1300. Sum of Mutated Array Closest to Target

Problem Description

Given an integer array arr and a target value target, the goal is to find an integer value such that when we replace all the integers in the array that are greater than value with value itself, the sum of the modified array is as close as possible to the target. In terms of absolute difference, we're looking to minimize the discrepancy between the sum of the modified array and the target. If there are multiple value that achieve the same minimum absolute difference, the problem asks us to return the smallest such integer. Also, it's important to note that the integer value we are looking for does not necessarily have to be an element already present in the array arr.

Intuition

The intuition behind the solution involves understanding that as we increase the value used to replace elements in the array, the sum of the array will increase in a step-wise fashion until it reaches a point where increasing value further won't change the sum since there won't be any larger elements left to replace. To find the integer value that brings the sum closest to the target, we can follow these steps:

1. Sort the Array: Sorting arr allows us to efficiently work through the array to find the point where changing greater elements to a certain value will affect the sum.

2. Accumulate the Sum: We use a prefix sum array s to keep a running total, which helps us quickly calculate the sum of the array when we hypothetically replace elements greater than the current value.

3. Iterate and Simulate: Iterate through potential values from 0 to the max(arr) and for each value, we simulate what the sum would be if all the elements larger than this value were replaced by it. We use binary search (bisect_right) to efficiently find the index where the elements start to be larger than the current value being considered.

4. Compare Differences: For each value, we compare the absolute difference between the target and the simulated sum. We keep track of the smallest difference and the associated value.

5. Return the Best Answer: Once we have considered all relevant values, we return the ans that corresponds to the smallest difference.

By iterating through the array and keeping track of the differences, we can find the best value that minimizes the absolute difference to the target. The choice of binary search for finding the right index is crucial for efficiency, and sorting the array beforehand makes such an approach possible.

Learn more about Binary Search and Sorting patterns.

Solution Approach

The solution approach implements the following steps in Python code, each relying on algorithms and data structures that ensure efficient execution:

1. Sort the Array: We begin by sorting the array arr in increasing order. This is done with arr.sort(), which uses an efficient sorting algorithm such as Timsort (Python's default sorting algorithm) that has a complexity of O(n log n) where n is the number of elements in arr.

2. Accumulate the Sum: We then create a prefix sum array s using the accumulate function with an initial value of 0 (initial=0). The prefix sum array holds the cumulative sum of elements, and by including initial=0, we ensure that s[0] is 0, representing that no elements have been summed yet. This array allows us to quickly calculate the sum of all elements up to a certain point.

3. Initialize Variables: Two variables are initialized: ans, which will hold the final answer, and diff, which is set to infinity inf as a placeholder for the smallest difference found so far.

4. Iteration and Value Simulation: We iterate through possible values for replacement value in the range from 0 to max(arr)+1 (inclusive). For each value, the code uses binary search with bisect_right to find the index i where elements in arr are greater than value. As arr is sorted, this tells us how many elements will be replaced with value.

5. Calculating the Difference: With the index i, the algorithm calculates the sum of the array up to i (the sum of elements not changed) and adds (len(arr) - i) * value (the sum if all larger elements are replaced by value). This gives us the simulated sum of the array after replacements. The difference d between the simulated sum and target is computed in absolute terms.

6. Track the Best Value: If the current difference d is smaller than the smallest difference found so far diff, we update diff to d and ans to the current value being considered. If two values result in the same diff, since we iterate in increasing order, ans will already hold the smaller value.

7. Return the Result: After the loop completes, the smallest ans is returned, representing the value that when used to replace elements in arr results in a sum closest to target.

The key to this solution is the combination of sorting, prefix sums, binary search, and careful tracking of the smallest absolute difference, which altogether yields an efficient approach to solving the problem.

Here is the solution code again for reference:

1class Solution:
2    def findBestValue(self, arr: List[int], target: int) -> int:
3        arr.sort()
4        s = list(accumulate(arr, initial=0))
5        ans, diff = 0, inf
6        for value in range(max(arr) + 1):
7            i = bisect_right(arr, value)
8            d = abs(s[i] + (len(arr) - i) * value - target)
9            if diff > d:
10                diff = d
11                ans = value
12        return ans

Example Walkthrough

Let's consider a small example to illustrate the solution approach. Suppose we have the following input:

• arr: [4, 9, 3]
• target: 10

Now let's walk through the solution step by step:

1. Sort the Array:

   • After sorting, arr becomes [3, 4, 9].

2. Accumulate the Sum:

   • We create a prefix sum array s which would be [0, 3, 7, 16]. Here, s[0] is 0, and other elements represent the sum up to that index in arr.

3. Initialize Variables:

   • We set ans to 0 and diff to infinity (inf) initially.

4. Iteration and Value Simulation:

   • We iterate value from 0 to max(arr) + 1, which is 10 in this case.
   • For each value, say 2, we use binary search to find the first index i in arr where the element is greater than 2.
   • In this case, for value equals 2, i is 1 because arr[1] is the first element greater than 2.

5. Calculating the Difference:

   • For value equals 2, the sum of the array if all elements larger than 2 are replaced will be s[i] + (len(arr) - i) * value. Here, it will be 0 + (3 - 1) * 2 which equals 4.
   • The difference d between this simulated sum (4) and the target (10) is 6. We compare this to diff and since 6 is less than inf, we update diff to 6 and ans to 2.

6. Track the Best Value:

   • We continue this process for each value up to 10. For each value, we find a new d and update ans and diff accordingly if we find a smaller d.

7. Return the Result:

   • Once we go through all possible values, let's imagine we found the best value to be 3 which gives us the least diff of 1 (the sum of the array becomes 3 + 3 + 3 = 9 which is closest to the target 10).
   • Thus, ans is set to 3, and that will be the result returned.

Through the above steps, the findBestValue function would eventually return 3 as the smallest integer we can replace the elements greater than value with to make the sum of the array closest to the target 10.

Solution Implementation

Time and Space Complexity

The given Python code aims to find an integer value that, when used to replace larger elements in an array, results in the sum of the array being as close as possible to a given target. The time complexity and space complexity of the code are analyzed as follows:

Time Complexity

• The arr.sort() operation has a time complexity of O(n log n), where n is the length of the array.
• The list(accumulate(arr, initial=0)) operation constructs a prefix sum array in O(n) time.
• The for loop runs for a maximum of max(arr) + 1 iterations. In the worst case, this is also O(n) since max(arr) is less than or equal to the sum of all n elements.
• Inside the loop, the bisect_right function is called, which operates in O(log n) time.
• The remaining operations within the loop (calculating the difference d and comparing/updating diff and ans) are all constant time operations, i.e., O(1).

Combining these, the overall time complexity is dominated by the sorting operation and the for loop that includes the binary search. Therefore, the time complexity is O(n log n) + O(n) * O(log n), which simplifies to O(n log n) because both terms have log n and the larger factor n log n dominates.

Space Complexity

• The additional space used by the algorithm includes the space for the prefix sum array s which is O(n) and a constant amount of space for variables like ans, diff, i, d, and value.

Therefore, the overall space complexity of the code is O(n) due to the use of the prefix sum array.

Learn more about how to find time and space complexity quickly using problem constraints.