3040. Maximum Number of Operations With the Same Score II
Description
Given an array of integers called nums, you can perform any of the following operation while nums contains at least 2 elements:
- Choose the first two elements of
numsand delete them. - Choose the last two elements of
numsand delete them. - Choose the first and the last elements of
numsand delete them.
The score of the operation is the sum of the deleted elements.
Your task is to find the maximum number of operations that can be performed, such that all operations have the same score.
Return the maximum number of operations possible that satisfy the condition mentioned above.
Example 1:
Input: nums = [3,2,1,2,3,4] Output: 3 Explanation: We perform the following operations: - Delete the first two elements, with score 3 + 2 = 5, nums = [1,2,3,4]. - Delete the first and the last elements, with score 1 + 4 = 5, nums = [2,3]. - Delete the first and the last elements, with score 2 + 3 = 5, nums = []. We are unable to perform any more operations as nums is empty.
Example 2:
Input: nums = [3,2,6,1,4] Output: 2 Explanation: We perform the following operations: - Delete the first two elements, with score 3 + 2 = 5, nums = [6,1,4]. - Delete the last two elements, with score 1 + 4 = 5, nums = [6]. It can be proven that we can perform at most 2 operations.
Constraints:
2 <= nums.length <= 20001 <= nums[i] <= 1000
Solutions
Solution 1: Memorization Search
There are three possible values for the score $s$, which are $s = nums[0] + nums[1]$, $s = nums[0] + nums[n-1]$, and $s = nums[n-1] + nums[n-2]$. We can perform memorization search for these three cases separately.
We design a function $dfs(i, j)$, which represents the maximum number of operations from index $i$ to index $j$ when the score is $s$. The execution process of function $dfs(i, j)$ is as follows:
If $j - i < 1$, it means that the length of the interval $[i, j]$ is less than $2$, and no operation can be performed, so return $0$.
If $nums[i] + nums[i+1] = s$, it means that the elements at index $i$ and index $i+1$ can be deleted. In this case, the maximum number of operations is $1 + dfs(i+2, j)$.
If $nums[i] + nums[j] = s$, it means that the elements at index $i$ and index $j$ can be deleted. In this case, the maximum number of operations is $1 + dfs(i+1, j-1)$.
If $nums[j-1] + nums[j] = s$, it means that the elements at index $j-1$ and index $j$ can be deleted. In this case, the maximum number of operations is $1 + dfs(i, j-2)$.
Return the maximum of the above values.
Finally, we calculate the maximum number of operations for the three cases separately, and return the maximum value.
The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Here, $n$ is the length of the array $nums$.
Python3
class Solution:
def maxOperations(self, nums: List[int]) -> int:
@cache
def dfs(i: int, j: int, s: int) -> int:
if j - i < 1:
return 0
ans = 0
if nums[i] + nums[i + 1] == s:
ans = max(ans, 1 + dfs(i + 2, j, s))
if nums[i] + nums[j] == s:
ans = max(ans, 1 + dfs(i + 1, j - 1, s))
if nums[j - 1] + nums[j] == s:
ans = max(ans, 1 + dfs(i, j - 2, s))
return ans
n = len(nums)
a = dfs(2, n - 1, nums[0] + nums[1])
