Question

I found this question on an online forum: Really interested on how it can be solved:

Given an array A of positive integers. Convert it to a sorted array with minimum cost. The only valid operation are:
1) Decrement with cost = 1
2) Delete an element completely from the array with cost = value of element

This is an interview question asked for a tech company

Answer 1

NOTE : The original answer has been replaced with one in which I have a lot more confidence (and I can explain it, too). Both answers produced the same results on my set of test cases.

You can solve this problem using a dynamic programming approach. The key observation is that it never makes sense to decrement a number to a value not found in the original array. (Informal proof: suppose that you decremented a number O1 to a value X that is not in the original sequence in order to avoid removing a number O2 > X from the result sequence. Then you can decrement O1 to O2 instead, and reduce the cost by O2-X).

Now the solution becomes easy to understand: it is a DP in two dimensions. If we sort the elements of the distinct elements of the original sequence d into a sorted array s, the length of d becomes the first dimension of the DP; the length of s becomes the second dimension.

We declare dp[d.Length,s.Length]. The value of dp[i,j] is the cost of solving subproblem d[0 to i] while keeping the last element of the solution under s[j]. Note: this cost includes the cost of eliminating d[i] if it is less than s[j].

The first row dp[0,j] is computed as the cost of trimming d[0] to s[j], or zero if d[0] < s[j]. The value of dp[i,j] next row is calculated as the minimum of dp[i-1, 0 to j] + trim, where trim is the cost of trimming d[i] to s[j], or d[i] if it needs to be eliminated because s[j] is bigger than d[i].

The answer is calculated as the minimum of the last row dp[d.Length-1, 0 to s.Length].

Here is an implementation in C#:

static int Cost(int[] d) {
    var s = d.Distinct().OrderBy(v => v).ToArray();
    var dp = new int[d.Length,s.Length];
    for (var j = 0 ; j != s.Length ; j++) {
        dp[0, j] = Math.Max(d[0] - s[j], 0);
    }
    for (var i = 1; i != d.Length; i++) {
        for (var j = 0 ; j != s.Length ; j++) {
            dp[i, j] = int.MaxValue;
            var trim = d[i] - s[j];
            if (trim < 0) {
                trim = d[i];
            }
            dp[i, j] = int.MaxValue;
            for (var k = j ; k >= 0 ; k--) {
                dp[i, j] = Math.Min(dp[i, j], dp[i - 1, k] + trim);
            }
        }
    }
    var best = int.MaxValue;
    for (var j = 0 ; j != s.Length ; j++) {
        best = Math.Min(best, dp[d.Length - 1, j]);
    }
    return best;
}

This direct implementation has space complexity of O(N^2). You can reduce it to O(N) by observing that only two last rows are used at the same time.