Given an integer N(1<=N<=10^9). The task is to represent N as a sum of the maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings. There can be multiple queries
Examples:
Input : 12 Output : 3 Explanation : 12 can be written has 4 + 4 + 4 or 6 + 6 or 8 + 4 But, 4 + 4 + 4 has maximum number of summands. Input : 7 Output : -1
Approach : Note that minimal composite number is equal to 4. So it is quite logical that there will be a lot of 4 in a splitting of big numbers. Let’s write for small numbers (1<=M<=N) dpN be the number of composite summands in splitting of N.
Let’s find an answer for all numbers from 1 to 15. Several observations:
- Only 4, 6, 9 occurs in optimal splittings.
- It is not beneficial to use 6 or 9 more than once because 6 + 6 = 4 + 4 + 4, 9 + 9 = 6 + 6 + 6.
- 12, 13, 14, 15 have valid splittings.
Let’s prove that all numbers that are greater than 15 will have 4 in optimal splitting. Let’s guess that it is incorrect. If the minimal number in splitting is neither 4 nor 6 nor 9 then this number will have some non-trivial splitting by induction.
If this number either 6 or 9 and we will decrease query by this number then we will sooner or later get some small number (which is less or equal than 15). There is no splitting of small numbers or it contains 4 in splitting (and it contradicts with minimality of the first number) or it contains 6 and 9. So we have contradiction in all cases.
We can subtract 4 from any big query and our solution is correct.
If our query n is small number let’s print dpn. Else let’s find minimal number k such that n – 4·k is a small number. Then print k + dpn – 4·k.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std; const int maxn = 16; // Function to generate the dp array vector< int > precompute() { vector< int > dp(maxn, -1); dp[0] = 0; for ( int i = 1; i < maxn; ++i) { // combination of three integers for ( auto j : vector< int >{ 4, 6, 9 }) { // take the maximum number of summands if (i >= j && dp[i - j] != -1) { dp[i] = max(dp[i], dp[i - j] + 1); } } } return dp; } // Function to find the maximum number of summands int Maximum_Summands(vector< int > dp, int n) { // If n is a smaller number, less than 16 // return dp[n] if (n < maxn) return dp[n]; else { // Else, find a minimal number t // as explained in solution int t = (n - maxn) / 4 + 1; return t + dp[n - 4 * t]; } } // Driver code int main() { int n = 12; // Generate dp array vector< int > dp = precompute(); cout << Maximum_Summands(dp, n) << endl; return 0; } |
Java
// Java implementation of the above approach class GFG { static int maxn = 16 ; // Function to generate the dp array static int [] precompute() { int dp[] = new int [maxn], arr[]={ 4 , 6 , 9 }; // initialize for ( int i = 0 ; i < maxn; i++)dp[i] = - 1 ; dp[ 0 ] = 0 ; for ( int i = 1 ; i < maxn; ++i) { // combination of three integers for ( int k = 0 ; k < 3 ; k++) { int j = arr[k]; // take the maximum number of summands if (i >= j && dp[i - j] != - 1 ) { dp[i] = Math.max(dp[i], dp[i - j] + 1 ); } } } return dp; } // Function to find the maximum number of summands static int Maximum_Summands( int [] dp, int n) { // If n is a smaller number, less than 16 // return dp[n] if (n < maxn) return dp[n]; else { // Else, find a minimal number t // as explained in solution int t = (n - maxn) / 4 + 1 ; return t + dp[n - 4 * t]; } } // Driver code public static void main(String args[]) { int n = 12 ; // Generate dp array int [] dp = precompute(); System.out.println(Maximum_Summands(dp, n)); } } // This code is contributed by Arnab Kundu |
Python3
