70. Climbing Stairs

You are climbing a stair case. It takes n steps to reach to the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

Note: Given n will be a positive integer.

Approach : The standard recursive solution times out so the idea is to use the iterative solution to find the nth fibonacci number.

int climbStairs(int n) {
    int f1 = 2;
    int f2 = 1;
    if(n == 1) {
        return f2;
    } else if(n == 2) {
        return f1;
    }

    int fn;
    for(int i = 3; i <= n; i++) {
        fn = f1 + f2;
        f2 = f1;
        f1 = fn;
    }
    return fn;
}

Here is the link to the ideone solution : http://ideone.com/ErEU6E

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69. Sqrt(x)

Implement int sqrt(int x).

Return the floor if x is not a perfect square.

Approach :  Using binary search making sure that overflow does not happen by taking long long.

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class Solution { public: int mySqrt(int x) { if(x <= 1){ return x; } int low = 0, high = x / 2; //long to prevent overflow long long mid; while(low <= high){ mid = (low + high) / 2; //needs to be long to prevent overflow long long t = mid * mid; //second part is to return floor if x is not perfect square if( t == x || ((t < x) && ((mid + 1) * (mid + 1) > x))){ break; } else if(x < t){ high = mid - 1; }else{ low = mid + 1; } } return mid; } };

Here is the link to the ideone solution : http://ideone.com/taTCKu

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38. Count and Say

Problem: The count-and-say sequence is the sequence of integers with the first five terms as following:

1.     1

2.     11
3.     21

4.     1211

5.     111221

1 is read off as "one 1" or 11.

11 is read off as "two 1s" or 21.

21 is read off as "one 2, then one 1" or 1211.

Given an integer n, generate the nth term of the count-and-say sequence.

Note: Each term of the sequence of integers will be represented as a string.

Example 1:

Input: 1

Output: "1"

Example 2:

Input: 4

Output: "1211" 

Approach: The solution is straight forward. The appendNext method is run over n times where n is the input.

class Solution {
public:
    public:
 
    string appendNext(string str){
        string str1;
        char ch = str[0];
        int chCount = 1;
        for(int i = 1; i <= str.size(); i++){
            if (str[i] == ch){
                chCount++;
            }
            else {
                char chr = chCount + '0';
                str1 = str1 + chr;
                str1 = str1 + ch;
                ch = str[i];
                chCount = 1;
            }
        }
        return str1;
    }
    
    string countAndSay(int n) {
        if (n == 1) {
            return "1";
        }
        string str1 = "1";
        string strn;
        for (int i = 1; i < n; i++){
            strn = appendNext(str1);
            str1 = strn;
        }
        return strn;
    }
};

Here is the link to the ideone solution : http://ideone.com/tbYXUM

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67. Add Binary

Given two binary strings, return their sum (also a binary string).

For example,
a = "11"
b = "1"
Return "100"

Approach : The idea is to start from the right and keep adding digits and forwarding carry (if any). Also, take care of the case when one of them gets exhausted. For that, keep adding zero to the other one and forward carry if required. 

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string Solution::addBinary(string a, string b) { int lenA = a.length(), lenB = b.length(); int i = lenA - 1, j = lenB - 1; int carry = 0; stack<char> out; int sum = 0; while(i >= 0 || j >= 0){ if(i >= 0 && j >= 0){ //convert to int and add sum = a[i] - '0' + b[j] - '0' + carry; i--; j--; } else{ //string a is exhausted if(i < 0){ sum = b[j] - '0' + carry; j--; } //string b is exhausted else{ sum = a[i] - '0' + carry; i--; } } //setting the carry accordingly if(sum > 1){ carry = 1; }else{ carry = 0; } sum = sum % 2; //convert back to character char aChar = '0' + sum; out.push(aChar); } //for final carry if(carry){ out.push('1'); } string res = ""; //reverse the output while(!out.empty()){ res = res + out.top(); out.pop(); } return res; }

Here is the link to the ideone solution : http://ideone.com/o5r3Eq

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Data Science & Machine Learning - 6.1 Matplotlib & Its Installation

Hi friends,

Welcome to this post on Data Visualization under Data Science & Machine Learning. In the previous post, we performed some practical data analysis using Pandas on the Kaggle SF Salaries Dataset. In this post, we'll learn about Matplotlib, one of the most popular Python libraries for visualizing our results using various plots. 

About Matplotlib [1]

1. Matplotlib is a Python 2D plotting library which produces high quality figures in a variety of hardcopy formats and interactive environments across platforms
2. It allows us to generate plots directly from NumPy Arrays and Pandas DataFrames which makes it even more popular
3. It can be used in Python scripts, the Python and IPython shell, the jupyter notebook, web application servers, etc.
4. It is widely said that Matplotlib tries to make easy things easy and hard things possible

Here are some of the plots supported by the Matplotlib library:

Image source: Matlplotlib

Matplotlib Installation

Just like the installation of other Python libraries such as NumPy/Pandas, it is recommended to use the Anaconda distribution of Python in order to install Matlplotlib as well. You can see the installation of Anaconda distribution of Python here. Once you have that installed, you can install Matplotlib by running the following command in the command prompt:

conda install matplotlib

You can still install Matplotlib even if you don't have the recommended Anaconda distribution of Python (not recommended) using the following command:

pip install matplotlib

You can see the list of various plots supported by Matplotlib from this link. Now that we have installed Matplotlib successfully on our systems, we will start using the Matplotlib library starting with basic plots from the next post.

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