LeetCode #2222 — MEDIUM

Number of Ways to Select Buildings

Move from brute-force thinking to an efficient approach using dynamic programming strategy.

The Problem

Problem Statement

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 i^th building is an office and
s[i] = '1' denotes that the i^th 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 1^st, 3^rd, and 5^th 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.

Example 2:

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'.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Dynamic Programming

Example 1

"001101"

Example 2

"11100"

Step 02

Core Insight

What unlocks the optimal approach

There are only 2 valid patterns: ‘101’ and ‘010’. Think about how we can construct these 2 patterns from smaller patterns.
Count the number of subsequences of the form ‘01’ or ‘10’ first. Let n01[i] be the number of ‘01’ subsequences that exist in the prefix of s up to the ith building. How can you compute n01[i]?
Let n0[i] and n1[i] be the number of ‘0’s and ‘1’s that exists in the prefix of s up to i respectively. Then n01[i] = n01[i – 1] if s[i] == ‘0’, otherwise n01[i] = n01[i – 1] + n0[i – 1].
The same logic applies to building the n10 array and subsequently the n101 and n010 arrays for the number of ‘101’ and ‘010‘ subsequences.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2222: Number of Ways to Select Buildings
class Solution {
    public long numberOfWays(String s) {
        int n = s.length();
        int[] l = new int[2];
        int[] r = new int[2];
        for (int i = 0; i < n; ++i) {
            r[s.charAt(i) - '0']++;
        }
        long ans = 0;
        for (int i = 0; i < n; ++i) {
            int x = s.charAt(i) - '0';
            r[x]--;
            ans += 1L * l[x ^ 1] * r[x ^ 1];
            l[x]++;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2222: Number of Ways to Select Buildings
func numberOfWays(s string) (ans int64) {
	n := len(s)
	l := [2]int{}
	r := [2]int{}
	r[0] = strings.Count(s, "0")
	r[1] = n - r[0]
	for _, c := range s {
		x := int(c - '0')
		r[x]--
		ans += int64(l[x^1] * r[x^1])
		l[x]++
	}
	return
}

# Accepted solution for LeetCode #2222: Number of Ways to Select Buildings
class Solution:
    def numberOfWays(self, s: str) -> int:
        l = [0, 0]
        r = [s.count("0"), s.count("1")]
        ans = 0
        for x in map(int, s):
            r[x] -= 1
            ans += l[x ^ 1] * r[x ^ 1]
            l[x] += 1
        return ans

// Accepted solution for LeetCode #2222: Number of Ways to Select Buildings
/**
 * [2222] 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 i^th building is an office and
 * 	s[i] = '1' denotes that the i^th 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 = "0<u>0</u>1<u>1</u>0<u>1</u>", we cannot select the 1^st, 3^rd, and 5^th buildings as that would form "0<u>11</u>" 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 "<u>0</u>0<u>1</u>1<u>0</u>1" forms "010"
 * - [0,3,4] from "<u>0</u>01<u>10</u>1" forms "010"
 * - [1,2,4] from "0<u>01</u>1<u>0</u>1" forms "010"
 * - [1,3,4] from "0<u>0</u>1<u>10</u>1" forms "010"
 * - [2,4,5] from "00<u>1</u>1<u>01</u>" forms "101"
 * - [3,4,5] from "001<u>101</u>" forms "101"
 * No other selection is valid. Thus, there are 6 total ways.
 *
 * Example 2:
 *
 * Input: s = "11100"
 * Output: 0
 * Explanation: It can be shown that there are no valid selections.
 *
 *  
 * Constraints:
 *
 * 	3 <= s.length <= 10^5
 * 	s[i] is either '0' or '1'.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/number-of-ways-to-select-buildings/
// discuss: https://leetcode.com/problems/number-of-ways-to-select-buildings/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn number_of_ways(s: String) -> i64 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2222_example_1() {
        let s = "001101".to_string();

        let result = 6;

        assert_eq!(Solution::number_of_ways(s), result);
    }

    #[test]
    #[ignore]
    fn test_2222_example_2() {
        let s = "11100".to_string();

        let result = 0;

        assert_eq!(Solution::number_of_ways(s), result);
    }
}

// Accepted solution for LeetCode #2222: Number of Ways to Select Buildings
function numberOfWays(s: string): number {
    const n = s.length;
    const l: number[] = [0, 0];
    const r: number[] = [s.split('').filter(c => c === '0').length, 0];
    r[1] = n - r[0];
    let ans: number = 0;
    for (const c of s) {
        const x = c === '0' ? 0 : 1;
        r[x]--;
        ans += l[x ^ 1] * r[x ^ 1];
        l[x]++;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.