A brute force approach would simulate every possible flip. For each element, you could try flipping the suffix starting at that index and recursively compute the number of flips required for the rest of the array. This would guarantee correctness because it explicitly explores all sequences of operations. However, this approach is exponential in time complexity O(2^n), because each element gives two possibilities: flip or do not flip. It is infeasible for large arrays.

Optimal Approach

The key insight for an optimal solution is that we can solve this problem greedily by observing the cumulative effect of flips. We iterate from left to right, maintaining a variable that tracks the current "flip parity," i.e., whether the current element has been flipped an odd number of times. If the current element, after accounting for previous flips, is 0, we need to flip the suffix starting at this index, and we increment our operation counter. If it is already 1, we do nothing. This ensures each 0 that needs to be turned to 1 is flipped exactly once.

This approach works in O(n) time and O(1) space because it only requires a single pass through the array and a counter to track flip parity.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(2^n)	O(n)	Simulate every sequence of flips recursively
Optimal	O(n)	O(1)	Greedy pass tracking cumulative flip parity

Algorithm Walkthrough

1. Initialize ops = 0 to count the number of flips.

2. Initialize flip = 0 to track whether the current element is flipped an odd number of times.

3. Iterate through each element num in the array:

4. If flip is even, the element is in its original state; if flip is odd, the element is flipped.

5. If the current effective value of num is 0, it must be flipped:

6. Increment ops by 1.

7. Increment flip by 1 to account for the flip affecting the remaining elements.

8. Return ops as the minimum number of operations required.

Why it works: The invariant is that we track how many flips have affected the current element. Flipping only when necessary ensures that each 0 is turned into 1 with the minimal number of operations. Any earlier flip changes the state of all subsequent elements, and by using parity, we can efficiently track this effect without modifying the array.

Python Solution

from typing import List

class Solution:
    def minOperations(self, nums: List[int]) -> int:
        ops = 0
        flip = 0
        for num in nums:
            effective = num if flip % 2 == 0 else 1 - num
            if effective == 0:
                ops += 1
                flip += 1
        return ops

Explanation: We iterate through the array while maintaining a flip count. For each element, we calculate its effective value after previous flips. If it is 0, we flip the suffix, incrementing both the operation counter and the flip counter. The solution uses constant space and makes a single pass over the array.

Go Solution

func minOperations(nums []int) int {
    ops := 0
    flip := 0
    for _, num := range nums {
        effective := num
        if flip % 2 != 0 {
            effective = 1 - num
        }
        if effective == 0 {
            ops++
            flip++
        }
    }
    return ops
}

Go-specific notes: We use a range loop to iterate over the slice, and the flip parity is checked with flip % 2. The logic mirrors the Python solution exactly. No special handling for empty slices is needed, as the loop simply skips them.

Worked Examples

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

Index	num	flip	effective	ops
0	0	0	0	1
1	1	1	0	2
2	1	2	1	2
3	0	2	0	3
4	1	3	0	4

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

Index	num	flip	effective	ops
0	1	0	1	0
1	0	0	0	1
2	0	1	1	1
3	0	1	1	1

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Single pass through the array, n = length of nums
Space	O(1)	Only two integer counters are used, no additional data structures

The linear time complexity is optimal since each element must be considered at least once. Constant space is achieved by tracking flip parity rather than modifying the array.

Test Cases

# Provided examples
assert Solution().minOperations([0,1,1,0,1]) == 4  # Mixed 0s and 1s
assert Solution().minOperations([1,0,0,0]) == 1    # Leading 1 then 0s

# Edge cases
assert Solution().minOperations([1,1,1,1]) == 0    # Already all 1s
assert Solution().minOperations([0,0,0,0]) == 1    # All 0s require one flip
assert Solution().minOperations([0]) == 1          # Single 0
assert Solution().minOperations([1]) == 0          # Single 1
assert Solution().minOperations([1,0,1,0,1,0]) == 3 # Alternating 1 and 0

Test	Why
[0,1,1,0,1]	Typical mixed array
[1,0,0,0]	Single initial 1 followed by 0s
[1,1,1,1]	No flips needed
[0,0,0,0]	Single flip converts all 0s
[0]	Minimal array with 0
[1]	Minimal array with 1
[1,0,1,0,1,0]	Alternating pattern stresses flip tracking

Edge Cases

All elements are already 1: If the array contains only 1s, no operations are needed. The algorithm correctly returns 0 because the effective value of every element is already 1.

All elements are 0: If the array contains only 0s, a single flip at index 0 will convert the entire array to 1. The algorithm handles this because the first element triggers a flip, and no additional flips are required.

Alternating 0s and 1s: Sequences like [1,0,1,0,1,0] could confuse naive implementations that flip blindly. The flip parity ensures that we only flip when necessary, correctly producing the minimal number of operations.

Single element arrays: Arrays with one element, either 0 or 1, are handled gracefully. If it is 0, one flip is needed. If it is 1, zero flips are required. This tests boundary handling in both Python and Go implementations.