// Source : https://leetcode.com/problems/bulb-switcher-iii
// Author : Hao Chen
// Date   : 2021-03-29

/***************************************************************************************************** 
 *
 * There is a room with n bulbs, numbered from 1 to n, arranged in a row from left to right. 
 * Initially, all the bulbs are turned off.
 * 
 * At moment k (for k from 0 to n - 1), we turn on the light[k] bulb. A bulb change color to blue only 
 * if it is on and all the previous bulbs (to the left) are turned on too.
 * 
 * Return the number of moments in which all turned on bulbs are blue.
 * 
 * Example 1:
 * 
 * Input: light = [2,1,3,5,4]
 * Output: 3
 * Explanation: All bulbs turned on, are blue at the moment 1, 2 and 4.
 * 
 * Example 2:
 * 
 * Input: light = [3,2,4,1,5]
 * Output: 2
 * Explanation: All bulbs turned on, are blue at the moment 3, and 4 (index-0).
 * 
 * Example 3:
 * 
 * Input: light = [4,1,2,3]
 * Output: 1
 * Explanation: All bulbs turned on, are blue at the moment 3 (index-0).
 * Bulb 4th changes to blue at the moment 3.
 * 
 * Example 4:
 * 
 * Input: light = [2,1,4,3,6,5]
 * Output: 3
 * 
 * Example 5:
 * 
 * Input: light = [1,2,3,4,5,6]
 * Output: 6
 * 
 * Constraints:
 * 
 * 	n == light.length
 * 	1 <= n <= 5 * 10^4
 * 	light is a permutation of  [1, 2, ..., n]
 ******************************************************************************************************/

class Solution {
public:
    int numTimesAllBlue(vector<int>& light) {
        int n = light.size();
        vector<bool> on(n, false);
        int left = 0; //tracking the most left place that all bubls are truned on.
        int result = 0;
        for(int i=0; i<light.size(); i++){
            on[light[i]-1] = true;
            while (left < n && on[left]) left++;
            //if the bulbs are on left is equal to current bulbs we trun on.
            //then they all are blue.
            if (left == i+1) result++; 
        }
        return result;
    }
};