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LeetCode: Find First and Last Position of Element in Sorted Array

2020-02-03

Challenge Description

Given an array of integers nums sorted in ascending order, 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].

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

Naive Solution

Linear scan array from left to right or from right to left.

Find the first index with given value. A trivial optimization here is when we find current element is greater than target(search left to right), we won’t search the left elements.

Time complexity: $O(N)$

class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
vector<int> res;
int le = -1, ri = -1;

for(int i=0; i < nums.size(); i++) {
if(nums[i] == target) {
le = i;
break;
} else if(nums[i] > target) break;
}

for(int i=nums.size()-1; i>=0; i--) {
if(nums[i] == target) {
ri = i;
break;
} else if(nums[i] < target) break;
}

res.push_back(le);
res.push_back(ri);
return res;
}
};

Naive approach will not meet the challenge requirement. We can use the approach of binary search, but it’s a variant of original binary-search.

Take search first left position for example, When we got an index(suppose idx) where value equal target, we store the current position and safely ignore the right part [idx ~ ..] of the array, so we update the hi pivot.

Search last right position is similar.

file:img/2020_02_03_leetcode-find-first-and-last-position-of-element-in-sorted-array.org_20200203_172745.png

Time complexity: $O(logN)$.

#include <vector>
using namespace std;

class Solution {
public:
int search(vector<int>& A, int target, bool search_left) {
int lo = 0, hi = A.size()-1;
int ans = -1;
while(lo <= hi) {
int mid = (lo + hi) / 2;
if(A[mid] < target)
lo = mid + 1;
else if(A[mid] > target)
hi = mid - 1;
else {
ans = mid;
// find first left
if(search_left)
hi = mid - 1;
else
lo = mid + 1;
}
}
return ans;
}

vector<int> searchRange(vector<int>& nums, int target) {
vector<int> res;
int left = search(nums, target, true);
res.push_back(left);
if(left == -1) {
res.push_back(-1);
return res;
}
int right = search(nums, target, false);
res.push_back(right);
return res;
}

};

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