Pair with minimum absolute difference after solving each query

Given Q queries and an empty list.

The queries can be of two types:

  1. addToList(x) : Add x to your list.
  2. removeFromList(x) : Remove x from your list.

The task is, after each query, you have to print the minimum value of abs(list[i]-list[j]) where, 0<=i<=n, 0<=j<=n and i ≠ j and n is the total number of elements present in the list.

Examples:



Input : Q = 4
addToList(1), insert 1 in our set
addToList(5), insert 5 in our set
addToList(3), insert 3 in our set
removeFromList(3), delete 3 in our set
Output :
0, as we have only one element {1} in our set.
4, as we have {1, 5} minimum difference between all possible pairs is 4 (5-1)
2, as we have {1, 3, 5} minimum difference between all possible pairs is 2 (3-1)
4, as we have {1, 5} minimum difference between all possible pairs is 4 (5-1)

Method 1: The idea is to use set in C++ to store all of the elements so that insertion or deletion can be done in O(log(n)) with keeping the elements in a sorted order. Now, to find the minimum difference, we have to iterate over the whole set and find the difference between only adjacent elements as elements are in sorted order i.e, the minimum difference will always be contributed by adjacent elements. This can be done in O(n) time complexity. So, the overall time complexity of this approach will be (q*log(n)+q*n).

Method 2:
We can use multiset to store and maintain all of the elements and map to store the difference between the adjacent elements in sorted order, where, map’s key represents the difference between adjacent elements and the corresponding map’s value represent the frequency of occurrence of this difference.

Below is the complete algorithm:
First of all, insert two sentinel value in the multi-set. Let’s say (-10^9+7 and 10^9+7) and initialize the map with the difference of these sentinel value, i.e 2*10^7+7 and set its frequency to be 1.

Now we have two type of queries:

  1. Insertion Query: For an insertion query, first insert the value in set. After inserting the element, we will have to also update the differences of adjacent elements in the map. To do this, find the left and right values of the newly inserted element in the set. Earlier when this new value was not inserted in the set, abs(right-left) is contributing to the difference term in the map. To update the map after inserting the new element, first remove the previous difference abs(right-left) as a new value is inserted between the elements left and right and add two differences to the map. i.e, abs(right-x) and abs(x-left).
  2. Deletion Query: In case of deletion of a value from the set. First, find the left and right elements of the element needed to be deleted, say x. Then according to the above algorithm, reduce the frequency of abs(right-x) and abs(left-x) and increment the frequency of abs(right-left).

So, each query can be solved in log(n) time complexity. Therefore overall complexity will be O(q*log(n)).

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// C++ implementation of above approach
#include <bits/stdc++.h>
#define ll long long int
const ll MOD = 1e9 + 7;
using namespace std;
  
// Function to add an element into the list
void addToList(int x, multiset<ll>& s, map<ll, ll>& mp)
{
    // firstly insert value in set
    s.insert(x);
  
    // find left and right value of inserted value.
    ll left, right;
  
    // now here is logic explained below
    left = *(--s.find(x));
    right = *(++s.find(x));
    mp[abs(left - x)]++;
    mp[abs(right - x)]++;
    mp[abs(left - right)]--;
    if (mp[abs(left - right)] == 0)
        mp.erase(abs(left - right));
}
  
// Function to remove an element from the list
void removeFromList(int x, multiset<ll>& s, map<ll, ll>& mp)
{
    ll left, right;
  
    // Find left element of number that we want to delete.
    left = *(--s.find(x));
  
    // Find right element of number that we want to delete.
    right = *(++s.find(x));
  
    // remove x from set
    s.erase(s.find(x));
  
    // Again this will explain below in article.
    mp[abs(left - x)]--;
    if (mp[abs(left - x)] == 0)
        mp.erase(abs(left - x));
    mp[abs(right - x)]--;
    if (mp[abs(right - x)] == 0)
        mp.erase(abs(right - x));
  
    mp[abs(left - right)]++;
}
  
// Driver code
int main()
{
  
    int q = 4;
  
    // define set to store all values.
    multiset<ll> s;
  
    // define map to store difference in sorted
    map<ll, ll> mp;
  
    // initialize set with sentinel values
    s.insert(-MOD);
    s.insert(MOD);
  
    // maintain freq of difference in map so, here intially
    // difference b/w sentinel value and its freq is one.
    mp[2 * MOD] = 1;
  
    // 1st query 1 1 i.e, include 1 in our set
    addToList(1, s, mp);
  
    // As now we have only one element {-10^9+7, 1, 10^9+7}
    // (except two are sentinel values)
    // so minimum among all pair is zero
    cout << 0 << endl;
  
    // 2nd query 1 5 i.e, include 5 in our set.
    addToList(5, s, mp);
  
    // find smallest difference possible
    // as it should be in beginning of map
    cout << mp.begin()->first << endl;
  
    // 3rd query 1 3 i.e, include 3 in our set.
    addToList(3, s, mp);
  
    // find smallest difference possible
    // as it should be in beginning of map
    cout << mp.begin()->first << endl;
  
    // 4th query 2 3 i.e, remove 3 from our list
    removeFromList(3, s, mp);
  
    cout << mp.begin()->first << endl;
  
 return 0;
}

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Output:

0
4
2
4


Explanation of each query:

  1. addToList(1): As now we have only one element {-10^9+7, 1, 10^9+7} (except two are sentinel values), so minimum among all pair is zero.
  2. addToList(5): After inserting 1 and 5 to the set, set becomes {-10^9+7, 1, 5, 10^9+7} and map: mp[5-1]=1 where key represents difference and its value represents frequency.
  3. addToList(3): As here we are inserting 3. So, first of all, we find the left and right element of 3 in the set after insertion i.e. left = 1, right = 5. As 3 came in between 1 and 5 so we remove map[5-1]–. (as 3 came in between 1 and 5 so 5-1 will no longer minimum) And we include these differences map[3-1]++ and map[5-3]++ (because 3 came between 1 and 5, so possible minimum have two cases).
  4. removeFromList(3): Similar as of above query, here we are deleting element. So, first of all, find the left and right element to 3 i.e. left = 1, right = 5. As we going to delete 3 which is in between 1 & 5 so minimum difference will be affected only of 3 and 5,
    before deleting ==>{1, 3, 5}, 
    after deleting ==> {1, 5}. 
    So, map[3-1]--, map[5-3]-- and add map[5-1]++.


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