Day 1 | Array
509. Fibonacci Number
The Fibonacci numbers, commonly denoted F(n) form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones, starting from 0 and 1. That is,
F(0) = 0, F(1) = 1
F(n) = F(n - 1) + F(n - 2), for n > 1.
Given n, calculate F(n).
Recursion
class Solution {
public:
int fib(int n) {
if(!n) return 0;
if(n == 1 or n == 2) return 1;
vector<int> dp(n);
dp[0] = dp[1] = 1;
for(int i=2; i<n; i++)
dp[i] = dp[i-1] + dp[i-2];
return dp[n-1];
}
};
DP
int fib(int n) {
if(n <= 1)
return n;
return fib(n - 1) + fib(n - 2);
}
Binet’s Formula
Binet’s formula for the n’th Fibonacci number:
This formula can compute the solution in O(1) time and O(1) space.
This implementation breaks down after n > 71 because Math.pow uses floating point numbers to be able to do exponents in O(1) time.
class Solution {
public:
int fib(int n) {
double sqrt5 = sqrt(5);
return (pow(1 + sqrt5, n) - pow(1 - sqrt5, n)) / pow(2, n) / sqrt5;
}
};
Golden Ration
This formula can compute the solution in O(logn) time and O(1) space.
This implementation breaks down after n > 71 because Math.pow uses floating point numbers to be able to do exponents in O(logn) time.
int fib(int n) {
double goldenRatio = (1 + sqrt(5)) / 2;
return round(pow(goldenRatio, n) / sqrt(5));
}
1491. Average Salary Excluding the Minimum and Maximum Salary
The Tribonacci sequence Tn is defined as follows:
T0 = 0, T1 = 1, T2 = 1, and Tn+3 = Tn + Tn+1 + Tn+2 for n >= 0.
Given n, return the value of Tn.
Recursion
class Solution {
public:
int tribonacci(int n) {
if(n<2)return n;
if(n==2)return 1;
return tribonacci(n-1)+tribonacci(n-2)+tribonacci(n-3);
}
};
DP
class Solution {
public:
int tribonacci(int n) {
int dp[3] = {0, 1, 1}4;
for (int i = 3; i <= n; ++i)
dp[i%3] += dp[(i+1)%3] + dp[(i+2)%3];
return dp[n%3];
}
};
Day 2
70. Climbing Stairs
You are climbing a staircase. It takes n steps to reach the top.
Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
Recursion
class Solution {
public:
int climbStairs(int n) {
return n <= 2 ? n : climbStairs(n-1) + climbStairs(n-2);
}
};
DP
class Solution {
public:
int climbStairs(int n) {
if(n <= 2) return n;
int * dp = new int[n];
dp[0] = 1, dp[1] = 2;
for(int i = 2; i < n; i++)
dp[i] = dp[i-1] + dp[i-2];
return dp[n-1];
}
};
746. Min Cost Climbing Stairs
You are given an integer array cost where cost[i] is the cost of ith step on a staircase. Once you pay the cost, you can either climb one or two steps.
You can either start from the step with index 0, or the step with index 1.
Return the minimum cost to reach the top of the floor.
class Solution {
public:
int minCostClimbingStairs(vector<int>& cost) {
int n=cost.size();
int tmp, first=cost[0],second=cost[1];
for(int i=2;i<n;i++){
tmp=min(first,second)+cost[i];
first=second;
second=tmp;
}
return min(first,second);
}
};