2809. Minimum Time to Make Array Sum At Most x

Description

You are given two 0-indexed integer arrays nums1 and nums2 of equal length. Every second, for all indices 0 <= i < nums1.length, value of nums1[i] is incremented by nums2[i]. After this is done, you can do the following operation:

• Choose an index 0 <= i < nums1.length and make nums1[i] = 0.

You are also given an integer x.

Return the minimum time in which you can make the sum of all elements of nums1 to be less than or equal to x, or -1 if this is not possible.

Example 1:

Input: nums1 = [1,2,3], nums2 = [1,2,3], x = 4
Output: 3
Explanation:
For the 1st second, we apply the operation on i = 0. Therefore nums1 = [0,2+2,3+3] = [0,4,6].
For the 2nd second, we apply the operation on i = 1. Therefore nums1 = [0+1,0,6+3] = [1,0,9].
For the 3rd second, we apply the operation on i = 2. Therefore nums1 = [1+1,0+2,0] = [2,2,0].
Now sum of nums1 = 4. It can be shown that these operations are optimal, so we return 3.



Example 2:

Input: nums1 = [1,2,3], nums2 = [3,3,3], x = 4
Output: -1
Explanation: It can be shown that the sum of nums1 will always be greater than x, no matter which operations are performed.


Constraints:

• 1 <= nums1.length <= 103
• 1 <= nums1[i] <= 103
• 0 <= nums2[i] <= 103
• nums1.length == nums2.length
• 0 <= x <= 106

Solutions

Solution 1: Sorting + Dynamic Programming

We notice that if we operate on the same number multiple times, only the last operation is meaningful, and the rest of the operations on that number will only increase the other numbers. Therefore, we operate on each number at most once, that is to say, the number of operations is within $[0,..n]$.

Let’s assume that we have performed $j$ operations, and the indices of the numbers operated on are $i_1, i_2, \cdots, i_j$. For these $j$ operations, the value that each operation can reduce the sum of array elements is:

\begin{aligned} & d_1 = nums_1[i_1] + nums_2[i_1] \times 1 \\ & d_2 = nums_1[i_2] + nums_2[i_2] \times 2 \\ & \cdots \\ & d_j = nums_1[i_j] + nums_2[i_j] \times j \end{aligned}

From a greedy perspective, in order to maximize the reduction of the sum of array elements, we should let the larger elements in $nums_2$ be operated on as late as possible. Therefore, we can sort $nums_1$ and $nums_2$ in ascending order of the element values in $nums_2$.

Next, we consider the implementation of dynamic programming. We use $f[i][j]$ to represent the maximum value that can reduce the sum of array elements for the first $i$ elements of the array $nums_1$ with $j$ operations. We can get the following state transition equation:

$f[i][j] = \max \{f[i-1][j], f[i-1][j-1] + nums_1[i] + nums_2[i] \times j\}$

Finally, we enumerate $j$ and find the smallest $j$ that satisfies $s_1 + s_2 \times j - f[n][j] \le x$.

The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$, where $n$ is the length of the array.

We notice that the state $f[i][j]$ is only related to $f[i-1][j]$ and $f[i-1][j-1]$, so we can optimize the first dimension and reduce the space complexity to $O(n)$.

• class Solution {
public int minimumTime(List<Integer> nums1, List<Integer> nums2, int x) {
int n = nums1.size();
int[][] f = new int[n + 1][n + 1];
int[][] nums = new int[n][0];
for (int i = 0; i < n; ++i) {
nums[i] = new int[] {nums1.get(i), nums2.get(i)};
}
Arrays.sort(nums, Comparator.comparingInt(a -> a[1]));
for (int i = 1; i <= n; ++i) {
for (int j = 0; j <= n; ++j) {
f[i][j] = f[i - 1][j];
if (j > 0) {
int a = nums[i - 1][0], b = nums[i - 1][1];
f[i][j] = Math.max(f[i][j], f[i - 1][j - 1] + a + b * j);
}
}
}
int s1 = 0, s2 = 0;
for (int v : nums1) {
s1 += v;
}
for (int v : nums2) {
s2 += v;
}

for (int j = 0; j <= n; ++j) {
if (s1 + s2 * j - f[n][j] <= x) {
return j;
}
}
return -1;
}
}

