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+---
+id: ugly-number-III
+title: Ugly Number III
+sidebar_label: 1201. Ugly Number III
+tags:
+- Math
+- Binary Search
+- Combinatorics
+- Number Theory
+description: "Solution to Leetcode 1201. Ugly Number III"
+---
+
+## Problem Description
+
+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.
+ 
+### Examples
+
+**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 3rd 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 4th is 6.
+
+```
+
+
+
+### Constraints
+- `1 <= n, a, b, c <= 10^9`
+- `1 <= a * b * c <= 10^18`
+
+### Approach 
+The idea is, we search for the smallest K such that the number of numbers divisible by a or b or c in [1, K] is exactly N.
+
+We know the number of numbers divisible by x in [1, K] is K div x (or K // x, in Python).
+The number of numbers divisible by a or b or c in [1, K]:
+= #(divisible by a) + #(divisible by b) + #(divisible by c) - #(divisible by a, b) - #(divisible by (b, c)) - #(divisible by a, c) + #(divisible by a, b, c)
+= (K//a + K// b + K//c) - (K//lcm(a,b) + K//lcm(b, c) + K//lcm(c, a)) + K//lcm(a, b, c)
+
+where lcm is lowest common multiplier of the inside numeric arguments.
+
+Time Complexity: Note that the time complexity of computing lcm or gcd is O(log N). Thus, the overall time complexity is O((logN)^2)
+
+### Complexity
+
+Time complexity: $O((logN)^2)$
+Space complexity: $O(1)$
+
+### Solution
+
+#### Code in Different Languages
+
+#### C++
+
+ ```cpp
+typedef long long ll;
+#define MAX_ANS 2e9 
+
+class Solution {
+public:
+    int nthUglyNumber(int n, int a, int b, int c) {
+        int left = 0, right = MAX_ANS, result = 0;
+        while (left <= right) {
+            int mid = left + (right - left) / 2;
+            if (count(mid, a, b, c) >= n) { 
+                result = mid;
+                right = mid - 1;
+            } else {
+                left = mid + 1;
+            }
+        }
+        return result;
+    }
+    int count(ll num, ll a, ll b, ll c) {
+        return (int)(num / a + num / b + num / c
+            - num / lcm(a, b)
+            - num / lcm(b, c)
+            - num / lcm(a, c)
+            + num / (lcm(a, lcm(b, c))));
+    }
+    ll gcd(ll a, ll b) {
+        if (a == 0) return b;
+        return gcd(b % a, a);
+    }
+    ll lcm(ll a, ll b) {
+        return a * b / gcd(a, b);
+    }
+};
+ ```
+
+#### PYTHON
+
+```python
+    def nthUglyNumber(self, n: int, a: int, b: int, c: int) -> int:
+        def lcm(a, b):
+            prod = a*b
+            if a < b:
+                a, b = b, a
+            while b > 0:
+                a = a % b
+                a, b = b, a
+            return prod//a
+        def count(k):
+            #how many numbers divisible by a or b or c in [1, k]
+            na = k // a
+            nb = k // b
+            nc = k // c
+            nab = k // lcm(a, b)
+            nac = k // lcm(a, c)
+            nbc = k // lcm(b, c)
+            nabc = k // lcm(lcm(a, b), c)
+            return na + nb + nc - nab - nac - nbc + nabc
+        l, r = 0, 2*10**9
+        while l <= r:
+            mid = l + (r - l)//2
+            if count(mid) >= n:
+                if count(mid) == n and (mid % a == 0 or mid % b == 0 or mid % c == 0):
+                    return mid
+                r = mid - 1
+            else:
+                l = mid + 1
+        return l
+```
+
+
+
+### Complexity Analysis
+
+- Time Complexity: $O((logN)^2)$ 
+
+- Space Complexity: $O(1)$ 
+
+### References
+
+- **LeetCode Problem**: Ugly Number III
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