21(June) Compare two fractions

21. Compare Two Fractions

The problem can be found at the following link: Question Link

Problem Description

You are given a string str containing two fractions a/b and c/d. Compare these fractions and return the greater. If they are equal, then return "equal".

Note: The string str contains "a/b, c/d" (fractions are separated by a comma , and a space ).

Examples:

Input:

str = "5/6, 11/45"

Output:

5/6

Explanation: 5/6 = 0.8333 and 11/45 = 0.2444, so 5/6 is the greater fraction.

My Approach

1. Parse the Input String:

   • Split the input string str into two parts using ", " as the delimiter to separate the fractions.

   • Extract the numerator and denominator for each fraction by splitting each part using '/'.

2. Compare the Fractions:

   • Calculate the cross multiplication for comparison:

     • Compare a * d with b * c.

   • Based on the comparison:

     • Return the greater fraction as a string.

     • If they are equal, return "equal".

3. Return the Result:

   • Return the greater fraction in string format or "equal" if both fractions are the same.

Time and Auxiliary Space Complexity

• Expected Time Complexity: O(len(str)), as we perform operations proportional to the length of the input string.

• Expected Auxiliary Space Complexity: O(1), as we only use a constant amount of additional space.

Code (C++)

class Solution {
public:
    string compareFrac(string& s) {
        int n = s.length();
        int a = 0, b = 0, c = 0, d = 0;
        int i = 0;
        while (i < n && s[i] != '/') {
            a = a * 10 + s[i] - '0';
            i++;
        }
        i++;
        while (i < n && s[i] != ',') {
            b = b * 10 + s[i] - '0';
            i++;
        }
        i += 2; // Skip ", "
        while (i < n && s[i] != '/') {
            c = c * 10 + s[i] - '0';
            i++;
        }
        i++;
        while (i < n) {
            d = d * 10 + s[i] - '0';
            i++;
        }
        if ((a * d) > (b * c)) {
            return to_string(a) + "/" + to_string(b);
        } else if ((a * d) < (b * c)) {
            return to_string(c) + "/" + to_string(d);
        }
        return "equal";
    }
};

Code (Java)

class Solution {
    String compareFrac(String str) {
        int n = str.length();
        int a = 0, b = 0, c = 0, d = 0;
        int i = 0;
        while (i < n && str.charAt(i) != '/') {
            a = a * 10 + str.charAt(i) - '0';
            i++;
        }
        i++;
        while (i < n && str.charAt(i) != ',') {
            b = b * 10 + str.charAt(i) - '0';
            i++;
        }
        i += 2; // Skip ", "
        while (i < n && str.charAt(i) != '/') {
            c = c * 10 + str.charAt(i) - '0';
            i++;
        }
        i++;
        while (i < n) {
            d = d * 10 + str.charAt(i) - '0';
            i++;
        }
        if ((a * d) > (b * c)) {
            return a + "/" + b;
        } else if ((a * d) < (b * c)) {
            return c + "/" + d;
        }
        return "equal";
    }
}

Code (Python)

class Solution:
    def compareFrac(self, s):
        parts = s.split(", ")
        a, b = map(int, parts[0].split('/'))
        c, d = map(int, parts[1].split('/'))

        if a * d > b * c:
            return "{}/{}".format(a, b)
        elif a * d < b * c:
            return "{}/{}".format(c, d)
        else:
            return "equal"

Contribution and Support

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