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This approach iterates through each kid's candies and checks if adding the extraCandies to their current candies makes them equal to or greater than the maximum number of candies present initially. It involves finding the maximum candies first and then performing this computation.
Time Complexity: O(n), where n is the number of kids.
Space Complexity: O(1), not counting the output array.
1def kidsWithCandies(candies, extraCandies):
2 maxCandies = max(candies)
3 return [(candy + extraCandies) >= maxCandies for candy in candies]The algorithm identifies the maximum number of candies using the built-in max function. It then uses list comprehension to form the result by verifying if each kid's candies plus extra candies would at least equal the maximum.
This approach optimizes the process by combining the finding of the max candies and constructing the result array into a single pass by keeping track of the maximum with conditional updates.
Time Complexity: O(n)
Space Complexity: O(1) aside from the result.
1
std::vector<bool> kidsWithCandies(std::vector<int>& candies, int extraCandies) {
int maxCandies = *max_element(candies.begin(), candies.end());
for (int i = 0; i < candies.size(); ++i) {
candies[i] = (candies[i] + extraCandies >= maxCandies);
}
return std::vector<bool>(candies.begin(), candies.end());
}This C++ solution leverages the STL to calculate the max candies and uses a loop to overwrite the current candies values with boolean results based on the condition check.