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The following code is SIMILAR to binary search. However, the element selected for mid in this scenario is NOT the middle array/sub-array element. By looking at the following code, explain the search strategy used by this algorithm (note: the statement mid = low in this scenario). Analyze the performance/time complexity of this method. index location (index low, index high) {     index mid;     if (low > high)         return 0;     else     {         mid = low;         if (x == S[mid])             return mid;         else if (x < S[mid])             return location(low, mid - 1);         else             return location(mid + 1, high);     } }

The following code is SIMILAR to binary search. However, the element selected for mid in this scenario is NOT the middle array/sub-array element. By looking at the following code, explain the search strategy used by this algorithm (note: the statement mid = low in this scenario). Analyze the performance/time complexity of this method. index location (index low, index high) {     index mid;     if (low > high)         return 0;     else     {         mid = low;         if (x == S[mid])             return mid;         else if (x < S[mid])             return location(low, mid - 1);         else             return location(mid + 1, high);     } }

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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The following code is SIMILAR to binary search. However, the element selected for mid in this scenario is NOT the middle array/sub-array element. By looking at the following code, explain the search strategy used by this algorithm (note: the statement mid = low in this scenario). Analyze the performance/time complexity of this method.

index location (index low, index high)
{
    index mid;
    if (low > high)
        return 0;

    else
    {
        mid = low;
        if (x == S[mid])
            return mid;

        else if (x < S[mid])
            return location(low, mid - 1);

        else
            return location(mid + 1, high);

    }

}

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