2702. Minimum Operations to Make Numbers Non-positive - Detailed Explanation

Problem Statement

Given a 0-indexed array of positive integers nums and two positive integers x and y (with y < x), you can perform the following operation on the array: choose an index i and decrease nums[i] by x while decreasing every other element (all indices except i) by y. Your goal is to make all the numbers in nums non-positive (i.e. ≤ 0) using the minimum number of such operations. You need to return this minimum number of operations.

• Constraints:

  • 1 <= nums.length <= 100,000
  • 1 <= nums[i] <= 10^9 (initial values are positive)
  • 1 <= y < x <= 10^9 (note: y is strictly less than x)

• Example 1:

  • Input: nums = [3, 4, 1, 7, 6], x = 4, y = 2
  • Output: 3
  • Explanation: One optimal sequence of operations is:
    1. Choose index 3 (value 7). Subtract x=4 from nums[3] and y=2 from all other indices.
       Array becomes [1, 2, -1, 3, 4].
    2. Choose index 3 again (now value 3). Subtract 4 from it and 2 from others.
       Array becomes [-1, 0, -3, -1, 2].
    3. Choose index 4 (value 2). Subtract 4 from it and 2 from others.
       Array becomes [-3, -2, -5, -3, -2].
       Now all numbers are ≤ 0 (non-positive). The minimum operations required is 3.

• Example 2:

  • Input: nums = [1, 2, 1], x = 2, y = 1
  • Output: 1
  • Explanation: Perform one operation choosing index 1 (value 2). Subtract x=2 from nums[1] and y=1 from the others. The array becomes [0, 0, 0], and all numbers are now non-positive. So the answer is 1.

Hints

• Hint 1: Think about how each operation affects all the numbers. One number (the chosen index) gets decremented by a larger amount x, while every other number gets decremented by y. How would you decide which index to target in each operation?

• Hint 2: Since x > y, targeting a number gives it an extra reduction of (x - y) (on top of the y that every number gets). It might be wise to target the largest remaining number, because smaller numbers might be taken care of by the smaller decrements over multiple operations.

• Hint 3: Consider this: if you could achieve the goal in t operations, would it be possible in t+1 operations as well? (If yes, this suggests a monotonic relationship that could allow techniques like binary search on the answer.)

• Hint 4: Try to figure out, for a given number of operations t, how to check if it’s enough to make all numbers ≤ 0. For each number in nums, how much total decrement will it receive after t operations depending on whether it’s targeted or not?

Approaches

Brute Force Approach (Naive Simulation)

Idea: A straightforward (but inefficient) way is to simulate the process of operations. We could repeatedly perform operations until all numbers become ≤ 0. In each operation, we need to decide which index to target. A natural greedy choice is to always target the currently largest positive number, since it requires the most reduction. By targeting the largest number, we give it a big decrement of x while still decrementing others by y. This simulation would proceed as follows:

1. Find the largest positive number in the array (the one that is farthest from being non-positive).
2. Perform an operation on that index: subtract x from it, and subtract y from every other element.
3. Repeat this process (recomputing the largest remaining positive number each time) until all numbers become ≤ 0.

Using the example [3, 4, 1, 7, 6] with x=4, y=2:

• Operation 1 targets the largest value 7 (index 3). After subtracting 4 from 7 and 2 from others, the array becomes [1, 2, -1, 3, 4].
• Operation 2 then targets the new largest value 4 (now at index 4). After the operation, the array becomes [-1, 0, -3, -1, 2].
• Operation 3 targets the remaining positive value 2 (index 4 again). The array becomes [-3, -2, -5, -3, -2], and we are done in 3 operations.

Why this works (greedily targeting the largest): Each operation gives every number a decrease of y, but the targeted number gets an additional (x - y) decrease. If we don’t target a large number, it only goes down by y that round, which may not be enough if it’s significantly large. By always targeting the largest, we ensure we’re efficiently using the big decrement x where it’s most needed.

