Given Q queries of three types where every query consists of a number.
- Add element num on the left
- Add element num on the right
- 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:
CPP
// 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 1void 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 2void 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 3void 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 Codeint 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;} |
Java
// Java code implementation for the above approachimport java.io.*;import java.util.*;class GFG { static final int MAXN = 100005; // function to perform query 1 static void performQueryOne(int num, int[] left, int[] right, int[] position, int[] mark) { // count number of elements on left left[0] = left[0] + 1; // store the index number position[num] = left[0]; // mark the element is on left mark[num] = 1; } // function to perform query 2 static void performQueryTwo(int num, int[] left, int[] right, int[] position, int[] mark) { // count number of elements on right right[0] = right[0] + 1; // store the index number position[num] = right[0]; // mark the element is on right mark[num] = 2; } // function to perform query 3 static 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[0] - position[num]; toleft = position[num] - 1 + left[0]; } // if the element is on left else if (mark[num] == 1) { toleft = left[0] - position[num]; toright = position[num] - 1 + right[0]; } // not present else { System.out.println("The number is not there"); return; } System.out.println( "The number of elements to right of " + num + ": " + toright); System.out.println( "The number of elements to left of " + num + ": " + toleft); } public static void main(String[] args) { int[] left = { 0 }, right = { 0 }; // hashing arrays int[] position = new int[MAXN]; int[] mark = new int[MAXN]; Arrays.fill(position, 0); Arrays.fill(mark, 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); }}// This code is contributed by lokesh. |
Python3
# python program to print the number of elements# on right and left of a given elementMAXN = 100005# function to perform query 1def performQueryOne(num, position, mark): global left # 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 2def performQueryTwo(num, position, mark): global right # 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 3def performQueryThree(num, position, mark): toright = 0 toleft = 0 global left global right # 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 elif mark[num] == 1: toleft = left - position[num] toright = position[num] - 1 + right # not present else: print("The number is not there \n") return print("The number of elements to right of ", num, " : " , toright) print("\nThe number of elements to left of ", num, " : ", toleft) # Driver Codeleft = 0right = 0# hashing arraysposition = [0]*MAXNmark = [0]*MAXNnum = 3# query type-1performQueryOne(num, position, mark)# query type-2num = 5performQueryTwo(num, position, mark)# query type-2num = 2performQueryOne(num, position, mark)# query type-2num = 4performQueryOne(num, position, mark)# query type-3num = 3performQueryThree(num, position, mark)# The code is contributed by Gautam goel. |
C#
using System;class GFG { // function to perform query 1 static void performQueryOne(int num, int[] left, int[] right, int[] position, int[] mark) { // count number of elements on left left[0] = left[0] + 1; // store the index number position[num] = left[0]; // mark the element is on left mark[num] = 1; } // function to perform query 2 static void performQueryTwo(int num, int[] left, int[] right, int[] position, int[] mark) { // count number of elements on right right[0] = right[0] + 1; // store the index number position[num] = right[0]; // mark the element is on right mark[num] = 2; } // function to perform query 3 static 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[0] - position[num]; toleft = position[num] - 1 + left[0]; } // if the element is on left else if (mark[num] == 1) { toleft = left[0] - position[num]; toright = position[num] - 1 + right[0]; } // not present else { Console.WriteLine("The number is not there\n"); return; } Console.WriteLine( "The number of elements to right of " + num + ": " + toright); Console.WriteLine("\nThe number of elements to left of " + num + ": " + toleft); } static void Main() { int MAXN = 100005; int[] left = {0}; int[] right = {0}; // hashing arrays int[] position = new int[MAXN]; int[] mark = new int[MAXN]; for(int i = 0; i < MAXN; i++){ position[i] = 0; mark[i] = 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); }}// the code is contributed by Nidhi goel. |
Javascript
<script> // JavaScript program to print the number of elements // on right and left of a given element const MAXN = 100005; let left = 0, right = 0; // hashing arrays let position = new Array(MAXN).fill(0); let mark = new Array(MAXN).fill(0); // function to perform query 1 const performQueryOne = (num) => { // 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 const performQueryTwo = (num) => { // 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 const performQueryThree = (num) => { let 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 { document.write("The number is not there<br/>"); return; } document.write(`The number of elements to right of ${num}: ${toright}`); document.write(`<br/>The number of elements to left of ${num}: ${toleft}`); } // Driver Code let num = 3; // query type-1 performQueryOne(num); // query type-2 num = 5; performQueryTwo(num); // query type-2 num = 2; performQueryOne(num); // query type-2 num = 4; performQueryOne(num); // query type-3 num = 3; performQueryThree(num);// This code is contributed by rakeshsahni</script> |
The number of elements to right of 3: 1 The number of elements to left of 3: 2
Time Complexity: O(1) for every query.
Auxiliary Space: O(MAXN), where MAXN is 105.
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