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Sum of Mutated Array Closest to Target

Difficulty: Medium

Problem Description

Given an integer array arr and a target value target, return the integer value such that when we change all the integers larger than value in the given array to be equal to value, the sum of the array gets as close as possible (in absolute difference) to target. In case of a tie, return the minimum such integer. The answer is not necessarily a number from arr.

Key Insights

  • The problem requires finding a threshold value to minimize the absolute difference between the sum of the mutated array and the target.
  • The mutation involves capping values in the array based on the chosen threshold.
  • A binary search approach is effective to efficiently find the optimal threshold value.
  • The answer should be the minimum integer that achieves the closest sum in case of multiple candidates.

Space and Time Complexity

Time Complexity: O(n log m), where n is the length of the array and m is the maximum value in the array.
Space Complexity: O(1), since we are using a constant amount of space.

Solution

To solve the problem, we can utilize a binary search approach. The idea is to perform a binary search over the possible threshold values (from 0 to the maximum value in the array). For each candidate threshold value, we calculate the sum of the mutated array where all values greater than the threshold are capped. We then compare the absolute difference of this sum from the target to find the closest match. The process is repeated until we find the optimal threshold.

Code Solutions

def findBestValue(arr, target):
    left, right = 0, max(arr)
    best_value = 0
    min_diff = float('inf')

    while left <= right:
        mid = (left + right) // 2
        current_sum = sum(min(x, mid) for x in arr)

        diff = abs(current_sum - target)
        if diff < min_diff:
            min_diff = diff
            best_value = mid
        elif diff == min_diff:
            best_value = min(best_value, mid)

        if current_sum < target:
            left = mid + 1
        else:
            right = mid - 1

    return best_value
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