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This approach involves sequentially counting how many coins are required to form a complete row one row at a time. While the number of available coins is greater than or equal to coins needed for the next row, continue adding rows and reducing the count of available coins accordingly.
Time Complexity: O(√n), as the sum of the first k natural numbers grows quadratically.
Space Complexity: O(1), constant space is used.
1#include <iostream>
2using namespace std;
3
4int arrangeCoins(int n) {
5 int k = 0;
6 while (n >= k + 1) {
7 k++;
8 n -= k;
9 }
10 return k;
11}
12
13int main() {
14 int n = 8;
15 cout << arrangeCoins(n) << endl; // Output: 3
16 return 0;
17}The same greedy logic is implemented in C++. The while loop efficiently counts complete rows by deducting the required coins for each row successively, with k representing the number of those complete rows.
Using a binary search, optimize the process of finding the maximum k where the sum of the first k natural numbers is less than or equal to n. Calculate mid values and determine if they result in a possible sum being less or equal to n and adjust the search space accordingly.
Time Complexity: O(log n)
Space Complexity: O(1)
1def
The Python solution utilizes binary search to explore possible values for k, stopping at the largest k where the complete sum of first k natural numbers does not exceed n.