LeetCode 35: Search Insert Position Solution

Problem Explanation

Explanation:

To solve this problem with a runtime complexity of O(log n), we can use a modified binary search algorithm. The idea is to find the target value in the sorted array. If found, return the index. If not found, return the index where it would be inserted while maintaining the sorted order.

Initialize two pointers, low and high, to the start and end of the array respectively.
While low <= high, calculate the middle index mid.
If the target is equal to the value at index mid, return mid.
If the target is less than the value at index mid, update high = mid - 1.
If the target is greater than the value at index mid, update low = mid + 1.
After the loop, return the low pointer as the position to insert the target.

Time Complexity: O(log n) Space Complexity: O(1)

Solution Code

class Solution {
    public int searchInsert(int[] nums, int target) {
        int low = 0, high = nums.length - 1;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (nums[mid] == target) {
                return mid;
            } else if (nums[mid] < target) {
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        return low;
    }
}

Frequently Asked Questions

How to solve LeetCode 35 (Search Insert Position)?

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The time complexity for LeetCode 35 (Search Insert Position) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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