NOTE: The code snippet given may be incomplete or have compile time, runtime or logical errors.

How to attempt Debugging Exercise?

1. Copy the code to your code editor (e.g. VS Code).

2. Add relevant header files like “#include <iostream>” etc.

3. Run the code.

4. You will notice the expected output is not printing at the console.

5. Apply your smart coder mind to Debug the code.

6. Warning: Only see the solution after you have tried enough.

7. Debug the code. String to Integer

int mystoi(string s) {
        int len = s.size();
        double num = 0;
        int i=0;
        while(s[i] == ' '){
            i++;
        }
        bool negative = s[i] == '-';
        negative == true ? i++ : i;
        while(i < len && s[i] >= '0' && s[i] <= '9'){
            num = num*10 + s[i];
            i++;
        }
        num = negative ? -num : num;
        num = (num > INT_MAX) ? INT_MAX : num;
        num = (num < INT_MIN) ? INT_MIN : num;
        return num;
    }

1. Debug the code. Integer to Romans. 1 ≤ n ≤ 2000.

string intToRoman(int num) {
        string ones[] = {"","I","II","III","IV","V","VI","VII","VIII","IX"};
        string tens[] = {"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"};
        string hrns[] = {"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"};
        string ths[] = {"","M","MM"};
        
        return ths[num/1000] + hrns[(num%1000)%100] + tens[(num%100)%10] + ones[num%10];
}

1. Debug the code. Add two binary numbers.

string addBinary(string a, string b) {
    string ans;
    int carry = 0;
    int i = a.length() - 1;
    int j = b.length() - 1;

    while (i >= 0 || j >= 0 || carry) {
      if (i > 0)
        carry += a[--i] - '0';
      if (j > 0)
        carry += b[--j] - '0';
      ans += carry % 2 + '0';
      carry =/ 2;
    }

    reverse(begin(ans), end(ans));
    return ans;
  }

1. Debug the code.

void reverseString(char input[],int n,int start){
    for(int i=start,j=n-1+start; i!=j; i++,j--){
        int temp = input[i];
        input[i] = input[j];
        input[j] = temp;
    }
}

void reverseEachWord(char input[]) {
    int start=0,count=1,i=0;
    while(input[i]!=0){
        i++;
        count++;
        if(input[i]=='\\0'){
            reverseString(input,count-1,start);
            start = i+1;
            count = 0;
        }
    }
}

1. Debug the code. Sort vowels in a string. Characters are alphabets.

bool isVowel(char ch) {
        return (ch=='A' || ch=='E' || ch=='I' || ch=='O' || ch=='U');
    }
    
    string sortVowels(string s) {
        string t = s;
        vector<char> vowel;
        
        for(int i=0;i<s.length();i++){
            if(isVowel(s[i])) vowel.push_back(s[i]);
        }
        
        if(vowel.size()==0) return s;
        
        sort(vowel.begin(), vowel.end());
        
        int j=0;
        for(int i=0;i<t.length();i++){
            if(isVowel(t[i])) {
                t[i]=vowel[++j];
            }
        }
        
        return t;
    }

1. Debug the code.

/*
Test Case:
Input: word1 = "abc", word2 = "pqr"
Output: "apbqcr"
*/
string mergeAlternately(string word1, string word2) {
        int m = word1.size();
        int n = word2.size();
        string result = "";
        int i = 0, j = 0;

        while (i < m || j < n) {
            if (i < m) {
                result.push_back(word1[i++]);
            }
            else if (j < n) {
                result.push_back(word2[j++]);
            }
        }

        return result;
    }

solution-1.txt

solution-2.txt

solution-3.txt

solution-4.txt

solution-5.txt

solution-6.txt