New Problem Solution -"Maximize Number of Nice Divisors"

Author: haoel

README.md

@@ -9,6 +9,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|1808|[Maximize Number of Nice Divisors](https://leetcode.com/problems/maximize-number-of-nice-divisors/) | [C++](./algorithms/cpp/maximizeNumberOfNiceDivisors/MaximizeNumberOfNiceDivisors.cpp)|Hard|
 |1807|[Evaluate the Bracket Pairs of a String](https://leetcode.com/problems/evaluate-the-bracket-pairs-of-a-string/) | [C++](./algorithms/cpp/evaluateTheBracketPairsOfAString/EvaluateTheBracketPairsOfAString.cpp)|Medium|
 |1806|[Minimum Number of Operations to Reinitialize a Permutation](https://leetcode.com/problems/minimum-number-of-operations-to-reinitialize-a-permutation/) | [C++](./algorithms/cpp/minimumNumberOfOperationsToReinitializeAPermutation/MinimumNumberOfOperationsToReinitializeAPermutation.cpp)|Medium|
 |1805|[Number of Different Integers in a String](https://leetcode.com/problems/number-of-different-integers-in-a-string/) | [C++](./algorithms/cpp/numberOfDifferentIntegersInAString/NumberOfDifferentIntegersInAString.cpp)|Easy|

algorithms/cpp/maximizeNumberOfNiceDivisors/MaximizeNumberOfNiceDivisors.cpp

@@ -0,0 +1,140 @@
+// Source : https://leetcode.com/problems/maximize-number-of-nice-divisors/
+// Author : Hao Chen
+// Date   : 2021-03-28
+
+/***************************************************************************************************** 
+ *
+ * You are given a positive integer primeFactors. You are asked to construct a positive integer n that 
+ * satisfies the following conditions:
+ * 
+ *   The number of prime factors of n (not necessarily distinct) is at most primeFactors.
+ *   The number of nice divisors of n is maximized. Note that a divisor of n is nice if it is 
+ * divisible by every prime factor of n. For example, if n = 12, then its prime factors are [2,2,3], 
+ * then 6 and 12 are nice divisors, while 3 and 4 are not.
+ * 
+ * Return the number of nice divisors of n. Since that number can be too large, return it modulo 10^9 
+ * + 7.
+ * 
+ * Note that a prime number is a natural number greater than 1 that is not a product of two smaller 
+ * natural numbers. The prime factors of a number n is a list of prime numbers such that their product 
+ * equals n.
+ * 
+ * Example 1:
+ * 
+ * Input: primeFactors = 5
+ * Output: 6
+ * Explanation: 200 is a valid value of n.
+ * It has 5 prime factors: [2,2,2,5,5], and it has 6 nice divisors: [10,20,40,50,100,200].
+ * There is not other value of n that has at most 5 prime factors and more nice divisors.
+ * 
+ * Example 2:
+ * 
+ * Input: primeFactors = 8
+ * Output: 18
+ * 
+ * Constraints:
+ * 
+ * 	1 <= primeFactors <= 10^9
+ ******************************************************************************************************/
+
+/*
+    considering `primeFactors = 5`
+    
+    So, we can have the following options:
+    1) [2,3,5,7,11] - all of factors are different, then we only can have 1 nice divisor
+    2) [2,2,3,5,7] - we can have 2*1*1*1 = 2 nice divisors: 2*3*5*7 and 2*2*3*5*7
+    3) [2,2,3,3,5] - we can have 2*2*1 = 4 nice divisors: 2*3*5， 2*2*3*5， 2*3*3*5， 2*2*3*3*5
+    4) [2,2,3,3,3] - we can have 2*3 = 6 nice divisors
+    5）[2,2,2,2,3] - we can have 4*1 =4 nice divisors: 2*3, 2*2*3, 2*2*2*3, 2*2*2*2*3
+    6) [2,2,2,2,2] - we can have 5 nice divisors: 2, 2*2, 2*2*2, 2*2*2*2, 2*2*2*2*2
+    
+    So, we can see we must have some duplicated factors.
+    
+    And what is the best number of duplication ?
+    primeFactors = 1, then 1        - example: [2]
+    primeFactors = 2, then 2        - example: [2,2]
+    primeFactors = 3, then 3        - example: [5,5,5]
+    primeFactors = 4, then 4 = 2*2  - example: [2,2,5,5])
+    primeFactors = 5, then 5 = 2*3  - example: [3,3,3,5,5])
+    primeFactors = 5, then 6 = 3*3  - example: [3,3,3,5,5,5])
+    primeFactors = 7, then 3*4 = 12 - example: [3,3,3,5,5,5,5])
+    primeFactors = 8, then 3*3*2 = 18  whcih > (2*2*2*2,  2*4*2, 3*5)
+    primeFactors = 9, then 3*3*3 = 27  
+    primeFactors = 10, then 3*3*4 = 36
+    
+    So, we can see the '3' & '4' are  specifial,
+    - most of case, we can be greedy for `3`
+    - but if the final rest is 4, then we need take 4.
+    
+ */
+
+const int mod = 1000000007;
+
+class Solution {
+public:
+    int maxNiceDivisors(int primeFactors) {
+        return maxNiceDivisors_03(primeFactors);
+        return maxNiceDivisors_02(primeFactors); //TLE
+        return maxNiceDivisors_01(primeFactors); //TLE
+    }
+    
+    int maxNiceDivisors_01(int primeFactors) {
+        int result = 1;
+        while ( primeFactors > 4 ) {
+            primeFactors -= 3;
+            result = (result * 3l) % mod;
+        }
+        result = (result * (long)primeFactors) % mod;
+        return result;
+    }
+    
+    int maxNiceDivisors_02(int primeFactors) {
+        if (primeFactors <= 4 ) return primeFactors;
+        int result = 1;
+        for (int i = 4; i > 0; i-- ){
+            if ((primeFactors - i) % 3 == 0){ 
+                result = i;
+                primeFactors -= i;
+                // now, `primeFactors` is 3 times - 3X
+                // we need convert 3X to 3^X
+                for (int x = primeFactors/3; x > 0; x-- ) {
+                    result = (result * 3l) % mod;
+                }
+                break;
+            }
+        }
+        return result;
+    }
+    
+    int pow3(int x) {
+        long  result = 1;
+        long  factor = 3;
+        while(x > 0) {
+            if (x & 1) {
+                result = (result * factor) % mod;
+                
+            }
+            factor *= factor;
+            factor %= mod;
+            x /= 2;
+        }
+        return result % mod;
+    }
+    
+    int maxNiceDivisors_03(int primeFactors) {
+        
+        if (primeFactors <= 4 ) return primeFactors;
+        int result = 1;
+        for (int i = 4; i > 0; i-- ){
+            if ((primeFactors - i) % 3 == 0){ 
+                primeFactors -= i;
+                // now, `primeFactors` is 3 times - 3X
+                // we need convert 3X to 3^X
+                int x = primeFactors / 3;
+                result = (long(i) * pow3(x)) % mod;
+                break;
+            }
+        }
+        return result;
+    }
+};