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Formatted question description: https://leetcode.ca/all/2272.html

2272. Substring With Largest Variance

  • Difficulty: Hard.
  • Related Topics: Array, Dynamic Programming.
  • Similar Questions: Maximum Subarray.

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.

  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.

Solution

  • 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;
    }
}

############

class Solution {
    public int largestVariance(String s) {
        int n = s.length();
        int ans = 0;
        for (char a = 'a'; a <= 'z'; ++a) {
            for (char b = 'a'; b <= 'z'; ++b) {
                if (a == b) {
                    continue;
                }
                int[] f = new int[] {0, -n};
                for (int i = 0; i < n; ++i) {
                    if (s.charAt(i) == a) {
                        f[0]++;
                        f[1]++;
                    } else if (s.charAt(i) == b) {
                        f[1] = Math.max(f[0] - 1, f[1] - 1);
                        f[0] = 0;
                    }
                    ans = Math.max(ans, f[1]);
                }
            }
        }
        return ans;
    }
}

  • class Solution:
    def largestVariance(self, s: str) -> int:
        ans = 0
        for a, b in permutations(ascii_lowercase, 2):
            if a == b:
                continue
            f = [0, -inf]
            for c in s:
                if c == a:
                    f[0], f[1] = f[0] + 1, f[1] + 1
                elif c == b:
                    f[1] = max(f[1] - 1, f[0] - 1)
                    f[0] = 0
                if ans < f[1]:
                    ans = f[1]
        return ans

############

# 2272. Substring With Largest Variance
# https://leetcode.com/problems/substring-with-largest-variance

class Solution:
    def largestVariance(self, s: str) -> int:
        n = len(s)
        res = 0
        
        for a, b in permutations(set(s), 2):
            cnt = 0
            hasB = firstB = False
            
            for x in s:
                if x == a:
                    cnt += 1
                    
                if x == b:
                    hasB = True
                    
                    if firstB and cnt >= 0:
                        firstB = False
                    else:
                        cnt -= 1
                        
                        if cnt < 0:
                            firstB = True
                            cnt = -1
                
                if hasB and cnt > res:
                    res = cnt

        return res


  • class Solution {
public:
    int largestVariance(string s) {
        int n = s.size();
        int ans = 0;
        for (char a = 'a'; a <= 'z'; ++a) {
            for (char b = 'a'; b <= 'z'; ++b) {
                if (a == b) continue;
                int f[2] = {0, -n};
                for (char c : s) {
                    if (c == a) {
                        f[0]++;
                        f[1]++;
                    } else if (c == b) {
                        f[1] = max(f[1] - 1, f[0] - 1);
                        f[0] = 0;
                    }
                    ans = max(ans, f[1]);
                }
            }
        }
        return ans;
    }
};

  • func largestVariance(s string) int {
	ans, n := 0, len(s)
	for a := 'a'; a <= 'z'; a++ {
		for b := 'a'; b <= 'z'; b++ {
			if a == b {
				continue
			}
			f := [2]int{0, -n}
			for _, c := range s {
				if c == a {
					f[0]++
					f[1]++
				} else if c == b {
					f[1] = max(f[1]-1, f[0]-1)
					f[0] = 0
				}
				ans = max(ans, f[1])
			}
		}
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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