Binary Search - Question List

Example

Question 1: Search Insert Position

Easy Level

Question 1: First Bad Version

Medium Level

Question 1: Search in Rotated Sorted Array

Hard Level

Question 1: Split Array Largest Sum

Question 2: Minimum Window Subsequence

Question 3: Nth Magical Number

Interview Questions

Question 1: Minimum steps to find a baseball

Comments

Greedy Algorithm - Question List

Question List: Question 1: Non-overlapping Intervals Question 2: Longest Increasing Subsequence More Questions Upper Layer

Binary Search - Example

Binary Search - Example Leetcode 35 Search Insert Position Given a sorted array of distinct integers 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 must write an algorithm with O(log n) runtime complexity. Constraints: 1 <= nums.length <= 10^4 -10^4 <= nums[i] <= 10^4 nums contains distinct values sorted in ascending order. -10^4 <= target <= 10^4 Analysis: The array is sorted, so we just need to run a routine binary search. Make a guest first, then based on the guessed result, we can adjust the search range. See the code below: class Solution { public: int searchInsert(vector<int>& nums, int target) { int left = 0, right = nums.size(); while(left < right) { int mid = left + (right - left) / 2; if(nums[mid] < target) left = mid + 1; else right = mid; } return left; } }; If you know t...

Bit Manipulation - Medium Level

Leetcode 416 Partition Equal Subset Sum Given a non-empty array nums containing only positive integers, find if the array can be partitioned into two subsets such that the sum of elements in both subsets is equal. Constraints: 1 <= nums.length <= 200 1 <= nums[i] <= 100 Analysis: There are different ways to solve this problem, such as dp with a time complexity of O(N^2). Since N is small to this question, so it is Okay to pass OJ. Besides the small N, the value of each element is also very small. So this gives us some chance to use "space to trad off time". The data structure to be used is bitset. For bitset, each bit can be either 0 or 1. The index of that bit can be used as the corresponding sum. When the bit is 1, means there is a sum with the value of its index. When a new number comes, this number needs to be added to all the previous sums, to form new "previous" sums. Thus for each number, we need to go through all the previous sums, the time co...

Algo Advance

Search This Blog