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1830. Minimum Number of Operations to Make String Sorted

中文文档

Description

You are given a string s (0-indexed)​​​​​​. You are asked to perform the following operation on s​​​​​​ until you get a sorted string:

  1. Find the largest index i such that 1 <= i < s.length and s[i] < s[i - 1].
  2. Find the largest index j such that i <= j < s.length and s[k] < s[i - 1] for all the possible values of k in the range [i, j] inclusive.
  3. Swap the two characters at indices i - 1​​​​ and j​​​​​.
  4. Reverse the suffix starting at index i​​​​​​.

Return the number of operations needed to make the string sorted. Since the answer can be too large, return it modulo 109 + 7.

 

Example 1:

Input: s = "cba"
Output: 5
Explanation: The simulation goes as follows:
Operation 1: i=2, j=2. Swap s[1] and s[2] to get s="cab", then reverse the suffix starting at 2. Now, s="cab".
Operation 2: i=1, j=2. Swap s[0] and s[2] to get s="bac", then reverse the suffix starting at 1. Now, s="bca".
Operation 3: i=2, j=2. Swap s[1] and s[2] to get s="bac", then reverse the suffix starting at 2. Now, s="bac".
Operation 4: i=1, j=1. Swap s[0] and s[1] to get s="abc", then reverse the suffix starting at 1. Now, s="acb".
Operation 5: i=2, j=2. Swap s[1] and s[2] to get s="abc", then reverse the suffix starting at 2. Now, s="abc".

Example 2:

Input: s = "aabaa"
Output: 2
Explanation: The simulation goes as follows:
Operation 1: i=3, j=4. Swap s[2] and s[4] to get s="aaaab", then reverse the substring starting at 3. Now, s="aaaba".
Operation 2: i=4, j=4. Swap s[3] and s[4] to get s="aaaab", then reverse the substring starting at 4. Now, s="aaaab".

 

Constraints:

  • 1 <= s.length <= 3000
  • s​​​​​​ consists only of lowercase English letters.

Solutions

Python3

n = 3010
mod = 10**9 + 7
f = [1] + [0] * n
g = [1] + [0] * n

for i in range(1, n):
    f[i] = f[i - 1] * i % mod
    g[i] = pow(f[i], mod - 2, mod)


class Solution:
    def makeStringSorted(self, s: str) -> int:
        cnt = Counter(s)
        ans, n = 0, len(s)
        for i, c in enumerate(s):
            m = sum(v for a, v in cnt.items() if a < c)
            t = f[n - i - 1] * m
            for v in cnt.values():
                t = t * g[v] % mod
            ans = (ans + t) % mod
            cnt[c] -= 1
            if cnt[c] == 0:
                cnt.pop(c)
        return ans

Java

class Solution {
    private static final int N = 3010;
    private static final int MOD = (int) 1e9 + 7;
    private static final long[] f = new long[N];
    private static final long[] g = new long[N];

    static {
        f[0] = 1;
        g[0] = 1;
        for (int i = 1; i < N; ++i) {
            f[i] = f[i - 1] * i % MOD;
            g[i] = qmi(f[i], MOD - 2);
        }
    }

    public static long qmi(long a, int k) {
        long res = 1;
        while (k != 0) {
            if ((k & 1) == 1) {
                res = res * a % MOD;
            }
            k >>= 1;
            a = a * a % MOD;
        }
        return res;
    }

    public int makeStringSorted(String s) {
        int[] cnt = new int[26];
        int n = s.length();
        for (int i = 0; i < n; ++i) {
            ++cnt[s.charAt(i) - 'a'];
        }
        long ans = 0;
        for (int i = 0; i < n; ++i) {
            int m = 0;
            for (int j = s.charAt(i) - 'a' - 1; j >= 0; --j) {
                m += cnt[j];
            }
            long t = m * f[n - i - 1] % MOD;
            for (int v : cnt) {
                t = t * g[v] % MOD;
            }
            --cnt[s.charAt(i) - 'a'];
            ans = (ans + t + MOD) % MOD;
        }
        return (int) ans;
    }
}

C++

const int N = 3010;
const int MOD = 1e9 + 7;
long f[N];
long g[N];

long qmi(long a, int k) {
    long res = 1;
    while (k != 0) {
        if ((k & 1) == 1) {
            res = res * a % MOD;
        }
        k >>= 1;
        a = a * a % MOD;
    }
    return res;
}

int init = []() {
    f[0] = g[0] = 1;
    for (int i = 1; i < N; ++i) {
        f[i] = f[i - 1] * i % MOD;
        g[i] = qmi(f[i], MOD - 2);
    }
    return 0;
}();

class Solution {
public:
    int makeStringSorted(string s) {
        int cnt[26]{};
        for (char& c : s) {
            ++cnt[c - 'a'];
        }
        int n = s.size();
        long ans = 0;
        for (int i = 0; i < n; ++i) {
            int m = 0;
            for (int j = s[i] - 'a' - 1; ~j; --j) {
                m += cnt[j];
            }
            long t = m * f[n - i - 1] % MOD;
            for (int& v : cnt) {
                t = t * g[v] % MOD;
            }
            ans = (ans + t + MOD) % MOD;
            --cnt[s[i] - 'a'];
        }
        return ans;
    }
};

Go

const n = 3010
const mod = 1e9 + 7

var f = make([]int, n)
var g = make([]int, n)

func qmi(a, k int) int {
	res := 1
	for k != 0 {
		if k&1 == 1 {
			res = res * a % mod
		}
		k >>= 1
		a = a * a % mod
	}
	return res
}

func init() {
	f[0], g[0] = 1, 1
	for i := 1; i < n; i++ {
		f[i] = f[i-1] * i % mod
		g[i] = qmi(f[i], mod-2)
	}
}

func makeStringSorted(s string) (ans int) {
	cnt := [26]int{}
	for _, c := range s {
		cnt[c-'a']++
	}
	for i, c := range s {
		m := 0
		for j := int(c-'a') - 1; j >= 0; j-- {
			m += cnt[j]
		}
		t := m * f[len(s)-i-1] % mod
		for _, v := range cnt {
			t = t * g[v] % mod
		}
		ans = (ans + t + mod) % mod
		cnt[c-'a']--
	}
	return
}

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