How do you solve Sum of Mutated Array Closest to Target on LeetCode?

Sum of Mutated Array Closest to Target (LeetCode #1300) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Understanding how mutation affects the sum is crucial.

What is the time complexity of Sum of Mutated Array Closest to Target?

The optimal solution for Sum of Mutated Array Closest to Target runs in O(n log n) time and uses O(1) space. The sorting step takes O(n log n), and the binary search runs in O(log m) where m is the maximum value in arr, leading to an overall complexity of O(n log n).

What is the brute force solution for Sum of Mutated Array Closest to Target?

The brute force approach for Sum of Mutated Array Closest to Target is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible value that could be the limit for mutation. For each value, we calculate the sum of the mutated array and see how close it gets to the target. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Sum of Mutated Array Closest to Target?

Sum of Mutated Array Closest to Target uses the following concepts: Array, Binary Search, Sorting. The recommended patterns to study are: Binary Search, Sorting.

What are the interview tips for Sum of Mutated Array Closest to Target?

Always clarify the problem requirements before jumping into coding. Think about edge cases, such as when all elements are less than or equal to the target. Explain your thought process and reasoning as you code.

What are common mistakes when solving Sum of Mutated Array Closest to Target?

Not considering values outside the array range. Forgetting to check both low and low-1 after binary search.

#1300

Sum of Mutated Array Closest to Target

Medium

ArrayBinary SearchSortingBinary SearchSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

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Intuition

Time O(n log n)Space O(1)

Instead of checking every possible value, we can use binary search on the possible values. The idea is to find the value that minimizes the absolute difference between the mutated sum and the target efficiently.

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Algorithm

5 steps

1Step 1: Sort the array to allow binary search.
2Step 2: Use binary search to find the best value. Set low to 0 and high to the maximum value in arr.
3Step 3: For each mid value, calculate the mutated sum and check if it's less than or greater than the target.
4Step 4: Adjust the search range based on the comparison with the target.
5Step 5: After the search, check the closest values to determine the best one.

solution.py21 lines

1# Full working Python code
2
3def findBestValue(arr, target):
4    arr.sort()
5    low, high = 0, max(arr)
6    
7    while low < high:
8        mid = (low + high) // 2
9        mutated_sum = sum(min(x, mid) for x in arr)
10        if mutated_sum < target:
11            low = mid + 1
12        else:
13            high = mid
14    
15    # Check low and low-1 for the best value
16    mutated_sum_low = sum(min(x, low) for x in arr)
17    mutated_sum_low_minus_1 = sum(min(x, low - 1) for x in arr)
18    return low if abs(mutated_sum_low - target) < abs(mutated_sum_low_minus_1 - target) else low - 1
19
20# Example usage
21print(findBestValue([4,9,3], 10))

ℹ

Complexity note: The sorting step takes O(n log n), and the binary search runs in O(log m) where m is the maximum value in arr, leading to an overall complexity of O(n log n).

1Understanding how mutation affects the sum is crucial.
2Binary search can significantly reduce the number of calculations needed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.