[Leetcode] Search for a Range / Search Insert Position

Search for a Range Mar 3 '12 5093 / 13502

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        vector<int> res;
        res.push_back(lower(A, 0, n - 1, target));
        res.push_back(upper(A, 0, n - 1, target));
        return res;
    }
    
    int lower(int a[], int s, int e, int target) {
        while (s < e) {
            if (s + 1 == e) {
                break;
            }
            int mid = (s + e) / 2;
            if (a[mid] >= target) {
                e = mid;
            }
            else if (a[mid] < target) {
                s = mid;
            }
        } 
        if (a[s] == target) return s;
        if (a[e] != target) return -1;
        return e;
    }
    
    int upper(int a[], int s, int e, int target) {
        while (s < e) {
            if (s + 1 == e) {
                break;
            }
            int mid = (s + e) / 2;
            if (a[mid] > target) {
                e = mid;
            }
            else if (a[mid] <= target) {
                s = mid;
            }
        } 
        if (a[e] == target) return e;
        if (a[s] != target) return -1;
        return s;
    }
};

Search Insert Position Mar 3 '12 5036 / 10139

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

class Solution {
public:
    int searchInsert(int A[], int n, int target) {
        return lower(A, 0, n - 1, target);
    }
    
    int lower(int a[], int s, int e, int target) {
        while (s < e) {  
            if (s + 1 == e) {  
                break;  
            }  
            int mid = (s + e) / 2;  
            if (a[mid] >= target) {  
                e = mid;  
            }  
            else if (a[mid] < target) {  
                s = mid;  
            }  
        }  
        
        if (a[e] < target) return e + 1;
        if (target <= a[s]) return  s;
        return e;
    }
};