This problem was asked by Google.
Given an array of sorted integers in ascending order, and a target integer, write a function to search target in nums array. If target exists, return the index of the target in the array, otherwise return -1.
Example
function binarySearch(arr, target){
};
const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
const target = 5;
console.log(binarySearch(arr, target)); // 4
Solution
function binarySearch(arr, target) {
let left = 0;
let right = arr.length - 1;
while (left <= right) {
const mid = Math.floor((left + right) / 2);
if (arr[mid] === target) {
return mid;
} else if (arr[mid] < target) {
left = mid + 1;
} else {
right = mid - 1;
}
}
return -1;
}
Explanation
- Initialize left and right pointers representing search range
- Calculate the middle index
- If middle element matches target, return the index
- If middle element is less than target, update left pointer to mid + 1
- If middle element is greater than target, update right pointer to mid - 1
- Repeat steps 2-5 narrowing search range each iteration
- If target not found, return -1
- This implements a classic binary search algorithm to efficiently search a sorted array in O(log n) time complexity. The array is continuously divided in half after each comparison, eliminating half of the elements until the target is found.
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