2222. Number of Ways to Select Buildings LeetCode Solution
In this guide, you will get 2222. Number of Ways to Select Buildings LeetCode Solution with the best time and space complexity. The solution to Number of Ways to Select Buildings problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Number of Ways to Select Buildings
You are given a 0-indexed binary string s which represents the types of buildings along a street where:
s[i] = ‘0’ denotes that the ith building is an office and
s[i] = ‘1’ denotes that the ith building is a restaurant.
As a city official, you would like to select 3 buildings for random inspection. However, to ensure variety, no two consecutive buildings out of the selected buildings can be of the same type.
For example, given s = “001101”, we cannot select the 1st, 3rd, and 5th buildings as that would form “011” which is not allowed due to having two consecutive buildings of the same type.
Return the number of valid ways to select 3 buildings.
Example 1:
Input: s = “001101”
Output: 6
Explanation:
The following sets of indices selected are valid:
– [0,2,4] from “001101” forms “010”
– [0,3,4] from “001101” forms “010”
– [1,2,4] from “001101” forms “010”
– [1,3,4] from “001101” forms “010”
– [2,4,5] from “001101” forms “101”
– [3,4,5] from “001101” forms “101”
No other selection is valid. Thus, there are 6 total ways.
Input: s = “11100”
Output: 0
Explanation: It can be shown that there are no valid selections.
Constraints:
3 <= s.length <= 105
s[i] is either '0' or '1'.
Complexity Analysis
Time Complexity: O(n)
Space Complexity: O(1)
2222. Number of Ways to Select Buildings LeetCode Solution in C++
class Solution {
public:
long long numberOfWays(string s) {
long ans = 0;
// before[i] := the number of i before the current digit
vector<int> before(2);
// after[i] := the number of i after the current digit
vector<int> after(2);
after[0] = ranges::count(s, '0');
after[1] = s.length() - after[0];
for (const char c : s) {
const int num = c - '0';
--after[num];
if (num == 0)
ans += before[1] * after[1];
else
ans += before[0] * after[0];
++before[num];
}
return ans;
}
}; /* code provided by PROGIEZ */
2222. Number of Ways to Select Buildings LeetCode Solution in Java
class Solution {
public long numberOfWays(String s) {
long ans = 0;
// before[i] := the number of i before the current digit
int[] before = new int[2];
// after[i] := the number of i after the current digit
int[] after = new int[2];
after[0] = (int) s.chars().filter(c -> c == '0').count();
after[1] = s.length() - after[0];
for (final char c : s.toCharArray()) {
final int num = c - '0';
--after[num];
if (num == 0)
ans += before[1] * after[1];
else
ans += before[0] * after[0];
++before[num];
}
return ans;
}
} // code provided by PROGIEZ
2222. Number of Ways to Select Buildings LeetCode Solution in Python
class Solution:
def numberOfWays(self, s: str) -> int:
ans = 0
# before[i] := the number of i before the current digit
before = [0] * 2
# after[i] := the number of i after the current digit
after = [0] * 2
after[0] = s.count('0')
after[1] = len(s) - after[0]
for c in s:
num = int(c)
after[num] -= 1
if num == 0:
ans += before[1] * after[1]
else:
ans += before[0] * after[0]
before[num] += 1
return ans # code by PROGIEZ
