Given an array containing N distinct elements. There are M queries, each containing an integer X and asking for the index of X in the array. For each query, the task is to perform linear search X from left to right and count the number of comparisons it took to find X and do the same thing right to left. In the end, print the total number of comparisons in both directions among all the queries.
Examples:Â
Input: arr[] = {1, 2}, q[] = {1, 2}Â
Output: 3, 3Â
For 1-based indexingÂ
For 1st query : Number of comparisons from left to right is 1 and from right to left is 2Â
For 2nd query : Number of comparisons from left to right is 2 and from right to left is 1Input: arr[] = {-1, 2, 4, 5, 1}, q[] = {-1, 4, 2}Â
Output: 3, 7Â
Approach: Find the index at which X is present in the array say i (1-based indexing), the number of comparisons for left to right would be i and the number of comparisons for right to left would be (n – i + 1). All we need to do is to find the index quickly. It can be done by using a map in which key is the element’s value and value is the index.
Below is the implementation of the above approach:Â
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;Â
// Function to return the count of comparisons from left to right// and right to left in linear search among m queriespair<int, int> countCamparisions(int n, int arr[], int m, int qry[]){Â Â Â Â int i;Â Â Â Â unordered_map<int, int> index;Â Â Â Â for (i = 1; i <= n; i++) {Â
        // arr[i] occurs at i        index[arr[i]] = i;    }Â
    // Count of comparisons for left to right and right to left    int ltr = 0, rtl = 0;    for (i = 1; i <= m; i++) {        int x = qry[i];        ltr += index[x];        rtl += n - index[x] + 1;    }    return make_pair(ltr, rtl);}Â
// Driver Codeint main(){    // -1 will be ignored as it is 1-based indexing    int arr[] = { -1, 2, 4, 5, 1 };    int n = (sizeof(arr) / sizeof(arr[0])) - 1;Â
    int q[] = { -1, 4, 2 };    int m = (sizeof(q) / sizeof(q[0])) - 1;Â
    pair<int, int> res = countCamparisions(n, arr, m, q);    cout << res.first << " " << res.second;} |
Java
// Java implementation of the approachimport java.util.HashMap;import java.util.Map;Â
class GFG{Â
// Function to return the count of // comparisons from left to right // and right to left in linear // search among m queries static Pair<Integer, Integer> countCamparisions(int n, Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â int arr[], int m, int qry[]) { Â Â Â Â int i; Â Â Â Â HashMap<Integer,Integer> index = new HashMap<>(); Â Â Â Â for (i = 1; i <= n; i++) Â Â Â Â { Â
        // arr[i] occurs at i         index.put(arr[i], i);     } Â
    // Count of comparisons for left    // to right and right to left     int ltr = 0, rtl = 0;     for (i = 1; i <= m; i++)    {         int x = qry[i];         ltr += index.get(x);         rtl += n - index.get(x) + 1;     }          Pair<Integer, Integer> ans = new Pair<>(ltr, rtl);    return ans; }Â
    // Driver Code     public static void main(String []args)    {                 // -1 will be ignored as it is 1-based indexing         int arr[] = { -1, 2, 4, 5, 1 };         int n = arr.length - 1;              int q[] = { -1, 4, 2 };         int m = q.length - 1;              Pair<Integer, Integer> res = countCamparisions(n, arr, m, q);         System.out.println(res.first + " " + res.second);    }}Â
class Pair<A, B> {Â Â Â Â A first;Â Â Â Â B second;Â
    public Pair(A first, B second)    {        this.first = first;        this.second = second;    }}     // This code is contributed by Rituraj Jain |
Python3
# Python 3 implementation of the# above approachÂ
# Function to return the count of # comparisons from left to right # and right to left in linear search# among m queries def countCamparisions(n, arr, m, qry) :Â
    index = {}    for i in range(1, n + 1) :Â
        # arr[i] occurs at i         index[arr[i]] = i         # Count of comparisons for left to     # right and right to left     ltr, rtl = 0, 0    for i in range(1, m + 1) :        x = qry[i]        ltr += index[x]         rtl += n - index[x] + 1         return (ltr, rtl) Â
# Driver Codeif __name__ == "__main__" :Â
    # -1 will be ignored as it     # is 1-based indexing     arr = [ -1, 2, 4, 5, 1 ]     n = len(arr) - 1Â
    q = [ -1, 4, 2 ]     m = len(q) - 1Â
    res = countCamparisions(n, arr, m, q)     print(res[0], res[1])      # This code is contributed by Ryuga |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;Â
class GFG{Â
// Function to return the count of // comparisons from left to right // and right to left in linear // search among m queries static Pair<int,             int> countCamparisions(int n, int []arr,                                    int m, int []qry) {     int i;     Dictionary<int,               int> index = new Dictionary<int,                                           int>();     for (i = 1; i <= n; i++)     { Â
        // arr[i] occurs at i         index.Add(arr[i], i);     } Â
    // Count of comparisons for left    // to right and right to left     int ltr = 0, rtl = 0;     for (i = 1; i <= m; i++)    {         int x = qry[i];         ltr += index[x];         rtl += n - index[x] + 1;     }          Pair<int,          int> ans = new Pair<int,                              int>(ltr, rtl);    return ans; }Â
// Driver Code public static void Main(String []args){         // -1 will be ignored as     // it is 1-based indexing     int []arr = { -1, 2, 4, 5, 1 };     int n = arr.Length - 1; Â
    int []q = { -1, 4, 2 };     int m = q.Length - 1; Â
    Pair<int, int> res = countCamparisions(n, arr, m, q);     Console.WriteLine(res.first + " " + res.second);}}Â
class Pair<A, B> {Â Â Â Â public A first;Â Â Â Â public B second;Â Â Â Â Â Â Â Â Â public Pair(A first, B second)Â Â Â Â {Â Â Â Â Â Â Â Â this.first = first;Â Â Â Â Â Â Â Â this.second = second;Â Â Â Â }}Â
// This code is contributed by 29AjayKumar |
Javascript
<script>Â
// Javascript implementation of the approachÂ
// Function to return the count of comparisons from left to right// and right to left in linear search among m queriesfunction countCamparisions(n, arr, m, qry){Â Â Â Â var i;Â Â Â Â var index = new Map();Â Â Â Â for (i = 1; i <= n; i++) {Â
        // arr[i] occurs at i        index.set(arr[i], i);    }Â
    // Count of comparisons for left to right and right to left    var ltr = 0, rtl = 0;    for (i = 1; i <= m; i++) {        var x = qry[i];        ltr += index.get(x);        rtl += n - index.get(x) + 1;    }    return [ltr, rtl];}Â
// Driver Code// -1 will be ignored as it is 1-based indexingvar arr = [-1, 2, 4, 5, 1];var n = arr.length - 1;var q = [-1, 4, 2];var m = q.length - 1;var res = countCamparisions(n, arr, m, q);document.write( res[0] + " " + res[1]);Â
// This code is contributed by famously.</script> |
Output
3 7
Complexity Analysis:
- Time Complexity: O(N + M)
- Auxiliary Space: O(N)Â
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