• class Solution {
public:
int minimumTime(vector<int>& nums1, vector<int>& nums2, int x) {
int n = nums1.size();
vector<pair<int, int>> nums;
for (int i = 0; i < n; ++i) {
nums.emplace_back(nums2[i], nums1[i]);
}
sort(nums.begin(), nums.end());
int f[n + 1][n + 1];
memset(f, 0, sizeof(f));
for (int i = 1; i <= n; ++i) {
for (int j = 0; j <= n; ++j) {
f[i][j] = f[i - 1][j];
if (j) {
auto [b, a] = nums[i - 1];
f[i][j] = max(f[i][j], f[i - 1][j - 1] + a + b * j);
}
}
}
int s1 = accumulate(nums1.begin(), nums1.end(), 0);
int s2 = accumulate(nums2.begin(), nums2.end(), 0);
for (int j = 0; j <= n; ++j) {
if (s1 + s2 * j - f[n][j] <= x) {
return j;
}
}
return -1;
}
};

• class Solution:
def minimumTime(self, nums1: List[int], nums2: List[int], x: int) -> int:
n = len(nums1)
f = [[0] * (n + 1) for _ in range(n + 1)]
for i, (a, b) in enumerate(sorted(zip(nums1, nums2), key=lambda z: z[1]), 1):
for j in range(n + 1):
f[i][j] = f[i - 1][j]
if j > 0:
f[i][j] = max(f[i][j], f[i - 1][j - 1] + a + b * j)
s1 = sum(nums1)
s2 = sum(nums2)
for j in range(n + 1):
if s1 + s2 * j - f[n][j] <= x:
return j
return -1


• func minimumTime(nums1 []int, nums2 []int, x int) int {
n := len(nums1)
f := make([][]int, n+1)
for i := range f {
f[i] = make([]int, n+1)
}
type pair struct{ a, b int }
nums := make([]pair, n)
var s1, s2 int
for i := range nums {
s1 += nums1[i]
s2 += nums2[i]
nums[i] = pair{nums1[i], nums2[i]}
}
sort.Slice(nums, func(i, j int) bool { return nums[i].b < nums[j].b })
for i := 1; i <= n; i++ {
for j := 0; j <= n; j++ {
f[i][j] = f[i-1][j]
if j > 0 {
a, b := nums[i-1].a, nums[i-1].b
f[i][j] = max(f[i][j], f[i-1][j-1]+a+b*j)
}
}
}
for j := 0; j <= n; j++ {
if s1+s2*j-f[n][j] <= x {
return j
}
}
return -1
}

• function minimumTime(nums1: number[], nums2: number[], x: number): number {
const n = nums1.length;
const f: number[][] = Array(n + 1)
.fill(0)
.map(() => Array(n + 1).fill(0));
const nums: number[][] = [];
for (let i = 0; i < n; ++i) {
nums.push([nums1[i], nums2[i]]);
}
nums.sort((a, b) => a[1] - b[1]);
for (let i = 1; i <= n; ++i) {
for (let j = 0; j <= n; ++j) {
f[i][j] = f[i - 1][j];
if (j > 0) {
const [a, b] = nums[i - 1];
f[i][j] = Math.max(f[i][j], f[i - 1][j - 1] + a + b * j);
}
}
}
const s1 = nums1.reduce((a, b) => a + b, 0);
const s2 = nums2.reduce((a, b) => a + b, 0);
for (let j = 0; j <= n; ++j) {
if (s1 + s2 * j - f[n][j] <= x) {
return j;
}
}
return -1;
}