Number of comparisons in each direction for m queries in linear search

Last Updated : 09 Sep, 2022

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:

Input: arr[] = {1, 2}, q[] = {1, 2}
Output: 3, 3
For 1-based indexing
For 1st query : Number of comparisons from left to right is 1 and from right to left is 2
For 2nd query : Number of comparisons from left to right is 2 and from right to left is 1

Input: arr[] = {-1, 2, 4, 5, 1}, q[] = {-1, 4, 2}
Output: 3, 7

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:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the count of comparisons from left to right
// and right to left in linear search among m queries
pair<int, int> countCamparisions(int n, int arr[], int m, int qry[])
{
    int i;
    unordered_map<int, int> index;
    for (i = 1; i <= n; i++) {

        // arr[i] occurs at i
        index[arr[i]] = i;
    }

    // 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);
}

// Driver Code
int 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;

    int q[] = { -1, 4, 2 };
    int m = (sizeof(q) / sizeof(q[0])) - 1;

    pair<int, int> res = countCamparisions(n, arr, m, q);
    cout << res.first << " " << res.second;
}

Java

// Java implementation of the approach
import java.util.HashMap;
import java.util.Map;

class GFG
{

// 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, 
                            int arr[], int m, int qry[]) 
{ 
    int i; 
    HashMap<Integer,Integer> index = new HashMap<>(); 
    for (i = 1; i <= n; i++) 
    { 

        // arr[i] occurs at i 
        index.put(arr[i], i); 
    } 

    // 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; 
}

    // 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);
    }
}

class Pair<A, B> 
{
    A first;
    B second;

    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 approach

# 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) :

    index = {}
    for i in range(1, n + 1) :

        # 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) 

# Driver Code
if __name__ == "__main__" :

    # -1 will be ignored as it 
    # is 1-based indexing 
    arr = [ -1, 2, 4, 5, 1 ] 
    n = len(arr) - 1

    q = [ -1, 4, 2 ] 
    m = len(q) - 1

    res = countCamparisions(n, arr, m, q) 
    print(res[0], res[1]) 
    
# This code is contributed by Ryuga

// C# implementation of the approach
using System;
using System.Collections.Generic;

class GFG
{

// 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++) 
    { 

        // arr[i] occurs at i 
        index.Add(arr[i], i); 
    } 

    // 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; 
}

// 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<int, int> res = countCamparisions(n, arr, m, q); 
    Console.WriteLine(res.first + " " + res.second);
}
}

class Pair<A, B> 
{
    public A first;
    public B second;
    
    public Pair(A first, B second)
    {
        this.first = first;
        this.second = second;
    }
}

// This code is contributed by 29AjayKumar

JavaScript

<script>

// Javascript implementation of the approach

// Function to return the count of comparisons from left to right
// and right to left in linear search among m queries
function countCamparisions(n, arr, m, qry)
{
    var i;
    var index = new Map();
    for (i = 1; i <= n; i++) {

        // arr[i] occurs at i
        index.set(arr[i], i);
    }

    // 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];
}

// Driver Code
// -1 will be ignored as it is 1-based indexing
var 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]);

// This code is contributed by famously.
</script>

Output

3 7

Complexity Analysis:

Time Complexity: O(N + M)
Auxiliary Space: O(N)

Search an element in an unsorted array using minimum number of comparisons

Abdullah Aslam

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