Asked in: AMAZON
Image of the Question
All Testcases Passed ✔
Solution
def findMinimumSum(power):
last = 0
current = 0
for i in power:
// ... rest of solution available after purchaseExplanation
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This problem involves transforming an array representing computational powers of servers into a non-decreasing sequence by applying increments to contiguous segments. The increments to these segments are represented by values x, and the goal is to minimize the total sum of these x values applied to achieve a non-decreasing final array.
To approach this problem, first understand the key constraints and desired outcome:
- The initial array of server powers may have decreases or dips; the goal is to fix these so that each server's power is at least as large as the previous one.
- You can increase powers only by adding increments x to contiguous segments (subarrays).
- After all increments, the powers must be non-decreasing.
- The objective is to minimize the sum of all increments applied to these segments.
This problem is a variant of incrementing subarrays to create a non-decreasing sequence with minimum total increment cost. Consider the following steps and directions to think about the problem:
1. **Understand the non-decreasing property:**
A non-decreasing array means for every i, power[i] ≤ power[i+1]. If at any index i, power[i] > power[i+1], you need to fix that by increasing the powers of servers starting from some index j ≥ i+1 so that the condition holds.
2. **Why contiguous segment increments?**
You cannot increment servers individually; increments must be applied to a continuous block of servers. This adds complexity because a change affects multiple servers at once. The key is to find which segments to increment and by how much, to efficiently fix decreases.
3. **Think about the minimal increments needed locally:**
When power[i] > power[i+1], you must increase the servers starting at i+1 or later so that power[i+1] ≥ power[i]. But if you increment just power[i+1], that's one segment of length 1; sometimes it might be better to increment a longer segment covering multiple servers to avoid multiple overlapping increments.
4. **Decompose the problem into layers of increments:**
Visualize the server powers as heights. Increasing contiguous segments can be thought of as stacking horizontal layers to fill gaps and ensure the non-decreasing order. The sum of increments corresponds to the total height added across these segments.
5. **Relate to differences between adjacent elements:**
Calculate the "difference array" or the amount by which each element must be increased relative to the previous one to ensure non-decreasing order. For each i,
- if power[i+1] < power[i], the difference needed is power[i] - power[i+1].
These differences point to where increments must happen and their magnitudes.
6. **Optimal increments relate to the maximum needed increments at each point:**
Since increments are applied to contiguous segments, overlapping increments can cover multiple decreases in one go.
The minimum sum of increments required corresponds to the sum of positive differences in the array. Why? Because each positive difference corresponds to an amount by which the later server(s) must be increased to catch up.
7. **Apply the concept of prefix increments:**
To correct all decreases, you can think of incrementing from left to right.
For example:
- Start at index 0, no increments needed.
- For index i > 0, if power[i] < power[i-1], increase power[i] by the difference (power[i-1] - power[i]).
- These increments are cumulative and correspond to segments starting at index i and extending to some later point.
8. **Minimizing the sum of increments translates to summing these necessary positive differences:**
The sum of increments is then the sum of all positive differences where power[i] < power[i-1]. This is because you can think of each needed increment as one contiguous segment increment starting at i and possibly extending rightward.
9. **Check the example:**
For power = [3, 4, 1, 6, 2]:
- Differences between consecutive servers: (4-3) = +1 (no increment needed), (1-4) = -3 (need to increase), (6-1) = +5 (no increment), (2-6) = -4 (need to increase)
- The increments needed are 3 (to fix 1 up to 4) and 4 (to fix 2 up to 6)
- Total increments = 3 + 4 = 7, matching the problem's example.
10. **Summarize the approach:**
- Iterate through the array from left to right.
- Whenever you find power[i] < power[i-1], calculate the difference needed to fix it.
- Sum all these positive differences to get the minimum total increment sum.
- This works because each negative slope requires an increment that can be applied as a contiguous segment starting at the position where the drop happens.
11. **Algorithm complexity:**
This approach requires a single pass through the array, making it efficient and suitable for large input sizes (up to 10^5).
12. **Edge cases:**
- Array already non-decreasing: no increments needed, sum is zero.
- All elements equal: no increments needed.
- Strictly decreasing array: increments needed for every pair, total increments sum to the total drop amount.
By thinking through the problem with these directions, you reduce a complex segment increment problem into a simpler task of calculating where and how much the sequence dips and summing those dips. This insight leads to an elegant, efficient solution.
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