// 7. Efficient Search function efficientSearch(array, item) { let minIndex 0; let maxIndex = array.length let currentIndex; let currentElement; 1; while (minIndex <= maxIndex) { Math.floor((minIndex + maxIndex) / 2); array[currentIndex]; currentIndex currentElement %3D if (currentElement < item) { minIndex = currentIndex + 1; } else if (currentElement > item) { maxIndex = currentIndex 1; } else { return currentIndex; } } return -1;
// 7. Efficient Search function efficientSearch(array, item) { let minIndex 0; let maxIndex = array.length let currentIndex; let currentElement; 1; while (minIndex <= maxIndex) { Math.floor((minIndex + maxIndex) / 2); array[currentIndex]; currentIndex currentElement %3D if (currentElement < item) { minIndex = currentIndex + 1; } else if (currentElement > item) { maxIndex = currentIndex 1; } else { return currentIndex; } } return -1;
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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Question
Analyze the Time Complexity. Show your solution and the corresponding Big-O notation.
![// 7. Efficient Search
function efficientSearch(array, item) {
let minIndex
0;
let maxIndex
array.length
1;
let currentIndex;
let currentElement;
while (minIndex <= maxIndex) {
Math.floor((minIndex + maxIndex) / 2);
array[currentIndex];
currentIndex
currentElement =
if (currentElement < item) {
currentIndex + 1;
minIndex =
}
else if (currentElement > item) {
maxIndex
currentIndex
1;
}
else {
return currentIndex;
}
}
return -1;
}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc4089a18-5c61-47ed-ac6e-617b6c4c912d%2F59a93686-aef5-4f8c-94f4-2e4ed2eaf770%2Fnoho0v8_processed.png&w=3840&q=75)
Transcribed Image Text:// 7. Efficient Search
function efficientSearch(array, item) {
let minIndex
0;
let maxIndex
array.length
1;
let currentIndex;
let currentElement;
while (minIndex <= maxIndex) {
Math.floor((minIndex + maxIndex) / 2);
array[currentIndex];
currentIndex
currentElement =
if (currentElement < item) {
currentIndex + 1;
minIndex =
}
else if (currentElement > item) {
maxIndex
currentIndex
1;
}
else {
return currentIndex;
}
}
return -1;
}
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