Q1. Given an array of size n with integers in it. And given two numbers a and b, find the minimum possible score of ELIGIBLE sub-arrays of the given array.

i) The score of a sub-array is calculated as the number of distinct elements within that sub-array
ii) a sub-array is eligible if it contains atleast 1 occurence of a and b each.

#variable length sliding window.
code:

vector<int> arr, int n, int a, int b;

int ans = n+1;
unordered_map<int, int> mp;
int i=0;
for(int j=0;j<n;j++){
  mp[arr[j]]++;
  if(mp[a] > 0 && mp[b] > 0)
    ans = min(ans, (int)mp.size());
  
  while(mp[a]>0 && mp[b]>0){
	  ans = min(ans, (int)mp.size());
	  mp[arr[i]]--;
	  if(mp[arr[i]] == 0)
	  mp.erase(arr[i]);
	  i++;
  }
}
cout<<ans<<endl;

15/15 test cases passed

Q2 There are multiple delivery centers and delivery warehouses all over the world! The world is represented by a number line from -10⁹ to 10⁹. There are n delivery centers, with the i-th one at location center[i]. A location x is called a suitable location for a warehouse if it is possible to bring all the products to that point by traveling a distance of no more than d.

At any one time, products can be brought from one delivery center and placed at point x. Given the positions of n delivery centers, calculate the number of suitable locations in the world. That is, calculate the number of points x on the number line (-10⁹ ≤ x ≤ 10⁹) where the travel distance required to bring all the products to that point is less than or equal to d.

Note: The distance between point x and center[i] is |x - center[i]|, their absolute difference.

#binary search.
i was not able to solve this problem fully. i tried using binary search twice to find the minimum x (minx) which satisfies the constraints and then maximum x (maxx) which satisfies the constraints.
if minx>maxx then ans = 0
else ans = maxx-minx+1

only 12/15 test cases passed. Can someone please help with the complete solution and where exactly I am missing in this.

bool cal(vector<int> arr, int d, int x){
	int sum = 0;
	for(int i:arr){
	sum+=abs(x-i);
	}
	return sum<=d;
}

vector<int> arr, int n, int d;
int l1 = -1e9, h1 = arr[n-1]-d;
int minx = h1;
while(l1<=h1){
  int m = (l1+h1)/2;
  if(cal(arr, d, m)){
    minx = m;
	h1=m-1;
  }else{
    l1 = m+1;
  } 
}
l1=arr[0]+d, h1=1e9;
int maxx = l1;
while(l1<=h1){
  int m = (l1+h1)/2;
  if(cal(arr, d, m)){
    maxx = m;
	l1=m+1;
  }else{
    h1 = m-1;
  } 
}

if(maxx<mixx)return 0;
return (maxx-mix+1);

I was able to pass only 12/15 test cases. Can someone help me realize why this would fail and what would be the correct solution for this.