Description
You are given an integer array nums. You must perform exactly one operation where you can replace one element nums[i] with nums[i] * nums[i].
Return the maximum possible subarray sum after exactly one operation. The subarray must be non-empty.
Example 1:
Input: nums = [2,-1,-4,-3]
Output: 17
Explanation: You can perform the operation on index 2 (0-indexed) to make nums = [2,-1,16,-3]. Now, the maximum subarray sum is 2 + -1 + 16 = 17.
Example 2:
Input: nums = [1,-1,1,1,-1,-1,1]
Output: 4
Explanation: You can perform the operation on index 1 (0-indexed) to make nums = [1,1,1,1,-1,-1,1]. Now, the maximum subarray sum is 1 + 1 + 1 + 1 = 4.
Constraints:
<li><code>1 <= nums.length <= 10<sup>5</sup></code></li>
<li><code>-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup></code></li>
Solutions
Solution 1
PythonJavaC++GoRust
class Solution:
def maxSumAfterOperation(self, nums: List[int]) -> int:
f = g = 0
ans = -inf
for x in nums:
ff = max(f, 0) + x
gg = max(max(f, 0) + x * x, g + x)
f, g = ff, gg
ans = max(ans, f, g)
return ans(code-box)
class Solution {
public int maxSumAfterOperation(int[] nums) {
int f = 0, g = 0;
int ans = Integer.MIN_VALUE;
for (int x : nums) {
int ff = Math.max(f, 0) + x;
int gg = Math.max(Math.max(f, 0) + x * x, g + x);
f = ff;
g = gg;
ans = Math.max(ans, Math.max(f, g));
}
return ans;
}
}(code-box)
class Solution {
public:
int maxSumAfterOperation(vector<int>& nums) {
int f = 0, g = 0;
int ans = INT_MIN;
for (int x : nums) {
int ff = max(f, 0) + x;
int gg = max(max(f, 0) + x * x, g + x);
f = ff;
g = gg;
ans = max({ans, f, g});
}
return ans;
}
};(code-box)
func maxSumAfterOperation(nums []int) int {
var f, g int
ans := -(1 << 30)
for _, x := range nums {
ff := max(f, 0) + x
gg := max(max(f, 0)+x*x, g+x)
f, g = ff, gg
ans = max(ans, max(f, g))
}
return ans
}(code-box)
impl Solution {
#[allow(dead_code)]
pub fn max_sum_after_operation(nums: Vec<i32>) -> i32 {
// Here f[i] represents the value of max sub-array that ends with nums[i] with no substitution
let mut f = 0;
// g[i] represents the case with exact one substitution
let mut g = 0;
let mut ret = 1 << 31;
// Begin the actual dp process
for e in &nums {
// f[i] = MAX(f[i - 1], 0) + nums[i]
let new_f = std::cmp::max(f, 0) + *e;
// g[i] = MAX(MAX(f[i - 1], 0) + nums[i] * nums[i], g[i - 1] + nums[i])
let new_g = std::cmp::max(std::cmp::max(f, 0) + *e * *e, g + *e);
// Update f[i] & g[i]
f = new_f;
g = new_g;
// Since we start at 0, update answer after updating f[i] & g[i]
ret = std::cmp::max(ret, g);
}
ret
}
}(code-box)
