g2201_2300.s2272_substring_with_largest_variance.Solution Maven / Gradle / Ivy

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package g2201_2300.s2272_substring_with_largest_variance;

// #Hard #Array #Dynamic_Programming #2022_06_16_Time_469_ms_(23.66%)_Space_43.7_MB_(32.44%)

/**
 * 2272 - Substring With Largest Variance\.
 *
 * Hard
 *
 * 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.
 *
 * **Example 1:**
 *
 * **Input:** s = "aababbb"
 *
 * **Output:** 3
 *
 * **Explanation:** All possible variances along with their respective substrings are listed below: 
 *
 * - Variance 0 for substrings "a", "aa", "ab", "abab", "aababb", "ba", "b", "bb", and "bbb". 
 *
 * - Variance 1 for substrings "aab", "aba", "abb", "aabab", "ababb", "aababbb", and "bab". 
 *
 * - Variance 2 for substrings "aaba", "ababbb", "abbb", and "babb". 
 *
 * - Variance 3 for substring "babbb". 
 *   
 * Since the largest possible variance is 3, we return it.
 *
 * **Example 2:**
 *
 * **Input:** s = "abcde"
 *
 * **Output:** 0
 *
 * **Explanation:** No letter occurs more than once in s, so the variance of every substring is 0.
 *
 * **Constraints:**
 *
 * *   1 <= s.length <= 10⁴
 * *   `s` consists of lowercase English letters.
**/
public class Solution {
    public int largestVariance(String s) {
        int[] freq = new int[26];
        for (int i = 0; i < s.length(); i++) {
            freq[s.charAt(i) - 'a']++;
        }
        int maxVariance = 0;
        for (int a = 0; a < 26; a++) {
            for (int b = 0; b < 26; b++) {
                int remainingA = freq[a];
                int remainingB = freq[b];
                if (a == b || remainingA == 0 || remainingB == 0) {
                    continue;
                }
                int currBFreq = 0;
                int currAFreq = 0;
                for (int i = 0; i < s.length(); i++) {
                    int c = s.charAt(i) - 'a';
                    if (c == b) {
                        currBFreq++;
                    }
                    if (c == a) {
                        currAFreq++;
                        remainingA--;
                    }

                    if (currAFreq > 0) {
                        maxVariance = Math.max(maxVariance, currBFreq - currAFreq);
                    }
                    if (currBFreq < currAFreq && remainingA >= 1) {
                        currBFreq = 0;
                        currAFreq = 0;
                    }
                }
            }
        }
        return maxVariance;
    }
}