Substring With Largest Variance

Substring With Largest Variance - Problem

The variance of a string is defined as the largest difference between the number of occurrences of any 2 characters present in the string. Note the two characters may or may not be the same.

Given a string s consisting of lowercase English letters only, return the largest variance possible among all substrings of s.

A substring is a contiguous sequence of characters within a string.

Input & Output

Example 1 — Basic Case

$ Input: s = "aababbb"

› Output: 3

💡 Note: The substring "bbb" has variance 3-0=3 (b appears 3 times, a appears 0 times). This is the maximum variance possible.

Example 2 — All Same Character

$ Input: s = "abcde"

› Output: 1

💡 Note: The substring "a" has a=1 occurrences while b=0 occurrences, giving variance 1-0=1. Similarly for other single character substrings with respect to any other character.

Example 3 — Two Characters

$ Input: s = "aab"

› Output: 2

💡 Note: The substring "aa" has a=2 occurrences while b=0 occurrences, giving variance 2-0=2. This is the maximum variance possible.

Constraints

1 ≤ s.length ≤ 10⁴
s consists of lowercase English letters

Visualization

Tap to expand

Asked in

f Facebook 15 G Google 12 a Amazon 8

The key insight is transforming the variance problem into a maximum subarray problem using Kadane's algorithm. For each character pair, convert the string to +1/-1 values and find the maximum subarray sum. Best approach is Modified Kadane's Algorithm. Time: O(n), Space: O(1) per character pair.

Common Approaches

✓ Brute Force - Check All Substrings

⏱️ Time: O(n³) Space: O(1)

For every possible substring, count the frequency of each character and find the maximum difference between any two character counts. This approach examines all O(n²) substrings.

Modified Kadane's Algorithm

⏱️ Time: O(n) Space: O(1)

For each pair of characters (major and minor), we transform the string into +1/-1 array where major char = +1 and minor char = -1. Then use Kadane's algorithm to find maximum subarray sum, which represents the maximum difference between these two characters.

Brute Force - Check All Substrings — Algorithm Steps

Generate all possible substrings
For each substring, count character frequencies
Find maximum difference between any two character counts
Track the overall maximum variance

Visualization

Tap to expand

Step-by-Step Walkthrough

Generate Substrings

Create all O(n²) possible substrings

Count Frequencies

For each substring, count character occurrences

Calculate Variance

Find max - min frequency difference

Code -

solution.c — C

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

int solution(char* s) {
    int n = strlen(s);
    int maxVariance = 0;
    
    // Check all possible substrings
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
            // Count frequencies in current substring
            int freq[26] = {0};
            for (int k = i; k <= j; k++) {
                freq[s[k] - 'a']++;
            }
            
            // Find max and min frequencies (only consider non-zero)
            int maxFreq = 0;
            int minFreq = INT_MAX;
            int countChars = 0;
            
            for (int idx = 0; idx < 26; idx++) {
                if (freq[idx] > 0) {
                    if (freq[idx] > maxFreq) maxFreq = freq[idx];
                    if (freq[idx] < minFreq) minFreq = freq[idx];
                    countChars++;
                }
            }
            
            // Calculate variance if at least 2 different characters
            if (countChars >= 2) {
                int variance = maxFreq - minFreq;
                if (variance > maxVariance) {
                    maxVariance = variance;
                }
            }
        }
    }
    
    return maxVariance;
}

int main() {
    char s[1000];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;  // Remove newline
    int result = solution(s);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

O(n²) substrings × O(n) to count frequencies for each substring

✓ Linear Growth

Space Complexity

O(1)

Only using frequency counter array of size 26 (constant space)

✓ Linear Space

23.0K Views

Medium Frequency

~35 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check All Substrings — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler