66 lines
1.9 KiB
C++

// Source : https://leetcode.com/problems/bulb-switcher-iv/
// Author : Hao Chen
// Date : 2021-03-29
/*****************************************************************************************************
*
* There is a room with n bulbs, numbered from 0 to n-1, arranged in a row from left to right.
* Initially all the bulbs are turned off.
*
* Your task is to obtain the configuration represented by target where target[i] is '1' if the i-th
* bulb is turned on and is '0' if it is turned off.
*
* You have a switch to flip the state of the bulb, a flip operation is defined as follows:
*
* Choose any bulb (index i) of your current configuration.
* Flip each bulb from index i to n-1.
*
* When any bulb is flipped it means that if it is 0 it changes to 1 and if it is 1 it changes to 0.
*
* Return the minimum number of flips required to form target.
*
* Example 1:
*
* Input: target = "10111"
* Output: 3
* Explanation: Initial configuration "00000".
* flip from the third bulb: "00000" -> "00111"
* flip from the first bulb: "00111" -> "11000"
* flip from the second bulb: "11000" -> "10111"
* We need at least 3 flip operations to form target.
*
* Example 2:
*
* Input: target = "101"
* Output: 3
* Explanation: "000" -> "111" -> "100" -> "101".
*
* Example 3:
*
* Input: target = "00000"
* Output: 0
*
* Example 4:
*
* Input: target = "001011101"
* Output: 5
*
* Constraints:
*
* 1 <= target.length <= 10^5
* target[i] == '0' or target[i] == '1'
******************************************************************************************************/
class Solution {
public:
int minFlips(string target) {
//flip the target to initalization
int flip = 0;
for(auto state : target) {
if (state == '0' && flip % 2 == 1 ) flip++;
if (state == '1' && flip % 2 == 0 ) flip++;
}
return flip;
}
};