Binary Search

Python

def binary_search(arr, target):
    left = 0;
    right = len(arr) - 1
    while left <= right:
        mid = left + (right - left) // 2;
        if arr[mid] == target:
            return mid
        elif arr[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
    return -1

def bisect_left(arr, target):
    """Returns the leftmost position that `target` should
    go to such that the sequence remains sorted."""
    left = 0
    right = len(arr)
    while left < right:
        mid = (left + right) // 2
        if arr[mid] < target:
            left = mid + 1
        else:
            right = mid
    return left

def bisect_right(arr, target):
    """Returns the rightmost position that `target` should
    go to such that the sequence remains sorted."""
    left = 0
    right = len(arr)
    while left < right:
        mid = (left + right) // 2
        if arr[mid] > target:
            right = mid
        else:
            left = mid + 1
    return left

print(binary_search([1, 2, 3, 10], 1) == 0)
print(binary_search([1, 2, 3, 10], 2) == 1)
print(binary_search([1, 2, 3, 10], 3) == 2)
print(binary_search([1, 2, 3, 10], 10) == 3)
print(binary_search([1, 2, 3, 10], 9) == -1)
print(binary_search([1, 2, 3, 10], 4) == -1)
print(binary_search([1, 2, 3, 10], 0) == -1)
print(binary_search([1, 2, 3, 10], 11) == -1)
print(binary_search([5, 7, 8, 10], 3) == -1)

print(bisect_left([1, 2, 3, 3, 10], 1) == 0)
print(bisect_left([1, 2, 3, 3, 10], 2) == 1)
print(bisect_left([1, 2, 3, 3, 10], 3) == 2) # First "3" is at index 2
print(bisect_left([1, 2, 3, 3, 10], 10) == 4)

# These return a valid index despite target not being in array.
print(bisect_left([1, 2, 3, 3, 10], 9) == 4)
print(bisect_left([1, 2, 3, 3, 10], 0) == 0) # Insert "0" at front
print(bisect_left([1, 2, 3, 3, 10], 11) == 5) # Insert "5" at back

print(bisect_right([1, 2, 3, 3, 10], 1) == 1)
print(bisect_right([1, 2, 3, 3, 10], 2) == 2)
print(bisect_right([1, 2, 3, 3, 10], 3) == 4) # Last "3" is at index 3, so insert new "3" at index 4
print(bisect_right([1, 2, 3, 3, 10], 10) == 5)

# These return a valid index despite target not being in array.
print(bisect_right([1, 2, 3, 3, 10], 9) == 4)
print(bisect_right([1, 2, 3, 3, 10], 0) == 0) # Insert "0" at front
print(bisect_right([1, 2, 3, 3, 10], 11) == 5) # Insert "5" at back

JavaScript

function binarySearch(arr, target) {
    let left = 0;
    let right = arr.length - 1;
    while (left <= right) {
        const mid = left + Math.floor((right - left) / 2);
        if (arr[mid] === target) {
            return mid;
        }
        if (arr[mid] < target) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    return -1;
}

console.log(binarySearch([1, 2, 3, 10], 1) === 0);
console.log(binarySearch([1, 2, 3, 10], 2) === 1);
console.log(binarySearch([1, 2, 3, 10], 3) === 2);
console.log(binarySearch([1, 2, 3, 10], 10) === 3);
console.log(binarySearch([1, 2, 3, 10], 9) === -1);
console.log(binarySearch([1, 2, 3, 10], 4) === -1);
console.log(binarySearch([1, 2, 3, 10], 0) === -1);
console.log(binarySearch([1, 2, 3, 10], 11) === -1);
console.log(binarySearch([5, 7, 8, 10], 3) === -1);

def binary_search(arr, target):
    left = 0;
    right = len(arr) - 1
    while left <= right:
        mid = left + (right - left) // 2;
        if arr[mid] == target:
            return mid
        elif arr[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
    return -1

def bisect_left(arr, target):
    """Returns the leftmost position that `target` should
    go to such that the sequence remains sorted."""
    left = 0
    right = len(arr)
    while left < right:
        mid = (left + right) // 2
        if arr[mid] < target:
            left = mid + 1
        else:
            right = mid
    return left

def bisect_right(arr, target):
    """Returns the rightmost position that `target` should
    go to such that the sequence remains sorted."""
    left = 0
    right = len(arr)
    while left < right:
        mid = (left + right) // 2
        if arr[mid] > target:
            right = mid
        else:
            left = mid + 1
    return left

print(binary_search([1, 2, 3, 10], 1) == 0)
print(binary_search([1, 2, 3, 10], 2) == 1)
print(binary_search([1, 2, 3, 10], 3) == 2)
print(binary_search([1, 2, 3, 10], 10) == 3)
print(binary_search([1, 2, 3, 10], 9) == -1)
print(binary_search([1, 2, 3, 10], 4) == -1)
print(binary_search([1, 2, 3, 10], 0) == -1)
print(binary_search([1, 2, 3, 10], 11) == -1)
print(binary_search([5, 7, 8, 10], 3) == -1)

print(bisect_left([1, 2, 3, 3, 10], 1) == 0)
print(bisect_left([1, 2, 3, 3, 10], 2) == 1)
print(bisect_left([1, 2, 3, 3, 10], 3) == 2) # First "3" is at index 2
print(bisect_left([1, 2, 3, 3, 10], 10) == 4)

# These return a valid index despite target not being in array.
print(bisect_left([1, 2, 3, 3, 10], 9) == 4)
print(bisect_left([1, 2, 3, 3, 10], 0) == 0) # Insert "0" at front
print(bisect_left([1, 2, 3, 3, 10], 11) == 5) # Insert "5" at back

print(bisect_right([1, 2, 3, 3, 10], 1) == 1)
print(bisect_right([1, 2, 3, 3, 10], 2) == 2)
print(bisect_right([1, 2, 3, 3, 10], 3) == 4) # Last "3" is at index 3, so insert new "3" at index 4
print(bisect_right([1, 2, 3, 3, 10], 10) == 5)

# These return a valid index despite target not being in array.
print(bisect_right([1, 2, 3, 3, 10], 9) == 4)
print(bisect_right([1, 2, 3, 3, 10], 0) == 0) # Insert "0" at front
print(bisect_right([1, 2, 3, 3, 10], 11) == 5) # Insert "5" at back