Smallest Range II in C++

Suppose we have an array A of integers, for each integer A[i] we have to choose either x = -K or x = K, and add x to A[i] (only once). So after this process, we have some array B. We have to find the smallest possible difference between the maximum value of B and the minimum value of B. So if the input is A = [0,10], K = 2, then the output will be 6, as B = [2,8].

To solve this, we will follow these steps −

• set ret := 0, n := size of array A

• sort the array A

• set ret := Last element of A – first element of A

• right := Last element of A – K and left := first element of A + k

• for i in range 0 to n – 1

  • mx := max of A[i] + k and right

  • mn := min of A[i + 1] – k and left

  • ret := min of ret and (mx - min)

• return ret

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
   public:
   int smallestRangeII(vector<int>& A, int k) {
      int ret = 0;
      int n = A.size();
      sort(A.begin(), A.end());
      ret = A[n - 1] - A[0];
      int mx, mn;
      int right = A[n - 1] - k;
      int left = A[0] + k;
      for(int i = 0; i < n - 1; i++){
         mx = max(A[i] + k, right);
         mn = min(A[i + 1] - k, left);
         ret = min(ret, mx - mn);
      }
    return ret;
   }
};
main(){
   vector<int> v = {0, 10};
   Solution ob;
   cout << (ob.smallestRangeII(v, 2));
}

Input

[0,10]
2

Output

6