Consider the following program that attempts to locate an element x in a sorted array a[ ] using binary search. Assume N>1. The program is erroneous. Under what conditions does the program fail? C++ #include <iostream> #include <vector> using namespace std; int find(vector<int>a, int n) { int i = 1, j = n; int x; do { int k = (i + j) / 2; if (a[k] < x) i = k + 1; else j = k; } while (a[k] != x && i < j); if (a[k] == x) cout << "x is in the array" << endl; else cout << "x is not in the array" << endl; return 0; } C #include <stdio.h> #include <stdbool.h> int find(int a[], int n, int x) { int i = 1, j = n; int k; do { k = (i + j) / 2; if (a[k] < x) i = k + 1; else j = k; } while (a[k] != x && i < j); if (a[k] == x) printf("x is in the array "); else printf("x is not in the array "); return 0; } Java import java.util.List; public class Main { public static void find(int arr[], int n, int x) { int i = 0, j = n; int k; do { k = (i + j) / 2; if (arr[k] < x) i = k + 1; else j = k; } while (i < j && arr[k] != x); if (arr[k] == x) System.out.println("x is in the array"); else System.out.println("x is not in the array"); } } Python def find(a, n, x): i = 0 j = n while i < j: k = (i + j) // 2 if a[k] < x: i = k + 1 else: j = k if i < len(a) and a[i] == x: print("x is in the array") else: print("x is not in the array") JavaScript function find(a, n, x) { let i = 0, j = n; let k; do { k = Math.floor((i + j) / 2); if (a[k] < x) i = k + 1; else j = k; } while (a[k] !== x && i < j); if (a[k] === x) console.log("x is in the array"); else console.log("x is not in the array"); }

x is the last element of the array a[]

x is greater than all elements of the array a[]

x is less than the last element of the array a[]

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Top MCQs on Searching Algorithm with Answers

Last Updated : Mar 19, 2024

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Question 1

Linear search is also called------

A
Random Search
B
Sequential search
C
Perfect search
D
None

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Question 2

Given a sorted array of integers, what can be the minimum worst-case time complexity to find ceiling of a number x in given array? The ceiling of an element x is the smallest element present in array which is greater than or equal to x. Ceiling is not present if x is greater than the maximum element present in array. For example, if the given array is {12, 67, 90, 100, 300, 399} and x = 95, then the output should be 100.

A
O(loglogn)
B
O(n)
C
O(log(n))
D
O(log(n) * log(n))

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Question 3

The increasing order of performance of the searching algorithms are:

A
linear search < jump search < binary search
B
linear search > jump search < binary search
C
linear search < jump search > binary search
D
linear search > jump search > binary search

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Question 4

The average case occurs in the Linear Search Algorithm when:

A
The item to be searched is in some where middle of the Array
B
The item to be searched is not in the array
C
The item to be searched is in the last of the array
D
The item to be searched is either in the last or not in the array

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Question 5

Consider the C function given below. Assume that the array listA contains n (> 0) elements, sorted in ascending order.

C++

#include <iostream>
using namespace std;

int ProcessArray(int *listA, int x, int n)
{
    int i, j, k;
    i = 0;
    j = n-1;
    do
    {
        k = (i+j)/2;
        if (x <= listA[k])
            j = k-1;
        if (listA[k] <= x)
            i = k+1;
    }
    while (i <= j);
    if (listA[k] == x)
        return(k);
    else
        return -1;
}

#include <stdio.h>

int ProcessArray(int *listA, int x, int n)
{
    int i, j, k;
    i = 0;
    j = n-1;
    do
    {
        k = (i+j)/2;
        if (x <= listA[k])
            j = k-1;
        if (listA[k] <= x)
            i = k+1;
    }
    while (i <= j);
    if (listA[k] == x)
        return(k);
    else
        return -1;
}

Java

public class Main {
    public static int ProcessArray(int[] listA, int x, int n) {
        int i = 0, j = n - 1, k;
        do {
            k = (i + j) / 2;
            if (x <= listA[k])
                j = k - 1;
            if (listA[k] <= x)
                i = k + 1;
        } while (i <= j);
        if (listA[k] == x)
            return k;
        else
            return -1;
    }
}

Python

def ProcessArray(listA, x, n):
    i = 0
    j = n - 1
    while i <= j:
        k = (i + j) // 2
        if x <= listA[k]:
            j = k - 1
        if listA[k] <= x:
            i = k + 1
    if listA[k] == x:
        return k
    else:
        return -1

JavaScript

function ProcessArray(listA, x, n) {
    let i = 0;
    let j = n - 1;
    let k;
    do {
        k = Math.floor((i + j) / 2);
        if (x <= listA[k])
            j = k - 1;
        if (listA[k] <= x)
            i = k + 1;
    } while (i <= j);
    if (listA[k] === x)
        return k;
    else
        return -1;
}

Which one of the following statements about the function ProcessArray is CORRECT?

A
It will run into an infinite loop when x is not in listA.
B
It is an implementation of binary search.
C
It will always find the maximum element in listA.
D
It will return −1 even when x is present in listA.

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Question 6

The average number of key comparisons done in a successful sequential search in a list of length n is

A
log n
B
(n-1)/2
C
n/2
D
(n+1)/2

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Question 7

Consider the following program that attempts to locate an element x in a sorted array a[ ] using binary search. Assume N>1. The program is erroneous. Under what conditions does the program fail?

