E-Content for B.Sc)H) Computer Sc. Paper-Discrete Structure Sem. II CVS,DU
Powerful Integers by Bruteforce Algorithm using C++
Given two positive integers x and y, an integer is powerful if it is equal to x^i + y^j for some integers i >= 0 and j >= 0. Return a list of all powerful integers that have value less than or equal to bound.
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 = 2^0 + 3^0 3 = 2^1 + 3^0 4 = 2^0 + 3^1 5 = 2^1 + 3^1 7 = 2^2 + 3^1 9 = 2^3 + 3^0 10 = 2^0 + 3^2 Example 2:
Input: x = 3, y = 5, bound = 15 Output: [2,4,6,8,10,14]
Note:
1 <= x <= 100 1 <= y <= 100 0 <= bound <= 10^6
C++ Bruteforce Algorithm to Compute the Powerful Integers
The edge cases are when x and y are equal to 1. We can use a set to store the unique powerful integers within the bound. If X = 1 or Y = 1, the time
complexity is . If both are 1, then the complexity is O(1) – as there is only 1 powerful integer, which is 1+1=2.
E-Content for B.Sc)H) Computer Sc. Paper-Discrete Structure Sem. II CVS,DU If neither X or Y is 1, the time complexity is . The space complexity is the same the time complexity as each number to test may be a potential powerful integer.
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class Solution { public:
vector<int> powerfulIntegers(int x, int y, int bound) { unordered_set<int> data;
int a, b;
for (int i = 0; (a = pow(x, i)) <= bound; ++ i) {
for (int j = 0; (b = a + pow(y, j)) <= bound; ++ j) { if (b <= bound) {
data.insert(b);
} else break;
if (y == 1) break;
}
if (x == 1) break;
}
vector<int> res;
std::copy(data.begin(), data.end(), std::back_inserter(res));
return res;
} };
And, if either X or Y is 1, we can break the loop – as 1 to the power of any will be still one, otherwise, it will be an endless loop. At last, we need to convert the set to std::vector in C++ .
E-Content for B.Sc)H) Computer Sc. Paper-Discrete Structure Sem. II CVS,DU
