[LeetCode] 2712. Minimum Cost to Make All Characters Equal

You are given a 0-indexed binary string s of length n on which you can apply two types of operations:

Choose an index i and invert all characters from index 0 to index i (both inclusive), with a cost of i + 1
Choose an index i and invert all characters from index i to index n - 1 (both inclusive), with a cost of n - i
Return the minimum cost to make all characters of the string equal.

Invert a character means if its value is ‘0’ it becomes ‘1’ and vice-versa.

Example 1:
Input: s = “0011”
Output: 2
Explanation: Apply the second operation with i = 2 to obtain s = “0000” for a cost of 2. It can be shown that 2 is the minimum cost to make all characters equal.

Example 2:
Input: s = “010101”
Output: 9
Explanation: Apply the first operation with i = 2 to obtain s = “101101” for a cost of 3.
Apply the first operation with i = 1 to obtain s = “011101” for a cost of 2.
Apply the first operation with i = 0 to obtain s = “111101” for a cost of 1.
Apply the second operation with i = 4 to obtain s = “111110” for a cost of 2.
Apply the second operation with i = 5 to obtain s = “111111” for a cost of 1.
The total cost to make all characters equal is 9. It can be shown that 9 is the minimum cost to make all characters equal.

Constraints:
1 <= s.length == n <= 105
s[i] is either ‘0’ or ‘1’

使所有字符相等的最小成本。

给你一个下标从 0 开始、长度为 n 的二进制字符串 s ，你可以对其执行两种操作：

选中一个下标 i 并且反转从下标 0 到下标 i（包括下标 0 和下标 i ）的所有字符，成本为 i + 1 。
选中一个下标 i 并且反转从下标 i 到下标 n - 1（包括下标 i 和下标 n - 1 ）的所有字符，成本为 n - i 。
返回使字符串内所有字符 相等 需要的 最小成本 。

反转 字符意味着：如果原来的值是 ‘0’ ，则反转后值变为 ‘1’ ，反之亦然。

思路

思路是贪心。题目定义了在下标 i 处反转的成本是 i + 1 或 n - i。如果在某个下标 i 处我们发现他与前一个位置 i - 1 处的字符不一样，那么我们一定要做一次翻转操作。翻转的成本是翻转从下标 0 到下标 i与从下标 i 到下标 n - 1的最小值。

复杂度

时间O(n)
空间O(n) - char array

代码

Java实现

class Solution {
    public long minimumCost(String s) {
        char[] letters = s.toCharArray();
        int n = s.length();
        long res = 0;
        for (int i = 1; i < n; i++) {
            if (letters[i] != letters[i - 1]) {
                res += Math.min(i, n - i);
            }
        }
        return res;
    }
}