Minimum cost to build N blocks from one block
Given a number N, the task is to build N blocks from 1 block by performing following operation:
 Double the number of blocks present in the container and cost for this operation is
X
.  Increase the number of blocks present in the container by one and cost for this operation is
Y
.  Decrease the number of blocks present in the container by one and cost for this operation is
Z
.
Examples:
Input: N = 5, X = 2, Y = 1, Z = 3
Output: 4
Explanation:
In the first operation just increase the number of blocks by one, cost = 1 and block count = 2
In the second operation double the blocks, cost = 3 and block count = 4
In the third operation again increase the number_of_blocks by one, cost = 4 and block count = 5
So minimum cost = 4
Input: N = 7, X = 1, Y = 7, Z = 2
Output: 5
Approach:
 Let
f(i)
denotes minimum cost to buildi
blocks. Sof(i)
=min(f(2*i)+X, f(i1)+Y, f(i+1)+Z)
Here current value depends on its next value as well as its previous value but you know that in dynamic programming current value depends only on its previous value. So try to convert its next value to its previous value. We can writef(2*i)
asf(i/2)
because when we have already built i/2 blocks then we can either double the number of current blocks or increase number of blocks by 1 or decrease number of blocks by 1. Here no need to calculatef(2*i)
.
Nowf(i) = min(f(i/2)+X, f(i1)+Y, f(i+1)+Z)

If
i
is even then(i+1)
must be odd. We can build(i)
blocks only by performing the increment operation and the double operation, so you don’t need to perform the decrement operation why?
This is because(i+1)
is odd number.So if you build(i+1)
blocks by performing increment operation then it is confirmed that we have already builti
blocks which means we have the minimum cost of buildingi
blocks.If we perform any other operation it must increase your optimal cost which is irrelevant. So recurrence relation when i is even:f(i)=min(f(i/2)+X, f(i1)+Y)

If
i
is odd then(i+1)
must be even. We can build(i+1)
blocks only by performing double operation or decrement operation why not increment operation?
This is because if you have already builti
blocks then no need to build(i+1)
blocks because it will add more cost. So we can calculatef(i+1)=f((i+1)/2)+ X.
So recurrence relation when i is odd:f(i)=min(f(i1)+Y, f( (i+1)/2)+X+Z)
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Below is the implementation of the above approach:
CPP
// C++ program to Minimum cost // to build N blocks from one block #include <bits/stdc++.h> using namespace std; // Function to calculate // min cost to build N blocks int minCost( int n, int x, int y, int z) { int dp[n + 1] = { 0 }; // Initialize base case dp[0] = dp[1] = 0; for ( int i = 2; i <= n; i++) { // Recurence when // i is odd if (i % 2 == 1) { dp[i] = min( dp[(i + 1) / 2] + x + z, dp[i  1] + y); } // Recurence when // i is even else { dp[i] = min(dp[i / 2] + x, dp[i  1] + y); } } return dp[n]; } // Driver code int main() { int n = 5, x = 2, y = 1, z = 3; cout << minCost(n, x, y, z) << endl; return 0; } 
Java
// Java program to Minimum cost // to build N blocks from one block class GFG { // Function to calculate // min cost to build N blocks static int minCost( int n, int x, int y, int z) { int dp[] = new int [n + 1 ]; // Initialize base case dp[ 0 ] = dp[ 1 ] = 0 ; for ( int i = 2 ; i <= n; i++) { // Recurence when // i is odd if (i % 2 == 1 ) { dp[i] = Math.min( dp[(i + 1 ) / 2 ] + x + z, dp[i  1 ] + y); } // Recurence when // i is even else { dp[i] = Math.min(dp[i / 2 ] + x, dp[i  1 ] + y); } } return dp[n]; } // Driver code public static void main(String[] args) { int n = 5 , x = 2 , y = 1 , z = 3 ; System.out.print(minCost(n, x, y, z) + "\n" ); } } // This code is contributed by RajputJi 
Python
# python3 program to Minimum cost # to build N blocks from one block # Function to calculate # min cost to build N blocks def minCost(n, x, y, z): dp = [ 0 ] * (n + 1 ) # Initialize base case dp[ 0 ] = dp[ 1 ] = 0 for i in range ( 2 , n + 1 ): # Recurence when # i is odd if (i % 2 = = 1 ): dp[i] = min (dp[(i + 1 ) / / 2 ] + x + z, dp[i  1 ] + y) # Recurence when # i is even else : dp[i] = min (dp[i / / 2 ] + x, dp[i  1 ] + y) return dp[n] # Driver code n = 5 x = 2 y = 1 z = 3 print (minCost(n, x, y, z)) # This code is contributed by mohit kumar 29 
C#
// C# program to Minimum cost // to build N blocks from one block using System; class GFG { // Function to calculate // min cost to build N blocks static int minCost( int n, int x, int y, int z) { int []dp = new int [n + 1]; // Initialize base case dp[0] = dp[1] = 0; for ( int i = 2; i <= n; i++) { // Recurence when // i is odd if (i % 2 == 1) { dp[i] = Math.Min( dp[(i + 1) / 2] + x + z, dp[i  1] + y); } // Recurence when // i is even else { dp[i] = Math.Min(dp[i / 2] + x, dp[i  1] + y); } } return dp[n]; } // Driver code public static void Main(String[] args) { int n = 5, x = 2, y = 1, z = 3; Console.Write(minCost(n, x, y, z) + "\n" ); } } // This code is contributed by PrinciRaj1992 
4
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