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

Tap to expand

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

✓ Brute Force - Try All Combinations

⏱️ Time: O(2^n × n) Space: O(n)

Generate all possible combinations of operations and check which one achieves all 1s with minimum operations. This involves trying every subset of possible 3-element flips.

Greedy Left-to-Right

⏱️ Time: O(n) Space: O(1)

Scan the array from left to right. Whenever we encounter a 0, we must flip it (since we can't change previous elements). Count the number of flips needed.

Brute Force - Try All Combinations — Algorithm Steps

Generate all possible positions to flip (0 to n-3)
Try all combinations of these positions
For each combination, simulate flips and check if result is all 1s
Return minimum operations needed

Visualization

Tap to expand

Step-by-Step Walkthrough

Generate Positions

List all possible starting positions for 3-element flips

Try Combinations

Test every subset of positions to flip

Find Minimum

Return the combination with fewest operations that makes all 1s

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.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 minOps = INT_MAX;
    int maxPos = n - 2;
    
    for (int mask = 0; mask < (1 << maxPos); mask++) {
        int temp[n];
        memcpy(temp, nums, n * sizeof(int));
        int ops = __builtin_popcount(mask);
        
        for (int i = 0; i < maxPos; i++) {
            if (mask & (1 << i)) {
                temp[i] ^= 1;
                temp[i + 1] ^= 1;
                temp[i + 2] ^= 1;
            }
        }
        
        int allOnes = 1;
        for (int i = 0; i < n; i++) {
            if (temp[i] != 1) {
                allOnes = 0;
                break;
            }
        }
        
        if (allOnes && ops < minOps) {
            minOps = ops;
        }
    }
    
    return minOps == INT_MAX ? -1 : minOps;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

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(2^n × n)

2^n possible combinations of positions to flip, each taking O(n) time to simulate

⚠ Quadratic Growth

Space Complexity

O(n)

Space to store current state and track operations

⚡ Linearithmic Space

12.5K Views

Medium Frequency

~25 min Avg. Time

245 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Try All Combinations — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler