Tuesday, April 26, 2016

Count minimum steps to get the given desired array - GeeksforGeeks


Count minimum steps to get the given desired array - GeeksforGeeks
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.

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.
// Returns count of minimum operations to covert 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++;
           }
        }
    }
}

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