Minimum Operations to Make Binary Array Elements Equal to One I

Minimum Operations to Make Binary Array Elements Equal to One I - Problem

Array Bit Manipulation Queue Sliding Window Prefix Sum Medium

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.

Input & Output

Example 1 — Basic Case

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

› Output: 3

💡 Note: Start with [0,1,1,1,0,0]. Flip positions 0-2: [1,0,0,1,0,0]. Flip positions 1-3: [1,1,1,0,0,0]. Flip positions 3-5: [1,1,1,1,1,1]. Total: 3 operations.

Example 2 — Impossible Case

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

› Output: -1

💡 Note: Array length is 4. Last element is 1, but positions 1,2 are both 1. If we flip positions 1-3, we get [0,0,0,0]. Cannot make all 1s.

Example 3 — All Ones

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

› Output: 0

💡 Note: All elements are already 1, no operations needed.

Constraints

3 ≤ nums.length ≤ 10⁵
nums[i] is either 0 or 1

Visualization

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Understanding the Visualization

Input Array

Binary array with 0s and 1s that need to become all 1s

3-Element Flips

Can only flip 3 consecutive elements at once

Minimum Operations

Find fewest flips to make all elements 1, or return -1 if impossible

Key Takeaway

🎯 Key Insight: Process left to right - when you encounter a 0, you must flip it immediately since you can't go back

Asked in

G Google 15 f Meta 12 a Amazon 8

The key insight is to use a greedy left-to-right approach. When we encounter a 0, we must flip it with the next 2 elements since we can't change previous elements. Best approach is greedy with Time: O(n), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Greedy Left-to-Right	O(n)	O(1)	Process array left to right, flip whenever we encounter a 0
Brute Force - Try All Combinations	O(2^n × n)	O(n)	Try all possible sequences of operations until finding the minimum

Greedy Left-to-Right — Algorithm Steps

Scan array from left to right
When we find a 0, flip it with next 2 elements
Count total operations
Check if final array has all 1s

Visualization

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Step-by-Step Walkthrough

Scan Left

Start from leftmost element, look for zeros

Flip When Zero

When finding 0, flip it with next 2 elements

Count Operations

Track total flips needed to make all 1s

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums, int n) {
    if (n < 3) {
        for (int i = 0; i < n; i++) {
            if (nums[i] == 0) return -1;
        }
        return 0;
    }
    
    int operations = 0;
    
    // Process from left to right
    for (int i = 0; i <= n - 3; i++) {
        if (nums[i] == 0) {
            // Flip 3 consecutive elements starting at i
            nums[i] ^= 1;
            nums[i + 1] ^= 1;
            nums[i + 2] ^= 1;
            operations++;
        }
    }
    
    // Check if all elements are 1
    for (int i = 0; i < n; i++) {
        if (nums[i] != 1) {
            return -1;
        }
    }
    
    return operations;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int nums[100];
    int n = 0;
    char* token = strtok(line, "[],\n ");
    while (token != NULL) {
        nums[n++] = atoi(token);
        token = strtok(NULL, "[],\n ");
    }
    
    int result = solution(nums, n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through array, constant work per element

✓ Linear Growth

Space Complexity

O(1)

Only using counter variable, modifying input array in place

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Greedy Left-to-Right — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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