LeetCode #1201 — MEDIUM

Ugly Number III

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

An ugly number is a positive integer that is divisible by a, b, or c.

Given four integers n, a, b, and c, return the n^th ugly number.

Example 1:

Input: n = 3, a = 2, b = 3, c = 5
Output: 4
Explanation: The ugly numbers are 2, 3, 4, 5, 6, 8, 9, 10... The 3^rd is 4.

Example 2:

Input: n = 4, a = 2, b = 3, c = 4
Output: 6
Explanation: The ugly numbers are 2, 3, 4, 6, 8, 9, 10, 12... The 4^th is 6.

Example 3:

Input: n = 5, a = 2, b = 11, c = 13
Output: 10
Explanation: The ugly numbers are 2, 4, 6, 8, 10, 11, 12, 13... The 5^th is 10.

Constraints:

1 <= n, a, b, c <= 10⁹
1 <= a * b * c <= 10¹⁸
It is guaranteed that the result will be in range [1, 2 * 10⁹].

Step 01

Brute Force Baseline

Problem summary: An ugly number is a positive integer that is divisible by a, b, or c. Given four integers n, a, b, and c, return the nth ugly number.

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

Example 3

Core Insight

What unlocks the optimal approach

Write a function f(k) to determine how many ugly numbers smaller than k. As f(k) is non-decreasing, try binary search.
Find all ugly numbers in [1, LCM(a, b, c)] (LCM is Least Common Multiple). Use inclusion-exclusion principle to expand the result.

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 #1201: Ugly Number III
class Solution {
    public int nthUglyNumber(int n, int a, int b, int c) {
        long ab = lcm(a, b);
        long bc = lcm(b, c);
        long ac = lcm(a, c);
        long abc = lcm(ab, c);
        long l = 1, r = 2000000000;
        while (l < r) {
            long mid = (l + r) >> 1;
            if (mid / a + mid / b + mid / c - mid / ab - mid / bc - mid / ac + mid / abc >= n) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return (int) l;
    }

    private long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a % b);
    }

    private long lcm(long a, long b) {
        return a * b / gcd(a, b);
    }
}

// Accepted solution for LeetCode #1201: Ugly Number III
func nthUglyNumber(n int, a int, b int, c int) int {
	ab, bc, ac := lcm(a, b), lcm(b, c), lcm(a, c)
	abc := lcm(ab, c)
	var l, r int = 1, 2e9
	for l < r {
		mid := (l + r) >> 1
		if mid/a+mid/b+mid/c-mid/ab-mid/bc-mid/ac+mid/abc >= n {
			r = mid
		} else {
			l = mid + 1
		}
	}
	return l
}

func gcd(a, b int) int {
	if b == 0 {
		return a
	}
	return gcd(b, a%b)
}

func lcm(a, b int) int {
	return a * b / gcd(a, b)
}

# Accepted solution for LeetCode #1201: Ugly Number III
class Solution:
    def nthUglyNumber(self, n: int, a: int, b: int, c: int) -> int:
        ab = lcm(a, b)
        bc = lcm(b, c)
        ac = lcm(a, c)
        abc = lcm(a, b, c)
        l, r = 1, 2 * 10**9
        while l < r:
            mid = (l + r) >> 1
            if (
                mid // a
                + mid // b
                + mid // c
                - mid // ab
                - mid // bc
                - mid // ac
                + mid // abc
                >= n
            ):
                r = mid
            else:
                l = mid + 1
        return l

// Accepted solution for LeetCode #1201: Ugly Number III
struct Solution;

impl Solution {
    fn nth_ugly_number(n: i32, a: i32, b: i32, c: i32) -> i32 {
        let mut left = 0;
        let mut right = 2_000_000_000;
        while left < right {
            let mid = left + (right - left) / 2;
            if Self::count(mid, a as u64, b as u64, c as u64) < n as u64 {
                left = mid + 1;
            } else {
                right = mid
            }
        }
        left as i32
    }

    fn count(num: u64, a: u64, b: u64, c: u64) -> u64 {
        num / a + num / b + num / c
            - num / Self::lcm(a, b)
            - num / Self::lcm(b, c)
            - num / Self::lcm(a, c)
            + num / Self::lcm(a, Self::lcm(b, c))
    }

    fn lcm(a: u64, b: u64) -> u64 {
        a * b / Self::gcd(a, b)
    }

    fn gcd(a: u64, b: u64) -> u64 {
        if a == 0 {
            b
        } else {
            Self::gcd(b % a, a)
        }
    }
}

#[test]
fn test() {
    let n = 3;
    let a = 2;
    let b = 3;
    let c = 5;
    let res = 4;
    assert_eq!(Solution::nth_ugly_number(n, a, b, c), res);
    let n = 4;
    let a = 2;
    let b = 3;
    let c = 4;
    let res = 6;
    assert_eq!(Solution::nth_ugly_number(n, a, b, c), res);
    let n = 5;
    let a = 2;
    let b = 11;
    let c = 13;
    let res = 10;
    assert_eq!(Solution::nth_ugly_number(n, a, b, c), res);
    let n = 1000000000;
    let a = 2;
    let b = 217983653;
    let c = 336916467;
    let res = 1999999984;
    assert_eq!(Solution::nth_ugly_number(n, a, b, c), res);
}

// Accepted solution for LeetCode #1201: Ugly Number III
function nthUglyNumber(n: number, a: number, b: number, c: number): number {
    const ab = lcm(BigInt(a), BigInt(b));
    const bc = lcm(BigInt(b), BigInt(c));
    const ac = lcm(BigInt(a), BigInt(c));
    const abc = lcm(BigInt(a), bc);
    let l = 1n;
    let r = BigInt(2e9);
    while (l < r) {
        const mid = (l + r) >> 1n;
        const count =
            mid / BigInt(a) +
            mid / BigInt(b) +
            mid / BigInt(c) -
            mid / ab -
            mid / bc -
            mid / ac +
            mid / abc;
        if (count >= BigInt(n)) {
            r = mid;
        } else {
            l = mid + 1n;
        }
    }
    return Number(l);
}

function gcd(a: bigint, b: bigint): bigint {
    return b === 0n ? a : gcd(b, a % b);
}

function lcm(a: bigint, b: bigint): bigint {
    return (a * b) / gcd(a, b);
}

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.