You are given two 0-indexed integer arrays nums1
and nums2
of equal length. Every second, for all indices 0 <= i < nums1.length
, value of nums1[i]
is incremented by nums2[i]
. After this is done, you can do the following operation:
0 <= i < nums1.length
and make nums1[i] = 0
.You are also given an integer x
.
Return the minimum time in which you can make the sum of all elements of nums1
to be less than or equal to x
, or -1
if this is not possible.
Example 1:
Input: nums1 = [1,2,3], nums2 = [1,2,3], x = 4 Output: 3 Explanation: For the 1st second, we apply the operation on i = 0. Therefore nums1 = [0,2+2,3+3] = [0,4,6]. For the 2nd second, we apply the operation on i = 1. Therefore nums1 = [0+1,0,6+3] = [1,0,9]. For the 3rd second, we apply the operation on i = 2. Therefore nums1 = [1+1,0+2,0] = [2,2,0]. Now sum of nums1 = 4. It can be shown that these operations are optimal, so we return 3.
Example 2:
Input: nums1 = [1,2,3], nums2 = [3,3,3], x = 4 Output: -1 Explanation: It can be shown that the sum of nums1 will always be greater than x, no matter which operations are performed.
Constraints:
1 <= nums1.length <= 103
1 <= nums1[i] <= 103
0 <= nums2[i] <= 103
nums1.length == nums2.length
0 <= x <= 106
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
The brute force strategy is like trying every single possible way to solve the problem. We will explore every combination of actions until we find the one that works within the given limitations.
Here's how the algorithm would work step-by-step:
def min_time_to_make_array_sum_at_most_x_brute_force(array, operations, target_sum):
min_time = float('inf')
for i in range(1 << len(operations)):
current_sum = sum(array)
current_time = 0
operations_performed = []
# Iterate through each operation to determine if its included in the current subset
for operation_index in range(len(operations)):
if (i >> operation_index) & 1:
operations_performed.append(operation_index)
# Apply chosen operations
for operation_index in operations_performed:
current_sum -= operations[operation_index][0]
current_time += operations[operation_index][1]
# Check if the array sum is at most x
if current_sum <= target_sum:
# Update min_time if necessary
min_time = min(min_time, current_time)
# Handle the case where no combination meets the target sum
if min_time == float('inf'):
return -1
else:
return min_time
This problem asks us to find the least amount of time to make the total of numbers in a collection less than a certain value. The key insight is to prioritize reducing larger values first, especially since each value decreases over time.
Here's how the algorithm would work step-by-step:
def min_time_to_make_array_sum_at_most_x(numbers, reduction_rates, target_sum):
number_count = len(numbers)
combined_array = list(zip(numbers, reduction_rates))
# Sort based on reduction rate to prioritize larger reductions.
combined_array.sort(key=lambda x: x[1], reverse=True)
minimum_time = float('inf')
for i in range(number_count + 1):
# Iterate through all possible group sizes.
current_numbers = []
current_reduction_rates = []
for j in range(i):
current_numbers.append(combined_array[j][0])
current_reduction_rates.append(combined_array[j][1])
low = 0
high = 10**18
best_time = float('inf')
while low <= high:
mid = (low + high) // 2
current_sum = 0
for k in range(number_count):
initial_number = numbers[k]
reduction_rate = reduction_rates[k]
current_sum += max(0, initial_number - reduction_rate * mid)
# Check if the target sum condition is met.
if current_sum <= target_sum:
best_time = mid
high = mid - 1
else:
low = mid + 1
minimum_time = min(minimum_time, best_time)
return minimum_time
Case | How to Handle |
---|---|
nums1 or nums2 is null or empty | Return 0 if the initial sum of nums1 is already <= x, otherwise return -1 because there's nothing to increment. |
n = 1 (single element in the arrays) | Calculate the minimum time required for the single element to be <= x, handling the case where nums2[0] is zero. |
Large n (maximum size input array) - performance concern | The solution must use an efficient algorithm like dynamic programming or a greedy approach with sorting to avoid exceeding time limits for large inputs. |
All elements in nums2 are zero | If the sum of nums1 is initially <= x, return 0; otherwise, return -1 since the values in nums1 cannot be changed. |
Elements in nums1 and nums2 are large positive numbers, leading to potential integer overflow during summation | Use appropriate data types (e.g., long long in C++, long in Java/Python) to prevent integer overflow during sum calculations. |
Some elements in nums2 are negative | Elements in nums2 can be negative, however, we should avoid choosing such an index since this is a minimization problem, in practice, we can filter out such cases at the start, and solve the problem for all the other indices. |
Initial sum of nums1 is already less than or equal to x | Return 0 immediately because no operations are required. |
It is not possible to make the sum of nums1 less than or equal to x | Return -1 when all elements in nums2 are non-negative and the initial sum of nums1 is greater than x; or there are negative elements in nums2 but even by repeatedly applying the negative increments, it is still not possible to bring the sum below or equal to x. |