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Consider array has n elements and searching element X. What is the minimum and the maximum number of iterations are required to search an element by using binary search?
Note: Searching is successful.
  • a)
    0, n-1
  • b)
    1, log n
  • c)
    0, n
  • d)
    o, log n
Correct answer is option 'B'. Can you explain this answer?
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Consider array has n elements and searching element X. What is the min...
Concept:
Binary Search:

A Binary Search is a sorting method used to find a specific member in a sorted array. Because binary search works only on sorted arrays, an array must be sorted before using binary search on it. As the number of iterations in the binary search reduces, it is a superior searching approach to the liner search technique.
Algorithm:
int binarySearch(int a[], int beg, int end, int K)    
{    
    int mid;    
    if(end >= beg)     
    {   mid = (beg + end)/2;    
        if(a[mid] == K)                 
            return mid+1;    
        else if(a[mid] < K) 
            return binarySearch(a, mid+1, end, K);      
        else   
            return binarySearch(a, beg, mid-1, K);            
    }    
    return -1;     
}   
Minimum number of iterations:
The given data,
A[4]= {10, 16, 22, 25} 
Searching element= K= 16
beg= 0
end = 3
mid = 3/2 = 1
 if K with A[mid] and search is successful. 
Hence minimum number iterations are =1.
Maximum number of iterations:
The given data,
A[4]= {10, 16, 22, 25} 
Searching element= K= 22
beg= 0
end = 3
mid = 3/2 = 1.
Compare 22 with A[1] is not found and K is greater than A[mid]. So  call binarySearch(a, 1+1, 3, 2); 
beg= 2
end = 3
mid = 5/2 = 2.
Compare 22 with A[2] is found.
Explanation:
For in each step the array is being cut into two parts, each of the size (n-1)/2, we have a maximum of log2(n-1) steps. This leads to a total of 2*log2(n-1) comparisons, which asymptotically indeed equals O(log(n)).
Hence the correct answer is 1, log n.
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Consider array has n elements and searching element X. What is the min...
Minimum and Maximum Number of Iterations in Binary Search

Binary search is an efficient algorithm used to search for an element in a sorted array. It follows the divide and conquer approach, dividing the array in half at each iteration to narrow down the search range. The minimum and maximum number of iterations required to search for an element using binary search can be determined by considering the number of elements in the array.

Minimum Number of Iterations
The minimum number of iterations required to search for an element using binary search is 1. This occurs when the element being searched for is found in the middle of the array in the first iteration.

Maximum Number of Iterations
The maximum number of iterations required to search for an element using binary search is log n, where n is the number of elements in the array. This occurs when the element being searched for is either the first or last element in the array.

Explanation
Binary search works by repeatedly dividing the array in half until the element is found or the search range becomes empty. Each iteration reduces the search range by half, which is why binary search is highly efficient.

At each iteration, the middle element of the search range is compared with the element being searched for. If they are equal, the search is successful and the algorithm terminates. Otherwise, depending on whether the middle element is greater or smaller than the target element, the search range is further divided in half and the process continues.

Since the array is sorted, binary search can eliminate half of the remaining elements in each iteration. This logarithmic reduction in the search range leads to a time complexity of O(log n) for binary search.

Therefore, the minimum number of iterations required is 1 when the element is found in the first iteration (in the middle of the array), and the maximum number of iterations required is log n when the element is either the first or last element in the array.

Conclusion
In conclusion, the minimum number of iterations required to search for an element using binary search is 1, and the maximum number of iterations required is log n, where n is the number of elements in the array. Binary search is an efficient algorithm for searching in sorted arrays due to its logarithmic time complexity.
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Consider array has n elements and searching element X. What is the minimum and the maximum number of iterations are required to search an element by using binary search?Note: Searching is successful.a)0, n-1b)1, log nc)0, nd)o, log nCorrect answer is option 'B'. Can you explain this answer?
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