Leetcode 970: Powerful Integers
Given three integers x, y, and bound, find all powerful integers that are less than or equal to bound. A powerful integer can be expressed in the form x^i + y^j for non-negative integers i and j. Return the results as a list with no duplicates, and the order of the elements does not matter.
Problem
Approach
Steps
Complexity
Input: Three integers x, y, and bound.
Example: Input: x = 2, y = 3, bound = 20
Constraints:
• 1 <= x, y <= 100
• 0 <= bound <= 10^6
Output: A list of all unique powerful integers less than or equal to bound.
Example: Output: [2, 3, 4, 5, 9, 10, 17]
Constraints:
• Each powerful integer must appear only once in the list.
• The integers in the output can be in any order.
Goal: Compute all unique powerful integers less than or equal to the given bound.
Steps:
• Iterate over possible values of i such that x^i <= bound.
• For each i, iterate over possible values of j such that x^i + y^j <= bound.
• If x or y is 1, break early to avoid infinite loops.
• Store the resulting sums in a set to ensure uniqueness.
• Convert the set to a list and return it.
Goal: Ensure the inputs are valid and the solution operates within computational limits.
Steps:
• 1 <= x, y <= 100
• 0 <= bound <= 10^6
• The result must contain no duplicate integers.
Assumptions:
• x and y are positive integers.
• The values of x and y can be 1, which simplifies the calculation.
• The bound is non-negative.
• Input: Input: x = 3, y = 2, bound = 25
• Explanation: The powerful integers are calculated as follows:
3^0 + 2^0 = 2, 3^1 + 2^0 = 4, 3^0 + 2^1 = 3, 3^1 + 2^1 = 5, 3^2 + 2^0 = 9, 3^2 + 2^1 = 11, 3^0 + 2^2 = 6, 3^1 + 2^2 = 10, 3^2 + 2^2 = 13, ...
Output: [2, 3, 4, 5, 6, 9, 10, 11, 13]
• Input: Input: x = 1, y = 1, bound = 5
• Explanation: Since x and y are both 1, only one unique powerful integer can be formed: 1 + 1 = 2. Output: [2]
Approach: Use nested loops to calculate all possible values of x^i + y^j within the bound, ensuring no duplicates by using a set.
Observations:
• Both x and y can potentially grow exponentially, so the number of iterations depends on the logarithmic scale of the bound.
• Using a set ensures uniqueness of results.
• Iterate over all possible powers of x and y while ensuring their sum remains within the bound.
• Break early when either x or y equals 1 since their powers do not grow.
Steps:
• Initialize an empty set to store results.
• Iterate over possible values of i for x, stopping when x^i > bound.
• For each i, iterate over possible values of j for y, stopping when x^i + y^j > bound.
• Add the sum x^i + y^j to the set.
• Break the inner loop early if y == 1 to avoid infinite iteration.
• Break the outer loop early if x == 1 to avoid infinite iteration.
• Convert the set to a list and return it.
Empty Inputs:
• If bound is 0, the result should be an empty list.
Large Inputs:
• For large values of bound, ensure the algorithm terminates efficiently by breaking early when x == 1 or y == 1.
Special Values:
• If x == 1 and y == 1, the only valid output is [2] (if bound >= 2).
• If x or y is 1, the respective loop becomes constant.
Constraints:
• Ensure that x^i and y^j are calculated without overflow within the constraints of bound.
vector<int> powerfulIntegers(int x, int y, int bound) {
unordered_set<int> s;
for(int i = 1; i <= bound; i *= x) {
for(int j = 1; i + j <= bound; j *= y) {
s.insert(i + j);
if(y == 1) break;
}
if(x == 1) break;
}
return vector<int>(s.begin(), s.end());
}
1 : Function Definition
vector<int> powerfulIntegers(int x, int y, int bound) {
This is the function header where `x`, `y`, and `bound` are the input parameters. It defines the function that will generate all powerful integers.
2 : Set Initialization
unordered_set<int> s;
A set `s` is initialized to store the unique powerful integers found during the process.
3 : Outer Loop
for(int i = 1; i <= bound; i *= x) {
Start a loop with `i` initialized to 1. It will multiply `i` by `x` in each iteration to generate powers of `x` up to the bound.
4 : Inner Loop
for(int j = 1; i + j <= bound; j *= y) {
Start an inner loop with `j` initialized to 1. It multiplies `j` by `y` to generate powers of `y` while ensuring that the sum `i + j` does not exceed the bound.
5 : Set Insertion
s.insert(i + j);
Add the sum of `i + j` to the set `s`. This ensures all combinations of `x^i + y^j` are stored without duplicates.
6 : Break Condition
if(y == 1) break;
If `y` is 1, the loop breaks early because further powers of `y` will always be 1, leading to redundant results.
7 : Outer Loop Break
if(x == 1) break;
If `x` is 1, the outer loop breaks early because further powers of `x` will always be 1, leading to redundant results.
8 : Return Statement
return vector<int>(s.begin(), s.end());
Convert the set `s` into a vector and return it as the result, containing all the unique powerful integers.
Best Case: O(log(bound))
Average Case: O(log(bound)^2)
Worst Case: O(log(bound)^2)
Description: The worst case involves nested loops iterating logarithmically with respect to x and y.
Best Case: O(1)
Worst Case: O(n)
Description: Space complexity is determined by the size of the set storing results, which is O(n), where n is the number of unique results.
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