Greedy Algorithm - Question List

Dynamic Programming - Easy Level - Question 1

Dynamic Programming - Easy Level - Question 1 Leetcode 1646 Get Maximum in Generated Array You are given an integer n. An array nums of length n + 1 is generated in the following way: nums[0] = 0 nums[1] = 1 nums[2 * i] = nums[i] when 2 <= 2 * i <= n nums[2 * i + 1] = nums[i] + nums[i + 1] when 2 <= 2 * i + 1 <= n Return the maximum integer in the array nums. Constraints: 0 <= n <= 100 Analysis: This question is quick straightforward: the state and transitional formula are given; the initialization is also given. So we can just ready the code to iterate all the states and find the maximum. See the code below: class Solution { public: int getMaximumGenerated(int n) { int res = 0; if(n<2) return n; vector<int> f(n+1, 0); f[1] = 1; for(int i=2; i<=n; ++i) { if(i&1) f[i] = f[i/2] + f[i/2+1]; else f[i] = f[i/2]; // cout<<i<<" "<<f[i]<<endl; ...

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...

Algo Advance

Search This Blog