LeetCode 1775. 通过最少操作次数使数组的和相等

LeetCode 1775. 通过最少操作次数使数组的和相等

由于只出现正整数[1, 6], 辣么明显先用计数排序处理, 得到数组和的表达式

\[sum_1 = \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ a_3 \\ a_4 \\ a_5 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ 6 \end{bmatrix}^{-1} sum_2 = \begin{bmatrix} b_0 \\ b_1 \\ b_2 \\ b_3 \\ b_4 \\ b_5 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ 6 \end{bmatrix}^{-1} \]

朴素观察, 则题意为通过改变\(a_i\) \(b_i\), 使得两数组和相等, 即

\[\begin{cases} sum_1 = sum_2 \\ \sum_{i=0}^5 a_i = C_1 \\ \sum_{i=0}^5 b_i = C_2 \end{cases} \]

\(C_1\) \(C_2\)均为常量, 通过调整\(a_i\) \(b_i\)使得\(sum_1\) \(sum_2\)相等

假设初始状态下 \(sum_1 < sum_2\)

由于每次值变化最终是单调函数, 则\(a_i\)只向\(a_{[i+1, 5]}\)转移, 同理\(b_i\)只向\(b_{[i+1, 5]}\)转移

更进一步, 同时观察\(a_i\) \(b_i\), 变化值v如下

\[v_1 = \begin{bmatrix} a_i \rightarrow a_{i+5} \\ a_i \rightarrow a_{i+4} \\ a_i \rightarrow a_{i+3} \\ a_i \rightarrow a_{i+2} \\ a_i \rightarrow a_{i+1} \\ \end{bmatrix} \cdot \begin{bmatrix} 5 \\ 4 \\ 3 \\ 2 \\ 1 \end{bmatrix}^{-1} v_2 = \begin{bmatrix} b_i \rightarrow b_{i-5} \\ b_i \rightarrow b_{i-4} \\ b_i \rightarrow b_{i-3} \\ b_i \rightarrow b_{i-2} \\ b_i \rightarrow b_{i-1} \\ \end{bmatrix} \cdot \begin{bmatrix} -5 \\ -4 \\ -3 \\ -2 \\ -1 \end{bmatrix}^{-1} \]

在\(|sum_2 - sum_1|\)减少\(n\)时, \(a_i \rightarrow a_{i+n}\)等价\(b_i \rightarrow b_{i-n}\)

即\(a_0 \rightarrow a_5\)等价\(b_5 \rightarrow b_0\)

合并\(a_i\)与\(b_i\)表达式

\[\begin{bmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \\ c_4 \\ c_5 \end{bmatrix} = \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ a_3 \\ a_4 \\ a_5 \end{bmatrix} + \begin{bmatrix} b_5 \\ b_4 \\ b_3 \\ b_2 \\ b_1 \\ b_0 \end{bmatrix} = \begin{bmatrix} a_0 + b_5 \\ a_1 + b_4 \\ a_2 + b_3 \\ a_3 + b_2 \\ a_4 + b_1 \\ a_5 + b_0 \end{bmatrix} \]

问题转换为通过削减 \(c_i\) 使 \(|sum_2 - sum_1|\) 接近0, 则设\(c_i\)到最终值为\(c'_i\)

\[\begin{bmatrix} c'_0 \\ c'_1 \\ c'_2 \\ c'_3 \\ c'_4 \\ c'_5 \end{bmatrix} \cdot \begin{bmatrix} 5 \\ 4 \\ 3 \\ 2 \\ 1 \\ 0 \end{bmatrix} = |sum_2 - sum_1| \]

\(c'_i\)限制条件如下

\[c'_i \in \begin{cases} [0, c_0] &i=0 \\ [0, c_i+\sum_0^{i-1}c'_i] &i\gt 0 \end{cases} \]

最终结果

\[ans = min(\sum_{i=0}^4 c'_i) \]

接着贪心, 为使总步骤数最小, 每次尽可能削减较大的值, 从5到1依次遍历, 就能得出结果了

Go

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

func minOperations(nums1 []int, nums2 []int) int {
	diff, cntArr := 0, make([]int, 6)
	for _, v := range nums1 {
		diff -= v
		cntArr[v-1]++
	}
	for _, v := range nums2 {
		diff += v
		cntArr[6-v]++
	}
	if diff < 0 {
		diff *= -1
		for i, j := 0, len(cntArr)-1; i < j; i, j = i+1, j-1 {
			cntArr[i], cntArr[j] = cntArr[j], cntArr[i]
		}
	}

	opt := 0
	for i := 0; i < 5 && diff > 0; i++ {
		c := min(cntArr[i], diff/(5-i))
		opt += c

		diff -= c * (5 - i)
		cntArr[i+1] += cntArr[i] - c
	}

	if diff > 0 {
		return -1
	}
	return opt
}