# Minimum cost to reach a point N from 0 with two different operations allowed

Given integers N, P and Q where N denotes the destination position. The task is to move from position 0 to position N with minimum cost possible and print the calculated cost. All valid movements are:

1. From position X you can go to position X + 1 with a cost of P
2. Or, you can go to the position 2 * X with a cost of Q

Examples:

Input: N = 1, P = 3, Q = 4
Output: 3
Move from position 0 to 1st position with cost = 3.

Input: N = 9, P = 5, Q = 1
Output: 13
Move from position 0 to 1st position with cost = 5,
then 1st to 2nd with cost = 1,
then 2nd to 4th with cost = 1,
then 4th to 8th with cost = 1,
finally 8th to 9th with cost = 5.
Total cost = 5 + 1 + 1 + 1 + 5 = 13.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Instead of going from beginning to destination we can start moving from the destination to initial position and keep track of the cost of jumps.

• If N is odd then the only valid move that could lead us here is N-1 to N with a cost of P.
• If N is even then we calculate cost of going from N to N/2 position with both the moves and take the minimum of them.
• When N equals 0, we return our total calculated cost.

Below is the implementation of above approach:

 `// CPP implementation of above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to return minimum ` `// cost to reach destination ` `int` `minCost(``int` `N, ``int` `P, ``int` `Q) ` `{ ` `    ``// Initialize cost to 0 ` `    ``int` `cost = 0; ` ` `  `    ``// going backwards until we ` `    ``// reach initial position ` `    ``while` `(N > 0) { ` ` `  `        ``if` `(N & 1) { ` `            ``cost += P; ` `            ``N--; ` `        ``} ` `        ``else` `{ ` `            ``int` `temp = N / 2; ` ` `  `            ``// if 2*X jump is ` `            ``// better than X+1 ` `            ``if` `(temp * P > Q) ` `                ``cost += Q; ` `            ``// if X+1 jump is better ` `            ``else` `                ``cost += P * temp; ` ` `  `            ``N /= 2; ` `        ``} ` `    ``} ` ` `  `    ``// return cost ` `    ``return` `cost; ` `} ` ` `  `// Driver program ` `int` `main() ` `{ ` `    ``int` `N = 9, P = 5, Q = 1; ` ` `  `    ``cout << minCost(N, P, Q); ` ` `  `    ``return` `0; ` `} `

 `// Java implementation of above approach ` ` `  `class` `GFG{ ` `// Function to return minimum ` `// cost to reach destination ` `static` `int` `minCost(``int` `N, ``int` `P, ``int` `Q) ` `{ ` `    ``// Initialize cost to 0 ` `    ``int` `cost = ``0``; ` ` `  `    ``// going backwards until we ` `    ``// reach initial position ` `    ``while` `(N > ``0``) { ` ` `  `        ``if` `((N & ``1``)>``0``) { ` `            ``cost += P; ` `            ``N--; ` `        ``} ` `        ``else` `{ ` `            ``int` `temp = N / ``2``; ` ` `  `            ``// if 2*X jump is ` `            ``// better than X+1 ` `            ``if` `(temp * P > Q) ` `                ``cost += Q; ` `            ``// if X+1 jump is better ` `            ``else` `                ``cost += P * temp; ` ` `  `            ``N /= ``2``; ` `        ``} ` `    ``} ` ` `  `    ``// return cost ` `    ``return` `cost; ` `} ` ` `  `// Driver program ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `N = ``9``, P = ``5``, Q = ``1``; ` ` `  `    ``System.out.println(minCost(N, P, Q)); ` `} ` `} ` `// This code is contributed by mits `

 `# Python implementation of above approach  ` ` `  ` `  `# Function to return minimum  ` `# cost to reach destination ` ` `  `def` `minCost(N,P,Q):  ` `    ``# Initialize cost to 0  ` `    ``cost ``=` `0`  `   `  `    ``# going backwards until we  ` `    ``# reach initial position  ` `    ``while` `(N > ``0``):   ` `        ``if` `(N & ``1``):  ` `            ``cost ``+``=` `P ` `            ``N``-``=``1` `        ``else``: ` `            ``temp ``=` `N ``/``/` `2``;  ` `   `  `            ``# if 2*X jump is  ` `            ``# better than X+1  ` `            ``if` `(temp ``*` `P > Q): ` `                ``cost ``+``=` `Q  ` `            ``# if X+1 jump is better  ` `            ``else``: ` `                ``cost ``+``=` `P ``*` `temp  ` `            ``N ``/``/``=` `2`  `    ``return` `cost ` ` `  `   `  `# Driver program  ` `N ``=` `9` `P ``=` `5` `Q ``=` `1` `print``(minCost(N, P, Q)) ` `#this code is improved by sahilshelangia  `

 `// C# implementation of above approach ` ` `  `class` `GFG ` `{ ` `// Function to return minimum ` `// cost to reach destination ` `static` `int` `minCost(``int` `N, ``int` `P, ``int` `Q) ` `{ ` `    ``// Initialize cost to 0 ` `    ``int` `cost = 0; ` ` `  `    ``// going backwards until we ` `    ``// reach initial position ` `    ``while` `(N > 0)  ` `    ``{ ` ` `  `        ``if` `((N & 1) > 0)  ` `        ``{ ` `            ``cost += P; ` `            ``N--; ` `        ``} ` `        ``else` `        ``{ ` `            ``int` `temp = N / 2; ` ` `  `            ``// if 2*X jump is ` `            ``// better than X+1 ` `            ``if` `(temp * P > Q) ` `                ``cost += Q; ` `                 `  `            ``// if X+1 jump is better ` `            ``else` `                ``cost += P * temp; ` ` `  `            ``N /= 2; ` `        ``} ` `    ``} ` ` `  `    ``// return cost ` `    ``return` `cost; ` `} ` ` `  `// Driver Code ` `static` `void` `Main() ` `{ ` `    ``int` `N = 9, P = 5, Q = 1; ` ` `  `    ``System.Console.WriteLine(minCost(N, P, Q)); ` `} ` `} ` ` `  `// This code is contributed by mits `

 ` 0) ` `    ``{ ` ` `  `        ``if` `(``\$N` `& 1) ` `        ``{ ` `            ``\$cost` `+= ``\$P``; ` `            ``\$N``--; ` `        ``} ` `        ``else` `        ``{ ` `            ``\$temp` `= ``\$N` `/ 2; ` ` `  `            ``// if 2*X jump is ` `            ``// better than X+1 ` `            ``if` `(``\$temp` `* ``\$P` `> ``\$Q``) ` `                ``\$cost` `+= ``\$Q``; ` `                 `  `            ``// if X+1 jump is better ` `            ``else` `                ``\$cost` `+= ``\$P` `* ``\$temp``; ` ` `  `            ``\$N` `/= 2; ` `        ``} ` `    ``} ` ` `  `    ``// return cost ` `    ``return` `\$cost``; ` `} ` ` `  `// Driver Code ` `\$N` `= 9; ``\$P` `= 5; ``\$Q` `= 1; ` ` `  `echo` `minCost(``\$N``, ``\$P``, ``\$Q``); ` ` `  `// This code is contributed  ` `// by Akanksha Rai ` `?> `

Output:
```13
```

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