grid47 Exploring patterns and algorithms
Oct 1, 2024
5 min read
Solution to LeetCode 365: Water and Jug Problem Problem
You are given two jugs with capacities x and y liters, and an infinite supply of water. You need to determine whether you can measure exactly target liters of water using the following operations: fill, empty, and pour water between the jugs.
Problem
Approach
Steps
Complexity
Input: The input consists of three integers: x, y, and target, representing the capacities of the two jugs and the target amount of water to measure.
Example: x = 3, y = 5, target = 4
Constraints:
• 1 <= x, y, target <= 1000
Output: The output should be true if it is possible to measure exactly the target amount of water, otherwise false.
Example: Input: x = 3, y = 5, target = 4
Output: true
Constraints:
• The output should be true or false based on the feasibility of measuring the target amount of water.
Goal: The goal is to check if it is possible to measure exactly the target amount of water using the given operations on the two jugs.
Steps:
• 1. If the total capacity of both jugs is less than the target, return false.
• 2. If either jug has the exact target capacity, return true.
• 3. If the sum of the two jugs' capacities is equal to the target, return true.
• 4. Otherwise, check if the target is divisible by the greatest common divisor (gcd) of the two jug capacities.
Goal: The input constraints ensure that the capacities of the jugs and the target are within the specified range.
Steps:
• 1 <= x, y, target <= 1000
Assumptions:
• The solution assumes that the jugs can be manipulated by filling, emptying, and pouring until the target amount is reached or proven impossible.
• Input: Input: x = 3, y = 5, target = 4
Output: true
• Explanation: You can measure 4 liters by filling and transferring water between the two jugs. Follow the steps to achieve the target of 4 liters.
Approach: The solution uses the greatest common divisor (gcd) method to determine whether the target amount can be measured by checking the mathematical conditions for each jug capacity.
Observations:
• The gcd of the jug capacities plays a key role in determining whether it's possible to measure the target.
• We need to ensure the target is either directly achievable or divisible by the gcd of the two jug capacities.
Steps:
• 1. Check if the sum of the two jugs' capacities is less than the target, return false.
• 2. Check if either jug has the target amount directly, return true.
• 3. Calculate the gcd of the two jug capacities, and check if the target is divisible by this gcd.
Empty Inputs:
• If x, y, or target is 0, return false.
Large Inputs:
• Ensure the solution handles inputs up to 1000 efficiently.
Special Values:
• When x or y equals the target, return true.
Constraints:
• The constraints guarantee that inputs will always be within the range of 1 to 1000.
boolcanMeasureWater(int x, int y, int z) {
if(x + y < z) returnfalse;
if(x == z || y == z || x + y == z) returntrue;
return z % gcd(x, y) ==0;
}
intgcd(int a, int b) {
return b ==0?a: gcd(b, a%b);
}
1 : Method Declaration
boolcanMeasureWater(int x, int y, int z) {
This line declares the method `canMeasureWater`, which takes three integers `x`, `y`, and `z`, representing the capacities of the two jugs and the target amount of water to measure.
2 : Conditional Check
if(x + y < z) returnfalse;
Checks if the sum of the two jugs' capacities is less than the target `z`. If so, it returns `false` because it’s impossible to measure `z` units of water.
3 : Conditional Check
if(x == z || y == z || x + y == z) returntrue;
Checks if either jug can directly hold the target amount of water (`z`) or if the sum of both jugs' capacities equals `z`. In either case, it returns `true`.
4 : Mathematical Condition
return z % gcd(x, y) ==0;
Returns `true` if `z` is divisible by the greatest common divisor (GCD) of `x` and `y`, which is a key condition for measuring exactly `z` units using the two jugs.
5 : Blank Line
This is a blank line for readability, separating the main method from the helper function `gcd`.
6 : Helper Method Declaration
intgcd(int a, int b) {
This line declares a helper method `gcd` to calculate the greatest common divisor (GCD) of two integers `a` and `b` using the Euclidean algorithm.
7 : Recursive Call
return b ==0?a: gcd(b, a%b);
This line uses the Euclidean algorithm to calculate the GCD recursively. If `b` is 0, it returns `a` as the GCD; otherwise, it recursively calls `gcd` with `b` and `a % b`.
Best Case: O(log(min(x, y))) for calculating gcd.
Average Case: O(log(min(x, y))) for gcd and conditional checks.
Worst Case: O(log(min(x, y))) for gcd calculation.
Description: The gcd computation takes logarithmic time relative to the smaller jug capacity.
Best Case: O(1) as no extra space is used.
Worst Case: O(1) as no additional data structures are required.
Description: The space complexity is constant because only a few variables are used.
