702675. Count minimum right flips to set all values in an array
Faulty Wiring and Bulbs: Count Minimum Right Flips to Set All Values in an Array
Summary
We are given an array of bulbs, where each bulb can be either on (1) or off (0). The wiring is faulty, so we need to perform a minimum number of right flips to set all values in the array. This problem involves binary representation and arrays.
Detailed Explanation
The solution involves iterating through the array from left to right. For each bulb, check if it needs to be flipped. If a bulb needs to be flipped, increment the flip count. The key insight is that we only need to flip bulbs on the right side of an unflipped bulb to set all values in the array.
Here's the step-by-step breakdown:
- Initialize the flip count to 0.
- Iterate through the array from left to right.
- For each bulb, check if it needs to be flipped by comparing its value with the previous bulb's value (if it exists).
- If a bulb needs to be flipped, increment the flip count.
- Return the flip count as the minimum number of right flips required.
Time complexity: O(n), where n is the length of the array. We only need to iterate through the array once.
Space complexity: O(1), as we only use a constant amount of space to store the flip count and previous bulb's value (if it exists).
Optimized Solutions
Java
public int minFlips(int[] bulbs) {
int flips = 0;
for (int i = 0; i < bulbs.length; i++) {
if (i == 0 || bulbs[i] != bulbs[i - 1]) {
flips++;
}
}
return flips;
}