b = dfs(0, n - 3, nums[-1] + nums[-2])
c = dfs(1, n - 2, nums[0] + nums[-1])
return 1 + max(a, b, c)
Java
class Solution {
private Integer[][] f;
private int[] nums;
private int s;
private int n;
public int maxOperations(int[] nums) {
this.nums = nums;
n = nums.length;
int a = g(2, n - 1, nums[0] + nums[1]);
int b = g(0, n - 3, nums[n - 2] + nums[n - 1]);
int c = g(1, n - 2, nums[0] + nums[n - 1]);
return 1 + Math.max(a, Math.max(b, c));
}
private int g(int i, int j, int s) {
f = new Integer[n][n];
this.s = s;
return dfs(i, j);
}
private int dfs(int i, int j) {
if (j - i < 1) {
return 0;
}
if (f[i][j] != null) {
return f[i][j];
}
int ans = 0;
if (nums[i] + nums[i + 1] == s) {
ans = Math.max(ans, 1 + dfs(i + 2, j));
}
if (nums[i] + nums[j] == s) {
ans = Math.max(ans, 1 + dfs(i + 1, j - 1));
}
if (nums[j - 1] + nums[j] == s) {
ans = Math.max(ans, 1 + dfs(i, j - 2));
}
return f[i][j] = ans;
}
}
C++
class Solution {
public:
int maxOperations(vector<int>& nums) {
int n = nums.size();
int f[n][n];
auto g = [&](int i, int j, int s) -> int {
memset(f, -1, sizeof(f));
function<int(int, int)> dfs = [&](int i, int j) -> int {
if (j - i < 1) {
return 0;
}
if (f[i][j] != -1) {
return f[i][j];
}
int ans = 0;
if (nums[i] + nums[i + 1] == s) {
ans = max(ans, 1 + dfs(i + 2, j));
}
if (nums[i] + nums[j] == s) {
ans = max(ans, 1 + dfs(i + 1, j - 1));
}
if (nums[j - 1] + nums[j] == s) {
ans = max(ans, 1 + dfs(i, j - 2));
}
return f[i][j] = ans;
};
return dfs(i, j);
};
int a = g(2, n - 1, nums[0] + nums[1]);
int b = g(0, n - 3, nums[n - 2] + nums[n - 1]);
int c = g(1, n - 2, nums[0] + nums[n - 1]);
return 1 + max({a, b, c});
}
};
Go
func maxOperations(nums []int) int {
n := len(nums)
var g func(i, j, s int) int
g = func(i, j, s int) int {
f := make([][]int, n)
for i := range f {
f[i] = make([]int, n)
for j := range f {
f[i][j] = -1
}
}
var dfs func(i, j int) int
dfs = func(i, j int) int {
if j-i < 1 {
return 0
}
if f[i][j] != -1 {
return f[i][j]
}
ans := 0
if nums[i]+nums[i+1] == s {
ans = max(ans, 1+dfs(i+2, j))
}
if nums[i]+nums[j] == s {
ans = max(ans, 1+dfs(i+1, j-1))
}
if nums[j-1]+nums[j] == s {
ans = max(ans, 1+dfs(i, j-2))
}
f[i][j] = ans
return ans
}
return dfs(i, j)
}
a := g(2, n-1, nums[0]+nums[1])
b := g(0, n-3, nums[n-1]+nums[n-2])
c := g(1, n-2, nums[0]+nums[n-1])
return 1 + max(a, b, c)
}
TypeScript
function maxOperations(nums: number[]): number {
const n = nums.length;
const f: number[][] = Array.from({ length: n }, () => Array(n));
const g = (i: number, j: number, s: number): number => {
f.forEach(row => row.fill(-1));
const dfs = (i: number, j: number): number => {
if (j - i < 1) {
return 0;
}
if (f[i][j] !== -1) {
return f[i][j];
}
let ans = 0;
if (nums[i] + nums[i + 1] === s) {
ans = Math.max(ans, 1 + dfs(i + 2, j));
}
if (nums[i] + nums[j] === s) {
ans = Math.max(ans, 1 + dfs(i + 1, j - 1));
}
if (nums[j - 1] + nums[j] === s) {
ans = Math.max(ans, 1 + dfs(i, j - 2));
}
return (f[i][j] = ans);
};
return dfs(i, j);
};
const a = g(2, n - 1, nums[0] + nums[1]);
const b = g(0, n - 3, nums[n - 2] + nums[n - 1]);
const c = g(1, n - 2, nums[0] + nums[n - 1]);
return 1 + Math.max(a, b, c);
}