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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?
var i,j,k: integer;  x: integer;
    a: array; [1....N] of integer;
begin    i:= 1; j:= N;
repeat    
    k:(i+j) div 2;
    if a[k] < x then i:= k 
    else j:= k 
until (a[k] = x) or (i >= j);
    
if (a[k] = x) then
    writeln ('x is in the array')
else
    writeln ('x is not in the array')
end;
  • 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[]
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Consider the following program that attempts to locate an element x in...
The above program doesn’t work for the cases where element to be searched is the last element of a[] or greater than the last element (or maximum element) in a[]. For such cases, program goes in an infinite loop because i is assigned value as k in all iterations, and i never becomes equal to or greater than j. So while condition never becomes false.
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Most Upvoted Answer
Consider the following program that attempts to locate an element x in...
Is the size of the array and x is the element to be searched.

```
int binary_search(int a[], int N, int x)
{
int low = 0, high = N-1, mid;
while (low <=>
{
mid = (low + high) / 2;
if (a[mid] == x)
return mid;
else if (a[mid] < />
low = mid + 1;
else
high = mid - 1;
}
return -1;
}
```

This program first initializes two variables, `low` and `high`, to the beginning and end of the array, respectively. It then enters a loop that continues as long as the `low` index is less than or equal to the `high` index.

In each iteration, the program computes the middle index `mid` of the current subarray using integer division. If the element at `mid` is equal to `x`, the function immediately returns the index `mid`.

Otherwise, if the element at `mid` is less than `x`, the program updates `low` to `mid + 1`, indicating that the element `x` must be in the upper half of the subarray.

Similarly, if the element at `mid` is greater than `x`, the program updates `high` to `mid - 1`, indicating that the element `x` must be in the lower half of the subarray.

If the loop completes without finding the element `x`, the function returns `-1`.

Overall, this program efficiently searches for an element in a sorted array by repeatedly dividing the search space in half. Its worst-case time complexity is O(log N), which is much faster than linear search, making it ideal for searching large arrays.
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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?var i,j,k: integer; x: integer; a: array; [1....N] of integer;begin i:= 1; j:= N;repeat k:(i+j) div 2; if a[k] < x then i:= k else j:= kuntil (a[k] = x) or (i >= j); if (a[k] = x) then writeln (x is in the array)else writeln (x is not in the array)end;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 Aboved)x is less than the last element of the array a[]Correct answer is option 'C'. Can you explain this answer?
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