Drawback: While the greedy simulation seems logical, implementing it naively is extremely slow for the given constraints. Each operation requires scanning or maintaining the maximum, and in the worst case the number of operations needed could be very large. For example, if x is only slightly larger than y (e.g. x = 2, y = 1), a number as large as 10^9 might need nearly 10^9 operations to reduce to 0 by increments of (x-y)=1. Simulating billions of steps is infeasible. The time complexity of this brute force approach is essentially O(n * O), where O is the number of operations required. In the worst case, this could be on the order of 10^14 operations (for n ≈ 10^5 and O ≈ 10^9), which is completely impractical. Thus, we need a more efficient strategy.

Optimized Approach (Mathematical Reasoning + Binary Search)

Idea: Instead of simulating every operation, we can determine mathematically whether a given number of operations t is sufficient to make all numbers non-positive. Then, using the fact that if t operations is possible, any larger number of operations is also possible (more operations can only help or be wasted), we can use binary search to find the minimum t that works.

Key Observation: If it’s possible to finish the job in t operations, then t+1 (or any larger number) operations would also be enough (you could always perform an extra "wasted" operation that doesn’t hurt the condition). This means the property "can all numbers be ≤ 0 in at most t operations" is monotonic – once it becomes true at some t, it remains true for all larger values. Such monotonic scenarios are perfect for binary search.

Feasibility Check: How do we check if a given number of operations t is sufficient? We need to reason about how much each number can be reduced in t operations at best:

• In each operation, every number is reduced by y (either via the global effect or because if it’s targeted, we can imagine it still got the y reduction plus extra).

• If a particular index is targeted in some operation, it gets an additional (x - y) reduction that round (because it got x instead of just y).

• Over t operations, if a number at index i is targeted k times (and thus not targeted the other t-k times), its total reduction will be k*x + (t-k)*y. This simplifies to:
  Total reduction for element i = t*y + k*(x - y), where 0 <= k <= t.
  Here, t*y is the reduction it would get if it were never targeted (just losing y in all t operations), and k*(x-y) is the extra reduction from the k times it was targeted.

• In the best-case scenario for making a specific number as low as possible, you would target that number as many times as needed (up to t). But you cannot target one number more than t times obviously, and you have to consider all numbers collectively.

For a single element with initial value v:

• If you do not target it at all in t operations, its value will become v - t*y after all operations (since each operation it only gets the y decrement). To have v become ≤ 0 without targeting it, we need v - t*y <= 0, i.e. v <= t * y. If this condition holds, that number will be non-positive just from the global decrements alone.
• If v > t * y, then not targeting it t times is not enough — it would still be positive. In that case, it must be targeted some number of times to get extra (x-y) hits. Suppose it needs to be targeted k times out of t. We need:
  v - (t*y + k*(x - y)) <= 0.
  Rearranged, this is v <= t*y + k*(x - y). We already know v > t*y, so subtract that:
  v - t*y <= k*(x - y).
  This gives a minimum required k:
  k >= (v - t*y) / (x - y).
  Since k must be an integer, the smallest integer k that satisfies this is k = ceil((v - t*y) / (x - y)). This is the minimum number of times we must target that particular element within t operations to bring it to ≤ 0.

Now, given a candidate number of operations t, we can determine for each element nums[i] how many times it needs to be targeted (at minimum) if we have t operations total:

• If nums[i] <= t * y, then this element doesn’t need to be targeted at all (the baseline reduction of t*y is enough or more than enough to make it non-positive).
• If nums[i] > t * y, then calculate needed_i = ceil((nums[i] - t*y) / (x - y)). This is how many operations must directly target i.

After computing this for every element, let total_needed = needed_1 + needed_2 + ... + needed_n (summing only where needed). This total_needed is the total number of targeting operations required to handle all elements given that we plan to do t operations in total. For t operations to be sufficient, we must have total_needed <= t (because we only have t operations to allocate, and we can’t target more times than the number of operations available).

