Maximize array sum after K negations


http://www.geeksforgeeks.org/maximize-array-sun-after-k-negation-operations/
Given an array of size n and a number k. We must modify array K number of times. Here modify array means in each operation we can replace any array element arr[i] by -arr[i]. We need to perform this operation in such a way that after K operations, sum of array must be maximum?
we just have to replace the minimum element arr[i] in array by -arr[i] for current operation. In this way we can make sum of array maximum after K operations. Once interesting case is, once minimum element becomes 0, we don’t need to make any more changes.
Time Complexity : O(k*n)
// This function does k operations on array
// in a way that maximize the array sum.
// index --> stores the index of current minimum
//           element for j'th operation
int maximumSum(int arr[], int n, int k)
{
    // Modify array K number of times
    for (int i=1; i<=k; i++)
    {
        int min = INT_MAX;
        int index = -1;
        // Find minimum element in array for
        // current operation and modify it
        // i.e; arr[j] --> -arr[j]
        for (int j=0; j<n; j++)
        {
            if (arr[j] < min)
            {
                min = arr[j];
                index = j;
            }
        }
        // this the condition if we find 0 as
        // minimum element, so it will useless to
        // replace 0 by -(0) for remaining operations
        if (min == 0)
            break;
        // Modify element of array
        arr[index] = -arr[index];
    }
    // Calculate sum of array
    int sum = 0;
    for (int i=0; i<n; i++)
        sum += arr[i];
    return sum;
}

http://www.geeksforgeeks.org/maximize-array-sum-k-negations-set-2/
In this post an optimized solution is implemented that uses a priority queue (or binary heap) to find minimum element quickly.
Note that this optimized solution can be implemented in O(n + kLogn) time as we can create a priority queue (or binary heap) in O(n) time.

    public static int maxSum(int[] a, int k)
    {
        // Create a priority queue and insert all array elements
        // int
        PriorityQueue<Integer> pq = new PriorityQueue<>();
        for (int x : a)
            pq.add(x);
        // Do k negations by removing a minimum element k times
        while (k-- > 0)
        {
            // Retrieve and remove min element
            int temp = pq.poll();
            // Modify the minimum element and add back
            // to priority queue
            temp *= -1;
            pq.add(temp);
        }
        // Compute sum of all elements in priority queue.
        int sum = 0;
        for (int x : pq)
            sum += x;
        return sum;
    }



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