Equal Beauty CodeChef SnackDown 2021 Round 1A Question The beauty of an (non-empty) array of integers is defined as the difference between its largest and smallest element. For example, the beauty of the array [2,3,4,4,6] is 6−2=4. An array A is said to be good if it is possible to partition the elements of A into two non-empty arrays B1 and B2 such that B1 and B2 have the same beauty. Each element of array A should be in exactly one array: either in B1 or in B2. For example, the array [6,2,4,4,4] is good because its elements can be partitioned into two arrays B1=[6,4,4] and B2=[2,4], where both B1 and B2 have the same beauty (6−4=4−2=2). You are given an array A of length N. In one move you can: Select an index i (1≤i≤N) and either increase Ai by 1 or decrease Ai by 1. Find the minimum number of moves required to make the array A good. Input Format The first line of input contains a single integer T, denoting the number of test cases. The description of T test cases follow. Each test
Climbing Stairs LeetCode Solution
Question
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?
Example 1:
Input: n = 2 Output: 2 Explanation: There are two ways to climb to the top. 1. 1 step + 1 step 2. 2 steps
Example 2:
Input: n = 3 Output: 3 Explanation: There are three ways to climb to the top. 1. 1 step + 1 step + 1 step 2. 1 step + 2 steps 3. 2 steps + 1 step
Constraints:
1 <= n <= 45
Explanation
This problem is quite simple and can be solved in numerous ways, some of them are using Fibonacci series logic, Dynamic Programming, Recursion. Here explanation will be done on Fibonacci series logic and Dynamic Programming Approach. First this is clearly a simple Fibonacci series problem that answer will always be a term of Fibonacci Series. There is nothing more to explain in this logic.
Lets explain by dynamic programming logic. The last element has to be 1 for every n length string. We start building our string from last element. Now if we are at 1 the previous element can be either 0 or 1 both. So we need two recursive calls to do this. If we are at 0 the previous element can only be 1 so we need a recursive call. Also we keep track of the length of the string we are generating using depth. Then we just memorize it and the number of unique recursive calls becomes (2 * depth). The time complexity will be O(n). That's all now just code it. I recommend try to code it yourself if you are stuck then you visit the solution given below in Java, C++, Python.
Program Code
Java
class Solution {
public int climbStairs(int n) {
if(n==1 || n==2){
return n;
}
int previous1 = 1;
int previous2 = 2;
for(int i =2 ;i<n ;i++){
int temp = previous2;
previous2 = previous1+previous2;
previous1 = temp;
}
return previous2;
}
}
C++
class Solution {
public:
int climbStairs(int n) {
int first=0;
int second=1;
int third;
for(int i=0 ;i<n ; i++)
{
third=first+second;
first=second;
second=third;
}
return third;
}
};
C++ using Dynamic Progarmmig
class Solution {
public:
vector<vector<int>>dp;
int dfs( int val , int depth ){
if( depth == 0 ){
return 1;
}else if( dp[val][depth] != -1 ){
return dp[val][depth];
}
if( val == 1 ){
dp[val][depth] = dfs(0,depth-1) + dfs(1,depth-1);
return dp[val][depth];
}else {
dp[val][depth] = dfs(1,depth-1);
return dp[val][depth];
}
}
int climbStairs(int n) {
dp.resize(2,vector<int>(n,-1));
return dfs(1,n-1);
}
};
Python
class Solution:
def climbStairs(self, n: int) -> int:
dp = [-1] * (n + 2)
def ways(x):
if x == n: return 1
if x == n + 1: return 0
if dp[x+1] == -1: dp[x+1] = ways(x+1)
if dp[x+2] == -1: dp[x+2] = ways(x+2)
return dp[x+1] + dp[x+2]
return ways(0)
More information about LeetCode Daily Challenge
More Information About LeetCode Daily Challenge
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