Substring With Largest Variance

class Solution {
public:
    int findMaxVariance(string str) {
        vector<int> freq(26, 0);
        for (char c : str) {
            freq[c - 'a']++;
        }
        int maxVariance = 0;
            
        for (int primary = 0; primary < 26; primary++) {
            for (int secondary = 0; secondary < 26; secondary++) {
                // primary and secondary must be distinct and present in the string
                if (primary == secondary || freq[primary] == 0 || freq[secondary] == 0) {
                    continue;
                }
                    
                char highFreqChar = 'a' + primary;
                char lowFreqChar = 'a' + secondary;
                int highCount = 0;
                int lowCount = 0;
                    
                // Remaining instances of low frequency character
                int remainingLow = freq[secondary];
                    
                for (char element : str) {
                    if (element == highFreqChar) {
                        highCount++;
                    }
                    if (element == lowFreqChar) {
                        lowCount++;
                        remainingLow--;
                    }
                        
                    // Update maxVariance if any lowFreqChar seen so far
                    if (lowCount > 0)
                        maxVariance = max(maxVariance, highCount - lowCount);
                        
                    // Reset counts when high is less than low and there are remaining lows
                    if (highCount < lowCount && remainingLow > 0) {
                        highCount = 0;
                        lowCount = 0;
                    }
                }
            }
        }
            
        return maxVariance;
    }
};

This solution aims to solve the "Substring With Largest Variance" problem, which involves finding the maximum variance between the frequency counts of two distinct characters in any substring of a given string. The code is implemented in C++ and can be summarized by the following:

• Frequency Calculation: Initially, compute the frequency of each character in the string using an array freq, which counts occurrences of each character from 'a' to 'z'.

• Nested Iteration Over Characters: Use two nested loops to iterate over possible pairs of characters (denoted by indices primary and secondary representing characters from 'a' to 'z'). These characters represent potential high and low frequency characters in substrings.

• Condition Checks: For each pair, check whether they are distinct and both are present in the string. Skip the iteration if they are the same or if either character is not present in the string.

• Variance Calculation For Each Pair: Initialize counters for high frequency character (highCount) and low frequency character (lowCount), then iterate through the string, updating these counts whenever their corresponding characters are encountered.

  • Reset on Low Instances Remaining: If the currently considered substring has ended (determined by highCount being less than lowCount and the presence of remaining low frequency characters), reset both counts to zero to start counting for a new potential substring.

  • Updating Maximum Variance: If lowCount is positive, indicating that the low frequency character has been encountered, update maxVariance if the difference highCount - lowCount is greater than the current maxVariance.

• Return Result: After evaluating all possible pairs and substrings, return maxVariance, which contains the largest variance found.

This method efficiently identifies the substring that exhibits the highest variance between the counts of any two distinct characters, leveraging a combination of frequency counting and iteratively assessing potential substrings through character-specific counters.

class Solution {
    public int maxVariance(String str) {
        int[] freq = new int[26];
        for (char c : str.toCharArray()) {
            freq[c - 'a']++;
        }
        int maxVariance = 0;
    
        for (int p = 0; p < 26; p++) {
            for (int q = 0; q < 26; q++) {
                // Ensure p and q are different characters and both have occurrences in the string
                if (p == q || freq[p] == 0 || freq[q] == 0) continue;
    
                char primary = (char)('a' + p);
                char secondary = (char)('a' + q);
                int countPrimary = 0;
                int countSecondary = 0;
    
                // Track secondary characters left in the remainder of the string
                int remainingSecondary = freq[q];
                    
                for (char c : str.toCharArray()) {
                    if (c == primary) {
                        countPrimary++;
                    }
                    if (c == secondary) {
                        countSecondary++;
                        remainingSecondary--;
                    }
                        
                    // Calculate variance where secondary is at least 1
                    if (countSecondary > 0)
                        maxVariance = Math.max(maxVariance, countPrimary - countSecondary);
                        
                    // Reset counts if secondary can still form another segment
                    if (countPrimary < countSecondary && remainingSecondary > 0) {
                        countPrimary = 0;
                        countSecondary = 0;
                    }
                }
            }
        }
    
        return maxVariance;
    }
}

The Java solution provided solves the problem of finding the "Substring with Largest Variance". The variance in this context is defined based on the frequency differences between two characters in substrings of a given string.

Approach Overview:

• First, the solution counts the frequency of each letter in the string using an array freq of size 26 (representing the alphabet).
• The next step entails iterating over all possible pairs of characters (p, q) where p != q and both characters appear in the input string at least once.
• For each character pair, a new sweep of the string is performed using two counters, countPrimary and countSecondary, which track the occurrences of p and q respectively as the string is traversed.
• A variable remainingSecondary counts the remaining occurrences of the secondary character as the string is parsed.
• It then calculates the variance for segments where at least one occurrence of the secondary character exists, updating maxVariance if a higher variance is found using the formula countPrimary - countSecondary.
• If a segment is found where the occurrences of q exceed p, and if there are further occurrences of q ahead, both counters are reset to zero to start counting a new segment.

Steps Performed:

1. Initialize an array freq to count occurrences of each character.
2. Use two nested loops to consider all pairs of characters as primary (p) and secondary (q) respectively, ensuring different characters are selected.
3. Initialize counters and track the remaining secondary characters.
4. Traverse through the string, updating the counters based on whether the character at the current position corresponds to p or q.
5. Calculate and potentially update the maxVariance.
6. Reset both counters under specified conditions to handle subsequent segments of the string.
7. Return maxVariance once all pairs have been analyzed.

Key Optimization:

• This solution efficiently manages the search space by limiting variance calculations to character pairs that both exist in the string, and by resetting and recalculating segment variance dynamically as new potential high variance segments are encountered.