LeetCode - Search Insert Position

Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You may assume no duplicates in the array. Here are few examples.

[1,3,5,6], 5 → 2; [1,3,5,6], 2 → 1; [1,3,5,6], 7 → 4; [1,3,5,6], 0 → 0

http://oj.leetcode.com/problems/search-insert-position/

Solution:

O(n) method is obvious. We are talking about O(logn) method, the modified binary search method.

Just add a new if condition like A[mid-1] < target < A[mid]. And if there is no position in the array, we need to insert it at the start.

https://github.com/starcroce/leetcode/blob/master/search_insert_position.cpp

// 44 ms for 62 cases
// Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
// You may assume no duplicates in the array.
class Solution {
public:
    int searchInsert(int A[], int n, int target) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int left = 0, right = n-1;
        while(left <= right) {
            int mid = (left + right) / 2;
            if(A[mid] == target) {
                return mid;
            }
            if(A[mid] > target && A[mid-1] < target) {
                return mid;
            }
            if(A[mid] > target) {
                right = mid - 1;
            }
            else {
                left = mid + 1;
            }
        }
        return left;
    }
};