Build confidence with an intuition-first walkthrough focused on two pointers fundamentals.
Given a binary string s, return the number of non-empty substrings that have the same number of 0's and 1's, and all the 0's and all the 1's in these substrings are grouped consecutively.
Substrings that occur multiple times are counted the number of times they occur.
Example 1:
Input: s = "00110011" Output: 6 Explanation: There are 6 substrings that have equal number of consecutive 1's and 0's: "0011", "01", "1100", "10", "0011", and "01". Notice that some of these substrings repeat and are counted the number of times they occur. Also, "00110011" is not a valid substring because all the 0's (and 1's) are not grouped together.
Example 2:
Input: s = "10101" Output: 4 Explanation: There are 4 substrings: "10", "01", "10", "01" that have equal number of consecutive 1's and 0's.
Constraints:
1 <= s.length <= 105s[i] is either '0' or '1'.Problem summary: Given a binary string s, return the number of non-empty substrings that have the same number of 0's and 1's, and all the 0's and all the 1's in these substrings are grouped consecutively. Substrings that occur multiple times are counted the number of times they occur.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Two Pointers
"00110011"
"10101"
encode-and-decode-strings)number-of-substrings-with-fixed-ratio)count-the-number-of-substrings-with-dominant-ones)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #696: Count Binary Substrings
class Solution {
public int countBinarySubstrings(String s) {
int n = s.length();
int ans = 0;
int i = 0;
int pre = 0;
while (i < n) {
int j = i + 1;
while (j < n && s.charAt(j) == s.charAt(i)) {
j++;
}
int cur = j - i;
ans += Math.min(pre, cur);
pre = cur;
i = j;
}
return ans;
}
}
// Accepted solution for LeetCode #696: Count Binary Substrings
func countBinarySubstrings(s string) (ans int) {
n := len(s)
i := 0
pre := 0
for i < n {
j := i + 1
for j < n && s[j] == s[i] {
j++
}
cur := j - i
ans += min(pre, cur)
pre = cur
i = j
}
return
}
# Accepted solution for LeetCode #696: Count Binary Substrings
class Solution:
def countBinarySubstrings(self, s: str) -> int:
n = len(s)
ans = i = 0
pre = 0
while i < n:
j = i + 1
while j < n and s[j] == s[i]:
j += 1
cur = j - i
ans += min(pre, cur)
pre = cur
i = j
return ans
// Accepted solution for LeetCode #696: Count Binary Substrings
impl Solution {
pub fn count_binary_substrings(s: String) -> i32 {
let bytes = s.as_bytes();
let n: usize = bytes.len();
let mut ans: i32 = 0;
let mut i: usize = 0;
let mut pre: i32 = 0;
while i < n {
let mut j: usize = i + 1;
while j < n && bytes[j] == bytes[i] {
j += 1;
}
let cur: i32 = (j - i) as i32;
ans += pre.min(cur);
pre = cur;
i = j;
}
ans
}
}
// Accepted solution for LeetCode #696: Count Binary Substrings
function countBinarySubstrings(s: string): number {
const n = s.length;
let ans = 0;
let i = 0;
let pre = 0;
while (i < n) {
let j = i + 1;
while (j < n && s[j] === s[i]) {
j++;
}
const cur = j - i;
ans += Math.min(pre, cur);
pre = cur;
i = j;
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.
Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.
Review these before coding to avoid predictable interview regressions.
Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.
Usually fails on: A valid pair can be skipped when only one side should move.
Fix: Move exactly one pointer per decision branch based on invariant.