LeetCode #441 — EASY

Arranging Coins

Build confidence with an intuition-first walkthrough focused on math fundamentals.

The Problem

Problem Statement

You have n coins and you want to build a staircase with these coins. The staircase consists of k rows where the i^th 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^rd row is incomplete, we return 2.

Example 2:

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

Constraints:

1 <= n <= 2³¹ - 1

Step 01

Brute Force Baseline

Problem summary: You have n coins and you want to build a staircase with these coins. The staircase consists of k rows where the 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Binary Search

Example 1

Example 2

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #441: Arranging Coins
class Solution {
    public int arrangeCoins(int n) {
        return (int) (Math.sqrt(2) * Math.sqrt(n + 0.125) - 0.5);
    }
}

// Accepted solution for LeetCode #441: Arranging Coins
func arrangeCoins(n int) int {
	left, right := 1, n
	for left < right {
		mid := (left + right + 1) >> 1
		if (1+mid)*mid/2 <= n {
			left = mid
		} else {
			right = mid - 1
		}
	}
	return left
}

# Accepted solution for LeetCode #441: Arranging Coins
class Solution:
    def arrangeCoins(self, n: int) -> int:
        return int(math.sqrt(2) * math.sqrt(n + 0.125) - 0.5)

// Accepted solution for LeetCode #441: Arranging Coins
impl Solution {
    pub fn arrange_coins(n: i32) -> i32 {
        let (mut l, mut r) = (1, n);
        let mut res = 0;

        while l <= r {
            let mid = l + (r - l) / 2;
            let coins = (mid as i64 * (mid as i64 + 1)) / 2;

            if coins > n as i64 {
                r = mid - 1;
            } else {
                l = mid + 1;
                res = mid;
            }
        }

        res as i32

    }
}

// Accepted solution for LeetCode #441: Arranging Coins
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #441: Arranging Coins
// class Solution {
//     public int arrangeCoins(int n) {
//         return (int) (Math.sqrt(2) * Math.sqrt(n + 0.125) - 0.5);
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.