# Python 3 implementation of the above approach global maxn maxn = 16 # Function to generate the dp array def precompute(): dp = [ - 1 for i in range (maxn)] dp[ 0 ] = 0 v = [ 4 , 6 , 9 ] for i in range ( 1 , maxn, 1 ): # combination of three integers for k in range ( 3 ): j = v[k] # take the maximum number of summands if (i > = j and dp[i - j] ! = - 1 ): dp[i] = max (dp[i], dp[i - j] + 1 ) return dp # Function to find the maximum number of summands def Maximum_Summands(dp, n): # If n is a smaller number, # less than 16, return dp[n] if (n < maxn): return dp[n] else : # Else, find a minimal number t # as explained in solution t = int ((n - maxn) / 4 ) + 1 return t + dp[n - 4 * t] # Driver code if __name__ = = '__main__' : n = 12 # Generate dp array dp = precompute() print (Maximum_Summands(dp, n)) # This code is contributed by # Surendra_Gangwar |
C#
// C# implementation of the above approach using System; using System.Collections; class GFG { static int maxn = 16; static int [] dp = new int [maxn + 1]; // Function to generate the dp array static void precompute() { for ( int i = 0; i <= maxn; i++) dp[i] = -1; dp[0] = 0; int [] vec = { 4, 6, 9 }; for ( int i = 1; i < maxn; ++i) { // combination of three integers foreach ( int j in vec) { // take the maximum number of summands if (i >= j && dp[i - j] != -1) { dp[i] = Math.Max(dp[i], dp[i - j] + 1); } } } } // Function to find the maximum number of summands static int Maximum_Summands( int n) { // If n is a smaller number, less than 16 // return dp[n] if (n < maxn) return dp[n]; else { // Else, find a minimal number t // as explained in solution int t = (n - maxn) / 4 + 1; return t + dp[n - 4 * t]; } } // Driver code static void Main() { int n = 12; // Generate dp array precompute(); Console.WriteLine(Maximum_Summands(n)); } } // This code is contributed by chandan_jnu |
PHP
<?php // PHP implementation of the above approach $maxn = 16; // Function to generate the dp array function precompute() { $dp = array_fill (0, $GLOBALS [ 'maxn' ], -1); $dp [0] = 0; $v = array (4, 6, 9); for ( $i = 1; $i < $GLOBALS [ 'maxn' ]; ++ $i ) { // combination of three integers for ( $k = 0; $k <3 ; $k ++) { $j = $v [ $k ]; // take the maximum number of summands if ( $i >= $j && $dp [ $i - $j ] != -1) { $dp [ $i ] = max( $dp [ $i ], $dp [ $i - $j ] + 1); } } } return $dp ; } // Function to find the maximum // number of summands function Maximum_Summands( $dp , $n ) { // If n is a smaller number, // less than 16 return dp[n] if ( $n < $GLOBALS [ 'maxn' ]) return $dp [ $n ]; else { // Else, find a minimal number t // as explained in solution $t = ( $n - $GLOBALS [ 'maxn' ]) / 4 + 1; return $t + $dp [ $n - 4 * $t ]; } } // Driver code $n = 12; // Generate dp array $dp = precompute(); echo Maximum_Summands( $dp , $n ); // This code is contributed by Ryuga ?> |
Javascript
<script> // JavaScript implementation of the above approach let maxn = 16; // Function to generate the dp array function precompute() { let dp = new Array(maxn); let arr = [ 4, 6, 9 ]; // initialize for (let i = 0; i < maxn; i++) dp[i] = -1; dp[0] = 0; for (let i = 1; i < maxn; ++i) { // combination of three integers for (let k = 0; k < 3; k++) { let j = arr[k]; // take the maximum number of summands if (i >= j && dp[i - j] != -1) { dp[i] = Math.max(dp[i], dp[i - j] + 1); } } } return dp; } // Function to find the maximum number of summands function Maximum_Summands(dp, n) { // If n is a smaller number, less than 16 // return dp[n] if (n < maxn) return dp[n]; else { // Else, find a minimal number t // as explained in solution let t = parseInt((n - maxn) / 4, 10) + 1; return t + dp[n - 4 * t]; } } let n = 12; // Generate dp array let dp = precompute(); document.write(Maximum_Summands(dp, n)); </script> |
3
Time Complexity: O(maxn)
Auxiliary Space: O(maxn)
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