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