In summary, t operations is enough if and only if:
[ \sum_{\text{for each } i \text{ where } nums[i] > ty} \left\lceil \frac{nums[i] - ty}{,x - y,} \right\rceil ; \leq ; t. ]
If this holds, there is a way to schedule the operations (which index to target in each round) such that all numbers will be ≤ 0 after at most t operations.

Using this check, we can apply binary search on t:

1. Search space: The minimum t could be 0 (in theory, if all numbers are already ≤ 0, though not in our input constraints) and the maximum t needed could be as high as the maximum initial value in nums (worst case scenario where x-y is 1 and we effectively reduce one chosen number by 1 each time). So we can set low = 0 and high = max(nums) as initial bounds for binary search on the answer.
2. Binary search loop: For a mid-value m = (low + high)//2, use the above feasibility check to see if m operations can suffice.
   • If m operations can make all numbers non-positive, then we try to see if we can do even better (fewer operations), so set high = m.
   • If m operations cannot achieve the goal, it means we need more operations, so set low = m + 1.
3. Continue narrowing the range until low >= high. The value at low (or high, they converge) will be the minimum operations required.

This approach effectively finds the smallest t for which the condition is true. The check for a given t runs in O(n) time (just a single pass through the array to sum up needed targets for that t), and the binary search runs in O(log M) time where M is roughly max(nums). Since max(nums) can be up to 10^9, log2(10^9) is about 30. So overall this approach runs in about O(n * log(max(nums))), which for n = 100k is on the order of a few million operations – very efficient.

Detailed Reasoning with Example:
Let's walk through the check for a guess t on a small example to clarify. Take nums = [3, 4, 7], x = 5, y = 2. Suppose we want to check if t = 2 operations are enough.

• t * y = 2*2 = 4. Reduce each number by 4 as a baseline (this is what 2 global decrements would do if a number was never targeted):
  • For 3: after baseline 4 reduction, it would be 3 - 4 = -1 (already ≤ 0). So it needs 0 direct targets.
  • For 4: after baseline it’s 4 - 4 = 0 (exactly 0, which counts as non-positive). Needs 0 direct targets.
  • For 7: after baseline it’s 7 - 4 = 3 (still positive). It needs extra help. How many targets? We calculate (7 - 4) / (5 - 2) = 3 / 3 = 1 exactly (so ceil is 1). It needs at least 1 direct targeting.
  • Sum of needed targets = 0 + 0 + 1 = 1. Since 1 ≤ t (which is 2), it means in 2 operations we can allocate 1 of them to targeting the 7 once (and that would suffice to bring 7 down to 0 or below) and the other operation can target something else or even target nothing in particular (as the other numbers are already fine with baseline). So 2 operations is feasible.
• We would then try lower t to see if maybe 1 operation works:
  • t = 1: baseline reduction = 2 for each number.
    • 3 -> after baseline 2, value 1 (needs help: (3-2)/(5-2) = 1/3 = 0.34 → ceil = 1 target).
    • 4 -> after baseline 2, value 2 (needs help: (4-2)/3 = 2/3 = 0.67 → ceil = 1 target).
    • 7 -> after baseline 2, value 5 (needs help: (7-2)/3 = 5/3 = 1.67 → ceil = 2 targets).
    • Total needed = 1+1+2 = 4 targets, but we only have t=1 operation to allocate — impossible. So 1 operation is not enough.
  • This would tell us that the answer lies above 1. Our binary search would then check the range [2, ...] etc., eventually confirming 2 as the answer for this example.

Complexity Analysis: This optimized approach runs in O(n log C) time, where n is the length of nums and C is the maximum value in nums. The binary search will run at most ~30 iterations (since C ≤ 1e9), and each iteration does an O(n) pass. For n = 100,000, this is roughly 30 * 100k ≈ 3 million operations in the worst case, which is very fast. The space complexity is O(1) extra (we only use a few counters and variables), aside from the input array.

Code Implementations

Below are implementations of the optimized approach in both Python and Java. Each includes a simple main/main() method to demonstrate usage with the example cases.

Python Implementation