910. Smallest Range II
Given an array A
of integers, for each integer A[i]
we need to choose either x = -K
or x = K
, and add x
to A[i] (only once)
.
After this process, we have some array B
.
Return the smallest possible difference between the maximum value of B
and the minimum value of B
.
Example 1:
Input: A = [1], K = 0
Output: 0
Explanation: B = [1]
Example 2:
Input: A = [0,10], K = 2
Output: 6
Explanation: B = [2,8]
Example 3:
Input: A = [1,3,6], K = 3
Output: 3
Explanation: B = [4,6,3]
Note:
1 <= A.length <= 10000
0 <= A[i] <= 10000
0 <= K <= 10000
Approach #1: C++.
class Solution {
public:
int smallestRangeII(vector<int>& A, int K) {
int N = A.size();
sort(A.begin(), A.end());
int ans = A[N-1] - A[0]; for (int i = 0; i < A.size()-1; ++i) {
int a = A[i], b = A[i+1];
int high = max(A[N-1]-K, a+K);
int low = min(A[0]+K, b-K);
ans = min(ans, high-low);
} return ans;
}
};
Analysis:
We can formalize the above concept: if A[i] < A[j]
, we don't need to consider when A[i]
goes down while A[j]
goes up. This is because the interval (A[i] + K, A[j] - K)
is a subset of (A[i] - K, A[j] + K)
(here, (a, b)
for a > b
denotes (b, a)
instead.)
That means that it is never worse to choose (up, down)
instead of (down, up)
. We can prove this claim that one interval is a subset of another, by showing both A[i] + K
and A[j] - K
are between A[i] - K
and A[j] + K
.
For sorted A
, say A[i]
is the largest i
that goes up. Then A[0] + K, A[i] + K, A[i+1] - K, A[A.length - 1] - K
are the only relevant values for calculating the answer: every other value is between one of these extremal values.
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