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1714
Biweekly Contest 47 Q3
Hash Table
String
Counting

1781. Sum of Beauty of All Substrings

中文文档

Description

The beauty of a string is the difference in frequencies between the most frequent and least frequent characters.

  • For example, the beauty of "abaacc" is 3 - 1 = 2.

Given a string s, return the sum of beauty of all of its substrings.

 

Example 1:

Input: s = "aabcb"
Output: 5
Explanation: The substrings with non-zero beauty are ["aab","aabc","aabcb","abcb","bcb"], each with beauty equal to 1.

Example 2:

Input: s = "aabcbaa"
Output: 17

 

Constraints:

  • 1 <= s.length <= 500
  • s consists of only lowercase English letters.

Solutions

Solution 1: Enumeration + Counting

Enumerate the starting position $i$ of each substring, find all substrings with the character at this starting position as the left endpoint, then calculate the beauty value of each substring, and accumulate it to the answer.

The time complexity is $O(n^2 \times C)$, and the space complexity is $O(C)$. Here, $n$ is the length of the string, and $C$ is the size of the character set. In this problem, $C = 26$.

Python3

class Solution:
    def beautySum(self, s: str) -> int:
        ans, n = 0, len(s)
        for i in range(n):
            cnt = Counter()
            for j in range(i, n):
                cnt[s[j]] += 1
                ans += max(cnt.values()) - min(cnt.values())
        return ans

Java

class Solution {
    public int beautySum(String s) {
        int ans = 0;
        int n = s.length();
        for (int i = 0; i < n; ++i) {
            int[] cnt = new int[26];
            for (int j = i; j < n; ++j) {
                ++cnt[s.charAt(j) - 'a'];
                int mi = 1000, mx = 0;
                for (int v : cnt) {
                    if (v > 0) {
                        mi = Math.min(mi, v);
                        mx = Math.max(mx, v);
                    }
                }
                ans += mx - mi;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int beautySum(string s) {
        int ans = 0;
        int n = s.size();
        int cnt[26];
        for (int i = 0; i < n; ++i) {
            memset(cnt, 0, sizeof cnt);
            for (int j = i; j < n; ++j) {
                ++cnt[s[j] - 'a'];
                int mi = 1000, mx = 0;
                for (int& v : cnt) {
                    if (v > 0) {
                        mi = min(mi, v);
                        mx = max(mx, v);
                    }
                }
                ans += mx - mi;
            }
        }
        return ans;
    }
};

Go

func beautySum(s string) (ans int) {
	for i := range s {
		cnt := [26]int{}
		for j := i; j < len(s); j++ {
			cnt[s[j]-'a']++
			mi, mx := 1000, 0
			for _, v := range cnt {
				if v > 0 {
					if mi > v {
						mi = v
					}
					if mx < v {
						mx = v
					}
				}
			}
			ans += mx - mi
		}
	}
	return
}

JavaScript

/**
 * @param {string} s
 * @return {number}
 */
var beautySum = function (s) {
    let ans = 0;
    for (let i = 0; i < s.length; ++i) {
        const cnt = new Map();
        for (let j = i; j < s.length; ++j) {
            cnt.set(s[j], (cnt.get(s[j]) || 0) + 1);
            const t = Array.from(cnt.values());
            ans += Math.max(...t) - Math.min(...t);
        }
    }
    return ans;
};

Solution 2

Python3

class Solution:
    def beautySum(self, s: str) -> int:
        ans, n = 0, len(s)
        for i in range(n):
            cnt = Counter()
            freq = Counter()
            mi = mx = 1
            for j in range(i, n):
                freq[cnt[s[j]]] -= 1
                cnt[s[j]] += 1
                freq[cnt[s[j]]] += 1

                if cnt[s[j]] == 1:
                    mi = 1
                if freq[mi] == 0:
                    mi += 1
                if cnt[s[j]] > mx:
                    mx = cnt[s[j]]

                ans += mx - mi
        return ans

Java

class Solution {
    public int beautySum(String s) {
        int n = s.length();
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            int[] cnt = new int[26];
            Map<Integer, Integer> freq = new HashMap<>();
            int mi = 1, mx = 1;
            for (int j = i; j < n; ++j) {
                int k = s.charAt(j) - 'a';
                freq.merge(cnt[k], -1, Integer::sum);
                ++cnt[k];
                freq.merge(cnt[k], 1, Integer::sum);

                if (cnt[k] == 1) {
                    mi = 1;
                }
                if (freq.getOrDefault(mi, 0) == 0) {
                    ++mi;
                }
                if (cnt[k] > mx) {
                    mx = cnt[k];
                }
                ans += mx - mi;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int beautySum(string s) {
        int n = s.size();
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            int cnt[26]{};
            unordered_map<int, int> freq;
            int mi = 1, mx = 1;
            for (int j = i; j < n; ++j) {
                int k = s[j] - 'a';
                --freq[cnt[k]];
                ++cnt[k];
                ++freq[cnt[k]];

                if (cnt[k] == 1) {
                    mi = 1;
                }
                if (freq[mi] == 0) {
                    ++mi;
                }
                if (cnt[k] > mx) {
                    mx = cnt[k];
                }
                ans += mx - mi;
            }
        }
        return ans;
    }
};

Go

func beautySum(s string) (ans int) {
	n := len(s)
	for i := 0; i < n; i++ {
		cnt := [26]int{}
		freq := map[int]int{}
		mi, mx := 1, 1
		for j := i; j < n; j++ {
			k := int(s[j] - 'a')
			freq[cnt[k]]--
			cnt[k]++
			freq[cnt[k]]++

			if cnt[k] == 1 {
				mi = 1
			}
			if freq[mi] == 0 {
				mi++
			}
			if cnt[k] > mx {
				mx = cnt[k]
			}
			ans += mx - mi
		}
	}
	return
}

JavaScript

/**
 * @param {string} s
 * @return {number}
 */
var beautySum = function (s) {
    const n = s.length;
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        const cnt = Array(26).fill(0);
        const freq = new Map();
        let [mi, mx] = [1, 1];
        for (let j = i; j < n; ++j) {
            const k = s[j].charCodeAt() - 97;
            freq.set(cnt[k], (freq.get(cnt[k]) || 0) - 1);
            ++cnt[k];
            freq.set(cnt[k], (freq.get(cnt[k]) || 0) + 1);
            if (cnt[k] === 1) {
                mi = 1;
            }
            if (freq.get(mi) === 0) {
                ++mi;
            }
            if (cnt[k] > mx) {
                mx = cnt[k];
            }
            ans += mx - mi;
        }
    }
    return ans;
};