502. IPO #154
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Topics: Suppose LeetCode will start its IPO soon. In order to sell a good price of its shares to Venture Capital, LeetCode would like to work on some projects to increase its capital before the IPO. Since it has limited resources, it can only finish at most You are given Initially, you have Pick a list of at most The answer is guaranteed to fit in a 32-bit signed integer. Example 1:
Example 2:
Constraints:
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We need to pick the projects that will maximize the final capital after finishing at most Approach
Let's implement this solution in PHP: 1595. Minimum Cost to Connect Two Groups of Points <?php
function findMaximizedCapital($k, $w, $profits, $capital) {
// Combine the projects into a list of tuples (capital, profit)
$projects = [];
$n = count($profits);
for ($i = 0; $i < $n; $i++) {
$projects[] = [$capital[$i], $profits[$i]];
}
// Sort projects by the capital needed (ascending)
usort($projects, function($a, $b) {
return $a[0] - $b[0];
});
// A max heap to store the available profits
$maxHeap = new SplMaxHeap();
$index = 0;
for ($i = 0; $i < $k; $i++) {
// Push all projects that can be started with the current capital into the heap
while ($index < $n && $projects[$index][0] <= $w) {
$maxHeap->insert($projects[$index][1]);
$index++;
}
// If there are no available projects, break out of the loop
if ($maxHeap->isEmpty()) {
break;
}
// Pick the project with the maximum profit
$w += $maxHeap->extract();
}
return $w;
}
// Example Usage
//Example 1
$k = 2;
$w = 0;
$profits = [1, 2, 3];
$capital = [0, 1, 1];
echo findMaximizedCapital($k, $w, $profits, $capital); // Output: 4
//Example 2
$k = 3;
$w = 0;
$profits = [1, 2, 3];
$capital = [0, 1, 2];
echo findMaximizedCapital($k, $w, $profits, $capital); // Output: 6
?> Explanation:
This approach efficiently handles the constraints and ensures that we maximize the final capital. |
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We need to pick the projects that will maximize the final capital after finishing at most
k
projects, given the initial capitalw
.Approach