Longest Substring with At Most K Distinct Characters

Longest Substring with At Most K Distinct Characters - Problem

Given a string s and an integer k, return the length of the longest substring of s that contains at most k distinct characters.

A substring is a contiguous sequence of characters within a string. The goal is to find the maximum possible length while ensuring the substring has no more than k unique characters.

Example: If s = "eceba" and k = 2, the longest substring with at most 2 distinct characters is "ece" with length 3.

Input & Output

Example 1 — Basic Case

$ Input: s = "eceba", k = 2

› Output: 3

💡 Note: The longest substring with at most 2 distinct characters is "ece" with length 3. It contains only 'e' and 'c'.

Example 2 — All Same Characters

$ Input: s = "aaaa", k = 1

› Output: 4

💡 Note: The entire string contains only 1 distinct character 'a', so the answer is 4.

Example 3 — Edge Case k=0

$ Input: s = "abc", k = 0

› Output: 0

💡 Note: When k=0, no characters are allowed, so the longest valid substring has length 0.

Constraints

1 ≤ s.length ≤ 5 × 10⁴
0 ≤ k ≤ 50
s consists of lowercase English letters

Visualization

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Understanding the Visualization

Input

String s and integer k representing max distinct characters

Process

Find longest substring with ≤ k distinct characters

Output

Length of the longest valid substring

Key Takeaway

🎯 Key Insight: Use sliding window technique - expand when valid, shrink when too many distinct characters

Asked in

G Google 45 f Facebook 38 a Amazon 32 M Microsoft 28

The sliding window approach is optimal. Use two pointers and a hash map to maintain a window with at most k distinct characters. Key insight: expand right when possible, shrink left when necessary. Time: O(n), Space: O(k)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Check All Substrings	O(n³)	O(min(k, 26))	Check every possible substring and count distinct characters
Sliding Window with Hash Map	O(n)	O(min(k, 26))	Use two pointers and hash map to maintain a valid window

Brute Force - Check All Substrings — Algorithm Steps

Step 1: Generate all possible substrings using nested loops
Step 2: For each substring, count distinct characters using a set
Step 3: If distinct count ≤ k, update maximum length
Step 4: Return the maximum length found

Visualization

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Step-by-Step Walkthrough

Generate Substrings

Create all possible substrings using nested loops

Count Distinct

For each substring, count unique characters

Track Maximum

Keep the longest valid substring length

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(char* s, int k) {
    if (k == 0) return 0;
    
    int maxLen = 0;
    int n = strlen(s);
    
    // Check all possible substrings
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
            // Count distinct characters in substring s[i...j]
            int charCount[256] = {0}; // ASCII characters
            int distinctCount = 0;
            
            for (int p = i; p <= j; p++) {
                if (charCount[(unsigned char)s[p]] == 0) {
                    distinctCount++;
                }
                charCount[(unsigned char)s[p]]++;
            }
            
            // If distinct characters <= k, update max length
            if (distinctCount <= k) {
                int currentLen = j - i + 1;
                if (currentLen > maxLen) {
                    maxLen = currentLen;
                }
            }
        }
    }
    
    return maxLen;
}

int main() {
    char s[10001];
    int k;
    
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0; // Remove newline
    
    scanf("%d", &k);
    
    int result = solution(s, k);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Outer loop O(n) for start, inner loop O(n) for end, O(n) to count distinct chars in each substring

⚠ Quadratic Growth

Space Complexity

O(min(k, 26))

Set to store distinct characters, at most k characters or 26 for lowercase letters

⚡ Linearithmic Space

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High Frequency

~18 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check All Substrings — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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