# Count minimum steps to get the given desired array

Hello Everyone,

Consider an array with n elements and value of all the elements is zero. We can perform following operations on the array.

1. Incremental operations:Choose 1 element from the array and increment its value by 1.
2. Doubling operation: Double the values of all the elements of array.

We are given desired array target[] containing n elements. Compute and return the smallest possible number of the operations needed to change the array from all zeros to desired array.
Examples:

Input: target[] = {2, 3}
Output: 4
To get the target array from {0, 0}, we
first increment both elements by 1 (2
operations), then double the array (1
operation). Finally increment second
element (1 more operation)

Input: target[] = {2, 1}
Output: 3
One of the optimal solution is to apply the
incremental operation 2 times to first and
once on second element.

Input: target[] = {16, 16, 16}
Output: 7
The output solution looks as follows. First
apply an incremental operation to each element.
Then apply the doubling operation four times.
Total number of operations is 3+4 = 7

One important thing to note is that the task is to count the number of steps to get the given target array (not to convert zero array to target array).
The idea is to follow reverse steps, i.e. to convert target to array of zeros. Below are steps.

Take the target array first. Initialize result as 0. If all are even, divide all elements by 2 and increment result by 1. Find all odd elements, make them even by reducing them by 1. and for every reduction, increment result by 1. Finally, we get all zeros in target array.

Below is the implementation of above algorithm.

`/* C++ program to count minimum number of operations`

` ` `to get the given target array */`

`#include <bits/stdc++.h>`

`using` `namespace` `std;`

`// Returns count of minimum operations to convert a`

`// zero array to target array with increment and`

`// doubling operations.`

`// This function computes count by doing reverse`

`// steps, i.e., convert target to zero array.`

`int` `countMinOperations(unsigned ` `int` `target[], ` `int` `n)`

`{`

` ` `// Initialize result (Count of minimum moves)`

` ` `int` `result = 0;`

` ` `// Keep looping while all elements of target`

` ` `// don't become 0.`

` ` `while` `(1)`

` ` `{`

` ` `// To store count of zeroes in current`

` ` `// target array`

` ` `int` `zero_count = 0;`

` ` `int` `i; ` `// To find first odd element`

` ` `for` `(i=0; i<n; i++)`

` ` `{`

` ` `// If odd number found`

` ` `if` `(target[i] & 1)`

` ` `break` `;`

` ` `// If 0, then increment zero_count`

` ` `else` `if` `(target[i] == 0)`

` ` `zero_count++;`

` ` `}`

` ` `// All numbers are 0`

` ` `if` `(zero_count == n)`

` ` `return` `result;`

` ` `// All numbers are even`

` ` `if` `(i == n)`

` ` `{`

` ` `// Divide the whole array by 2`

` ` `// and increment result`

` ` `for` `(` `int` `j=0; j<n; j++)`

` ` `target[j] = target[j]/2;`

` ` `result++;`

` ` `}`

` ` `// Make all odd numbers even by subtracting`

` ` `// one and increment result.`

` ` `for` `(` `int` `j=i; j<n; j++)`

` ` `{`

` ` `if` `(target[j] & 1)`

` ` `{`

` ` `target[j]--;`

` ` `result++;`

` ` `}`

` ` `}`

` ` `}`

`}`

`/* Driver program to test above functions*/`

`int` `main()`

`{`

` ` `unsigned ` `int` `arr[] = {16, 16, 16};`

` ` `int` `n = ` `sizeof` `(arr)/` `sizeof` `(arr[0]);`

` ` `cout << ` `"Minimum number of steps required to "`

` ` `"get the given target array is "`

` ` `<< countMinOperations(arr, n);`

` ` `return` `0;`

`}`

Output :

Minimum number of steps required to get the given target array is 7