Number of Substrings Containing All Three Characters

Medium Arrays & Hashing Sliding Window · Hash Map Original on LeetCode

Number of Substrings Containing All Three Characters is a medium arrays & hashing problem solved with the sliding window · hash map pattern. The best approach, optimal (sliding window with counts), runs in O(n) time and O(1) space. Below are 2 approaches in Java, from brute force up.

Problem

Given a string s made of only a, b and c, count the substrings that contain at least one of each of the three letters.

Examples

Example 1

Input: s = "abcab"
Output: 6
Why: Starting at 0: abc, abca, abcab. At 1: bca, bcab. At 2: cab.

Example 2

Input: s = "aaacb"
Output: 3
Why: aaacb, aacb and acb.

Constraints

3 <= s.length <= 5 * 10^4
s contains only a, b and c.

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

Approach	Time	Space
Brute force	`O(n²)`	`O(1)`
Optimal (sliding window with counts)	`O(n)`	`O(1)`

1Brute force

TimeO(n²)

SpaceO(1)

For each start, extend until all three letters are present; every longer substring from that start also qualifies.

For each i, scan j until counts of a, b, c are all positive.
Add n - j to the answer.

Java

class Solution {
    public int numberOfSubstrings(String s) {
        int n = s.length(), total = 0;
        for (int i = 0; i < n; i++) {
            int[] c = new int[3];
            for (int j = i; j < n; j++) {
                c[s.charAt(j) - 'a']++;
                if (c[0] > 0 && c[1] > 0 && c[2] > 0) { total += n - j; break; }
            }
        }
        return total;
    }
}

2Optimal (sliding window with counts)

TimeO(n)

SpaceO(1)

Move r right, adding s[r]. While the window [l, r] has all three letters, every substring starting at l and ending at r or later is valid, so add n - r, then drop s[l] and move l.

Keep counts of a, b, c in the window.
For each r, add s[r].
While all counts > 0: total += n - r; remove s[l]; l++.

Java

class Solution {
    public int numberOfSubstrings(String s) {
        int n = s.length(), l = 0, total = 0;
        int[] c = new int[3];
        for (int r = 0; r < n; r++) {
            c[s.charAt(r) - 'a']++;
            while (c[0] > 0 && c[1] > 0 && c[2] > 0) {
                total += n - r;
                c[s.charAt(l++) - 'a']--;
            }
        }
        return total;
    }
}

Edge cases to test

One letter never appears (answer 0)
The minimal window is at the very end

Hints

Hint 1

If s[l..r] contains all three letters, then so does every extension s[l..r'] for r' >= r.

FAQ

What is the best time complexity for Number of Substrings Containing All Three Characters?

Optimal (sliding window with counts) runs in O(n) time and O(1) extra space.

Which pattern does Number of Substrings Containing All Three Characters use?

It is a arrays & hashing problem that uses the sliding window · hash map pattern. Other problems with the same pattern: Minimum Window Substring.

Is there a brute force solution for Number of Substrings Containing All Three Characters?

Yes. Brute force takes O(n²) time and O(1) space. For each start, extend until all three letters are present; every longer substring from that start also qualifies.

Which edge cases should I test for Number of Substrings Containing All Three Characters?

One letter never appears (answer 0); The minimal window is at the very end.