Queries for number of elements on right and left

Given Q queries of three types where every query consists of a number.

  1. Add element num on the left
  2. Add element num on the right
  3. Print the number of elements to the right an left of the given element num.

The task is to write a program that performs the above queries.

Note: Elements in queries 1 and 2 are distinct and 0 <= num <= 105.



Examples:

Input:
Q = 5
Query of type 1: num = 3
Query of type 2: num = 5
Query of type 1: num = 2
Query of type 1: num = 4
Query of type 3: num = 3
Output: element on the right: 1 element on the left: 2

After query 1, the element positioning is 3
After query 2, the element positioning is 35
After query 3, the element positioning is 235
After query 4, the element positioning is 4235
So there is 1 element to the right and 2 elements on the left of 3 when
query 3 is called.

The following steps can be followed to solve the above problem.

  • Initialize a variable left and right as 0, and two hash arrays position[] and mark[]. position[] is used to store the index of num on left and right and mark[] is used to mark if the num is on left or right.
  • For query-1, increase the count of left, and mark position[num] as left and mark[num] as 1.
  • For query-2, increase the count of right, and mark position[num] as right and mark[num] as 2
  • .

  • For query-3, check if the given number is present on right or left using mark[].
  • If the number is present on right, then the number of elements to the left of num will be (left-position[num]), and the number of elements to the right of num will be (position[num] – 1 + right).
  • If the number is present on left, then the number of elements to the right of num will be (right-position[num]), and the number of elements to the left of num will be (position[num] – 1 + *left).

Below is the implementation of the above approach:

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// C++ program to print the number of elements
// on right and left of a given element
#include <bits/stdc++.h>
using namespace std;
#define MAXN 100005
  
// function to perform query 1
void performQueryOne(int num, int* left, int* right,
                     int* position, int* mark)
{
    // count number of elements on left
    *left = *left + 1;
  
    // store the index number
    position[num] = *(left);
  
    // mark the element is on left
    mark[num] = 1;
}
  
// function to perform query 2
void performQueryTwo(int num, int* left, int* right,
                     int* position, int* mark)
{
    // count number of elements on right
    *right = *right + 1;
  
    // store the index number
    position[num] = *(right);
  
    // mark the element is on right
    mark[num] = 2;
}
  
// function to perform query 3
void performQueryThree(int num, int* left, int* right,
                       int* position, int* mark)
{
    int toright, toleft;
  
    // if the element is on right
    if (mark[num] == 2) {
        toright = *right - position[num];
        toleft = position[num] - 1 + *left;
    }
  
    // if the element is on left
    else if (mark[num] == 1) {
        toleft = *left - position[num];
        toright = position[num] - 1 + *right;
    }
  
    // not present
    else {
        cout << "The number is not there\n";
        return;
    }
  
    cout << "The number of elements to right of "
         << num << ": " << toright;
  
    cout << "\nThe number of elements to left of "
         << num << ": " << toleft;
}
  
// Driver Code
int main()
{
  
    int left = 0, right = 0;
  
    // hashing arrays
    int position[MAXN] = { 0 };
    int mark[MAXN] = { 0 };
  
    int num = 3;
  
    // query type-1
    performQueryOne(num, &left, &right, position, mark);
  
    // query type-2
    num = 5;
    performQueryTwo(num, &left, &right, position, mark);
  
    // query type-2
    num = 2;
    performQueryOne(num, &left, &right, position, mark);
  
    // query type-2
    num = 4;
    performQueryOne(num, &left, &right, position, mark);
  
    // query type-3
    num = 3;
    performQueryThree(num, &left, &right, position, mark);
  
    return 0;
}

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Time Complexity: O(1) for every query.
Auxiliary Space: O(MAXN), where MAXN is 105.



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Striver(underscore)79 at Codechef and codeforces D

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