LeetCode //C - 441. Arranging Coins

441. Arranging Coins

You have n coins and you want to build a staircase with these coins. The staircase consists of k rows where the i t h i^{th} ith row has exactly i coins. The last row of the staircase may be incomplete.

Given the integer n, return the number of complete rows of the staircase you will build.

Example 1:

Input: n = 5
Output: 2
Explanation: Because the 3 r d 3^{rd} 3rd row is incomplete, we return 2.

Example 2:

Input: n = 8
Output: 3
Explanation: Because the 4 t h 4^{th} 4th row is incomplete, we return 3.

Constraints:

• 1 < = n < = 2 31 − 1 1 <= n <= 2^{31} - 1 1<=n<=231−1

From: LeetCode

Link: 441. Arranging Coins

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Solution:

Ideas:

1. Binary Search: This is an efficient algorithm used to find an item in a sorted list. The algorithm repeatedly divides the range of possible solutions in half, choosing the half where the item must be, until it narrows down to the exact location.

2. Using the Formula for the Sum of Natural Numbers: To find how many coins are needed to completely fill k rows, the formula k * (k + 1) / 2 is used. This is the sum of the first k natural numbers, and in the context of this problem, it represents the total number of coins needed to fill k rows of the staircase.

3. Finding the Maximum k: The goal is to find the largest k such that k * (k + 1) / 2 is less than or equal to n. This k represents the number of complete rows you can build with n coins.

4. Binary Search Implementation:

• The search starts with the full range of possible row counts (left at 0 and right at n).
  It repeatedly calculates the middle point (mid) and the number of coins needed to fill that many rows (curr).
• If curr equals n, we've found the exact number of rows we can fill and return mid.
• If curr is greater than n, we have too many rows, so we reduce the search range to the left half.
• If curr is less than n, we have too few rows, so we increase the search range to the right half.
• The search stops when left exceeds right, meaning we've narrowed down the range as much as possible. At this point, right will be the greatest number of complete rows that can be built with n coins.

Code:

int arrangeCoins(int n) {
    long left = 0, right = n;
    long mid, curr;
    
    while (left <= right) {
        mid = left + (right - left) / 2;
        curr = mid * (mid + 1) / 2;
        
        if (curr == n) return (int)mid;
        
        if (n < curr) {
            right = mid - 1;
        } else {
            left = mid + 1;
        }
    }
    
    return (int)right;
}