3357. Minimize the Maximum Adjacent Element Difference
You are given an array of integers nums. Some values in nums are missing and are denoted by -1.
You can choose a pair of positive integers (x, y) exactly once and replace each missing element with either x or y.
You need to minimize the maximum absolute difference between adjacent elements of nums after replacements.
Return the minimum possible difference.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
| class Solution {
bool helper(vector<long long>& nums, long long k) {
vector<pair<long long, long long>> A;
vector<long long> fl(nums.size());
for(int i = 0; i < nums.size(); i++) {
if(nums[i] != -1) continue;
long long l = INT_MIN, r = INT_MAX;
if(i and nums[i-1] != -1) {
l = max(l, nums[i-1] - k);
r = min(r, nums[i-1] + k);
}
if(i + 1 < nums.size() and nums[i+1] != -1) {
l = max(l, nums[i+1] - k);
r = min(r, nums[i+1] + k);
}
if(l > r) return false;
bool ok = false;
for(int j = 0; j < A.size() and !ok; j++) {
long long mi = max(l, A[j].first), ma = min(r, A[j].second);
if(mi <= ma) {
A[j] = {mi,ma};
ok = true;
fl[i] = j + 1;
}
}
if(!ok) {
A.push_back({l,r});
fl[i] = A.size();
}
if(A.size() > 2) return false;
}
for(int i = 0; i < nums.size() - 1; i++) {
if(nums[i] == -1 and nums[i+1] == -1) {
if(fl[i] != fl[i+1]) {
sort(begin(A), end(A));
return A[1].first - A[0].second <= k;
}
}
}
return true;
}
public:
int minDifference(vector<int>& nums) {
long long l = 0, r = INT_MAX, res = r;
vector<long long> A;
for(int i = 0; i < nums.size(); i++) {
if(i + 1 != nums.size() and nums[i] != -1 and nums[i+1] != -1) l = max(l, 1ll * abs(nums[i] - nums[i+1]));
if(nums[i] != -1) A.push_back(nums[i]);
else {
if(A.size() == 0 or A.back() != -1) A.push_back(nums[i]);
else if(A.size() == 1 and A.back() == -1) {}
else {
if(A[A.size() - 2] == -1) continue;
A.push_back(nums[i]);
}
}
}
if(A.size() >= 2 and A.back() == -1 and A[A.size() - 2] == -1) A.pop_back();
if(A.size() <= 1) return 0;
while(l <= r){
int m = l + (r - l) / 2;
bool ok = helper(A,m);
if(ok) {
res = m;
r = m - 1;
} else l = m + 1;
}
return res;
}
};
|