C++

#include <iostream>
#include <vector>
using namespace std;

int find(vector<int>a, int n) {
    int i = 1, j = n;
    int x;
    do {
        int k = (i + j) / 2;
        if (a[k] < x) i = k + 1;
        else j = k;
    } while (a[k] != x && i < j);
    
    if (a[k] == x)
        cout << "x is in the array" << endl;
    else
        cout << "x is not in the array" << endl;
    return 0;
}

#include <stdio.h>
#include <stdbool.h>

int find(int a[], int n, int x) {
    int i = 1, j = n;
    int k;
    do {
        k = (i + j) / 2;
        if (a[k] < x) i = k + 1;
        else j = k;
    } while (a[k] != x && i < j);
    
    if (a[k] == x)
        printf("x is in the array\n");
    else
        printf("x is not in the array\n");
    return 0;
}

Java

import java.util.List;

public class Main {
    public static void find(int arr[], int n, int x) {
        int i = 0, j = n;
        int k;
        do {
            k = (i + j) / 2;
            if (arr[k] < x) i = k + 1;
            else j = k;
        } while (i < j && arr[k] != x);

        if (arr[k] == x)
            System.out.println("x is in the array");
        else
            System.out.println("x is not in the array");
    }
}

Python

def find(a, n, x):
    i = 0
    j = n
    while i < j:
        k = (i + j) // 2
        if a[k] < x:
            i = k + 1
        else:
            j = k
    
    if i < len(a) and a[i] == x:
        print("x is in the array")
    else:
        print("x is not in the array")

JavaScript

function find(a, n, x) {
    let i = 0, j = n;
    let k;
    do {
        k = Math.floor((i + j) / 2);
        if (a[k] < x) i = k + 1;
        else j = k;
    } while (a[k] !== x && i < j);
    
    if (a[k] === x)
        console.log("x is in the array");
    else
        console.log("x is not in the array");
}

A
x is the last element of the array a[]
B
x is greater than all elements of the array a[]
C
Both of the Above
D
x is less than the last element of the array a[]

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Question 8

The necessary condition for using binary search in an array is :-

A
The array should not be too long
B
The array should of more size
C
The array should be sorted
D
None of these

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Question 9

what is the name of the below searching Algorithm?

C++

int function(vector<int> arr,int x){
	int n = arr.size();
	
	if(n == 0)
		return -1;

	// Find range for binary search by repeatedly doubling i
	int i = 1;
	while(i < n and arr[i] < x)
		i *= 2;

	// Perform binary search on the range [i/2, min(i, n-1)]
	int left = i /2;
	int right = min(i, n-1);

	while(left <= right){
		int mid = (left + right)/2;
		
		if(arr[mid] == x) return mid;
		else if(arr[mid] < x) left = mid + 1;
		else right = mid - 1;
	}

	return -1;
}

#include <stdio.h>
#include <stddef.h>

int binary_search(int arr[], int n, int x) {
    if(n == 0)
        return -1;

    // Find range for binary search by repeatedly doubling i
    int i = 1;
    while(i < n && arr[i] < x)
        i *= 2;

    // Perform binary search on the range [i/2, min(i, n-1)]
    int left = i / 2;
    int right = (i < n) ? i : n - 1;

    while(left <= right) {
        int mid = (left + right) / 2;
        
        if(arr[mid] == x) return mid;
        else if(arr[mid] < x) left = mid + 1;
        else right = mid - 1;
    }

    return -1;
}

Java

import java.util.*;

public class Main {
    public static int function(List<Integer> arr, int x) {
        int n = arr.size();
        
        if (n == 0)
            return -1;

        // Find range for binary search by repeatedly doubling i
        int i = 1;
        while (i < n && arr.get(i) < x)
            i *= 2;

        // Perform binary search on the range [i/2, min(i, n-1)]
        int left = i / 2;
        int right = Math.min(i, n - 1);

        while (left <= right) {
            int mid = (left + right) / 2;

            if (arr.get(mid) == x) return mid;
            else if (arr.get(mid) < x) left = mid + 1;
            else right = mid - 1;
        }

        return -1;
    }
}

Python

def function(arr, x):
    n = len(arr)
    
    if n == 0:
        return -1

    # Find range for binary search by repeatedly doubling i
    i = 1
    while i < n and arr[i] < x:
        i *= 2

    # Perform binary search on the range [i/2, min(i, n-1)]
    left = i // 2
    right = min(i, n - 1)

    while left <= right:
        mid = (left + right) // 2

        if arr[mid] == x:
            return mid
        elif arr[mid] < x:
            left = mid + 1
        else:
            right = mid - 1

    return -1

JavaScript

function functionSearch(arr, x) {
    let n = arr.length;
    
    if (n === 0)
        return -1;

    // Find range for binary search by repeatedly doubling i
    let i = 1;
    while (i < n && arr[i] < x)
        i *= 2;

    // Perform binary search on the range [i/2, Math.min(i, n-1)]
    let left = Math.floor(i / 2);
    let right = Math.min(i, n - 1);

    while (left <= right) {
        let mid = Math.floor((left + right) / 2);

        if (arr[mid] === x) return mid;
        else if (arr[mid] < x) left = mid + 1;
        else right = mid - 1;
    }

    return -1;
}

A
Linear Search
B
Sequential Search
C
Jump Search
D
Exponential Search

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Question 10

What is the time complexity for performing Jump search?

A
O(LogN)
B
O(N)
C
O(N^2)
D
O(√N)

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