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Binary Search

  • List must be sorted
  • Divide and conquer
  • Eliminates half the elements every time
  • Maximum steps --> log2nlog_2 n
  • As input doubles, the maximum number of steps increases by one

Complexity

Time Space
O(log(n))O(log(n))
Split the search space by two on every iteration

Process

  • Start by searching in the middle
  • If what you are looking for is smaller than what you find in the middle, restrict search up to the middle
  • If what you are looking for is bigger than what you find in the middle, restrict search space from middle on
  • Repeat until you find what you're looking for or your search space has nothing in it

Implementations

JavaScript

function binarySearch(target, sortedArray) {
  // Start the search space with the entire array
  let lowIndex = 0;
  let highIndex = sortedArray.length - 1;

  // While the search space has elements in it...
  while (lowIndex <= highIndex) {
    // Grab the middle, acounting for a min (lowIndex)
    const middleIndex = lowIndex + Math.floor((highIndex - lowIndex) / 2);
    const guess = sortedArray[middleIndex];

    // Target found!
    if (guess === target) {
      return middleIndex;
    }

    if (guess < target) {
      // Middle is less than target
      // Adjust search space to second half of current search space
      lowIndex = middleIndex + 1;
    } else {
      // Middle is greater than target
      // Adjust search space to first half of current space
      highIndex = middleIndex - 1;
    }
  }

  return -1;
}

Python

def binary_search(target, sortedArray):
  lowIndex = 0
  highIndex = len(list) - 1

  while lowIndex <= highIndex:
    middleIndex = (lowIndex + highIndex) / 2
    guess = sortedArray[middleIndex]

    if guess == target:
      return middleIndex

    if guess < target:
      lowIndex = middleIndex + 1
    else:
      highIndex = middleIndex - 1

  return None

Computer Science Algorithm

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