Took me a while to realize that sorting is the key.
First, clarify the problem and identify that the variation at each index depends on the running min and max. Then, derive that sorting the array and adding elements in a specific order (e.g., alternating smallest and largest) minimizes the sum, and prove it with an exchange argument. Finally, implement the solution in O(n log n) time by sorting and computing the sum.
Pro tip: Mention that this problem is equivalent to minimizing the sum of ranges of prefixes, and that sorting is optimal because it groups similar values together, reducing the impact of new extremes. Also, note that the optimal order is to add elements in increasing order of their distance from the median, but a simpler alternating approach works.
Restate the problem in your own words: we need to reorder the array to minimize the sum over all prefixes of (max - min). Clarify that the variation at index i is the difference between the maximum and minimum of the first i+1 elements.
Note that the total sum depends only on the order in which elements are added. The first element contributes 0. Each subsequent element may increase the current range if it is outside the current min-max interval.
Argue that to minimize the sum, we should avoid introducing new extremes early. Sorting the array and then adding elements in an order that keeps the range small initially (e.g., start from the median and expand outward) minimizes the sum. Alternatively, prove that any order that is not sorted (or reverse sorted) can be improved by swapping adjacent out-of-order elements.
Use an exchange argument: if there is an inversion (a larger element before a smaller one), swapping them does not increase the total sum. Thus, the sorted order (either ascending or descending) is optimal. However, note that ascending and descending give the same sum? Actually, ascending gives sum of (a[i] - a[0]) for i>=1, which is sum of a[i] - n*a[0]. Descending gives sum of (a[n-1] - a[i]) for i<=n-2, which is n*a[n-1] - sum of a[i]. These are not equal in general. So we need to choose the better of the two? Wait, is that true? Let's test: array [1,2,3]. Ascending: prefixes: [1] range 0, [1,2] range 1, [1,2,3] range 2 => sum=3. Descending: [3] 0, [3,2] range 1, [3,2,1] range 2 => sum=3. Same. For [1,5,10]: ascending: 0 +4 +9 =13. Descending: 0 +5 +9 =14. So ascending is better. So sorted ascending is optimal? But wait, what about order [5,1,10]? prefixes: 0,4,9 =>13. Same as ascending. So many orders give same? Actually, the sum is sum_{i=1}^{n-1} (max_i - min_i). If we sort ascending, max_i = a[i], min_i = a[0], so sum = sum_{i=1}^{n-1} (a[i] - a[0]) = sum_{i=1}^{n-1} a[i] - (n-1)a[0]. If we sort descending, max_i = a[n-1], min_i = a[i], sum = sum_{i=1}^{n-1} (a[n-1] - a[i]) = (n-1)a[n-1] - sum_{i=1}^{n-1} a[i]. Which is smaller? Depends on the array. But is there an even better order? Consider [1,10,5]: prefixes: 0,9,9 =>18. Worse. So sorted ascending seems good. But is it always optimal? Let's test [1,2,100,101]. Ascending: 0,1,99,100 =>200. Descending: 0,99,100,100 =>299. So ascending better. What about [1,50,51,100]? Ascending: 0,49,50,99 =>198. Descending: 0,50,99,99 =>248. So ascending better. But is there a case where descending is better? Suppose array [1,2,3,100]. Ascending: 0,1,2,99 =>102. Descending: 0,98,99,99 =>296. So ascending better. It seems ascending is always better? Let's check [1,100,101,102]. Ascending: 0,99,100,101 =>300. Descending: 0,1,2,101 =>104. Oh! Descending is better here. So the optimal order depends on the array. Actually, the optimal order is to sort and then add elements in an order that minimizes the sum. The known solution is to sort the array and then the minimum sum is sum_{i=1}^{n-1} (a[i] - a[i-1]) * i? No, that's for something else. Wait, there is a known problem: "Minimize sum of max-min over all prefixes" and the optimal is to sort and then the sum is sum_{i=1}^{n-1} (a[i] - a[i-1]) * (n-i)? Let's derive: If we sort ascending, the sum is sum_{i=1}^{n-1} (a[i] - a[0]) = sum_{i=1}^{n-1} a[i] - (n-1)a[0]. If we sort descending, sum = (n-1)a[n-1] - sum_{i=1}^{n-1} a[i]. But we can also interleave. Actually, the optimal order is to sort and then add elements in increasing order of their distance from the median? Not exactly. There is a known result: the minimum sum is sum_{i=1}^{n-1} (a[i] - a[i-1]) * min(i, n-i)? Let's test with [1,100,101,102] sorted: differences: 99,1,1. min(i, n-i) for i=1: min(1,3)=1 =>99; i=2: min(2,2)=2 =>2; i=3: min(3,1)=1 =>1; total=102. But we found descending gives 104, ascending gives 300. So 102 is better? Can we achieve 102? Let's try order: start with 100? [100] 0, then 101: range 1, then 1: range 100, then 102: range 101 => total 202. Not 102. Try [100,101,102,1]: 0,1,2,101 =>104. Try [101,100,102,1]: 0,1,2,101 =>104. Try [100,1,101,102]: 0,99,100,101 =>300. So 102 seems not achievable. So the formula is not that. Actually, the known solution for this problem (often called "minimize sum of ranges of prefixes") is to sort the array and then the minimum sum is sum_{i=1}^{n-1} (a[i] - a[i-1]) * i? Let's test: [1,100,101,102]: diffs: 99,1,1. i=1:99*1=99; i=2:1*2=2; i=3:1*3=3; total=104. That matches descending! And for [1,2,3]: diffs:1,1. i=1:1*1=1; i=2:1*2=2; total=3. Matches. For [1,5,10]: diffs:4,5. i=1:4*1=4; i=2:5*2=10; total=14. But ascending gave 13. So 14 is not minimal. So that formula gives 14, but we found 13. So the formula is not correct. Actually, the correct minimum sum is sum_{i=1}^{n-1} (a[i] - a[i-1]) * min(i, n-i)? For [1,5,10]: diffs:4,5. i=1: min(1,2)=1 =>4; i=2: min(2,1)=1 =>5; total=9. But we found 13. So no. Let's think differently. The problem is: given a set of numbers, we want to order them to minimize sum_{k=1}^n (max_{i<=k} a_i - min_{i<=k} a_i). This is equivalent to: we start with an empty set, and we add elements one by one. The cost of adding an element is the increase in the range if the new element is outside
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