Given three integers x, y, and bound, return a list of all the powerful integers that have a value less than or equal to bound.
An integer is powerful if it can be represented as xi + yj for some integers i >= 0 and j >= 0.
You may return the answer in any order. In your answer, each value should occur at most once.
Example 1:
Input: x = 2, y = 3, bound = 10 Output: [2,3,4,5,7,9,10] Explanation: 2 = 20 + 30 3 = 21 + 30 4 = 20 + 31 5 = 21 + 31 7 = 22 + 31 9 = 23 + 30 10 = 20 + 32
Example 2:
Input: x = 3, y = 5, bound = 15 Output: [2,4,6,8,10,14]
Constraints:
1 <= x, y <= 1000 <= bound <= 106This approach involves iterating over possible powers of x and y, and calculating their sums while keeping the sum within the bound. We use two nested loops: one to iterate over powers of x and the other to iterate over powers of y. The results are stored in a set to ensure uniqueness.
The function powerful_integers calculates all powerful integers within the given bound. It uses nested loops where the outer loop iterates over powers of x and the inner loop iterates over powers of y. It calculates the sum x^i + y^j and checks if it's within the bound. If so, it adds the sum to a set. This set ensures all elements are unique.
C++
Time Complexity: O(log(bound)^2) since the loops iterate logarithmically with respect to the bound.
Space Complexity: O(n) where n is the number of unique powerful integers within the bound.
In this approach, we optimize the early termination condition by ensuring that we stop iterating as soon as additional powers do not contribute to any new sums within the bound. This prevents unnecessary iterations and computations.
This JavaScript function employs a similar logic to previous approaches but uses multiplicative updates instead of powers for progression through x^i and y^j. It utilizes Set for uniqueness and converts the result to an array for the final output.
Java
Time Complexity: O(log(bound)^2) as it bounds progression over the log of the bound value.
Space Complexity: O(n) where n is the number of unique powerful integers.
| Approach | Complexity |
|---|---|
| Approach 1: Iterative Power Combination | Time Complexity: O(log(bound)^2) since the loops iterate logarithmically with respect to the bound. |
| Approach 2: Optimized Early Termination | Time Complexity: O(log(bound)^2) as it bounds progression over the log of the bound value. |
Count the Number of Powerful Integers | Detailed Thought Process | Leetcode 2999 | codestorywithMIK • codestorywithMIK • 11,301 views views
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