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 = 4Input: 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(i-1)+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(i-1)+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(i-1)+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(i-1)+Y, f( (i+1)/2)+X+Z)
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 Rajput-Ji
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
Output:4
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