LeetCode #1300 — MEDIUM

Sum of Mutated Array Closest to Target

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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.

Notice that the answer is not neccesarilly a number from arr.

Example 1:

Input: arr = [4,9,3], target = 10
Output: 3
Explanation: When using 3 arr converts to [3, 3, 3] which sums 9 and that's the optimal answer.

Example 2:

Input: arr = [2,3,5], target = 10
Output: 5

Example 3:

Input: arr = [60864,25176,27249,21296,20204], target = 56803
Output: 11361

Constraints:

1 <= arr.length <= 10⁴
1 <= arr[i], target <= 10⁵

Step 01

Brute Force Baseline

Problem summary: 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. Notice that the answer is not neccesarilly a number from arr.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search

Example 1

[4,9,3]
10

Example 2

[2,3,5]
10

Example 3

[60864,25176,27249,21296,20204]
56803

Step 02

Core Insight

What unlocks the optimal approach

If you draw a graph with the value on one axis and the absolute difference between the target and the array sum, what will you get?
That graph is uni-modal.
Use ternary search on that graph to find the best value.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1300: Sum of Mutated Array Closest to Target
class Solution {
    public int findBestValue(int[] arr, int target) {
        Arrays.sort(arr);
        int n = arr.length;
        int[] s = new int[n + 1];
        int mx = 0;
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + arr[i];
            mx = Math.max(mx, arr[i]);
        }
        int ans = 0, diff = 1 << 30;
        for (int value = 0; value <= mx; ++value) {
            int i = search(arr, value);
            int d = Math.abs(s[i] + (n - i) * value - target);
            if (diff > d) {
                diff = d;
                ans = value;
            }
        }
        return ans;
    }

    private int search(int[] arr, int x) {
        int left = 0, right = arr.length;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (arr[mid] > x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

// Accepted solution for LeetCode #1300: Sum of Mutated Array Closest to Target
func findBestValue(arr []int, target int) (ans int) {
	sort.Ints(arr)
	n := len(arr)
	s := make([]int, n+1)
	mx := slices.Max(arr)
	for i, x := range arr {
		s[i+1] = s[i] + x
	}
	diff := 1 << 30
	for value := 0; value <= mx; value++ {
		i := sort.SearchInts(arr, value+1)
		d := abs(s[i] + (n-i)*value - target)
		if diff > d {
			diff = d
			ans = value
		}
	}
	return
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #1300: Sum of Mutated Array Closest to Target
class Solution:
    def findBestValue(self, arr: List[int], target: int) -> int:
        arr.sort()
        s = list(accumulate(arr, initial=0))
        ans, diff = 0, inf
        for value in range(max(arr) + 1):
            i = bisect_right(arr, value)
            d = abs(s[i] + (len(arr) - i) * value - target)
            if diff > d:
                diff = d
                ans = value
        return ans

// Accepted solution for LeetCode #1300: Sum of Mutated Array Closest to Target
struct Solution;

impl Solution {
    fn find_best_value(mut arr: Vec<i32>, mut target: i32) -> i32 {
        arr.sort_unstable();
        let n = arr.len();
        let mut i = 0;
        while i < n && target > arr[i] * (n - i) as i32 {
            target -= arr[i];
            i += 1;
        }
        if i == n {
            return arr[n - 1];
        }
        let res = target / (n - i) as i32;
        if (res + 1) * (n - i) as i32 - target < target - res * (n - i) as i32 {
            res + 1
        } else {
            res
        }
    }
}

#[test]
fn test() {
    let arr = vec![4, 9, 3];
    let target = 10;
    let res = 3;
    assert_eq!(Solution::find_best_value(arr, target), res);
    let arr = vec![2, 3, 5];
    let target = 10;
    let res = 5;
    assert_eq!(Solution::find_best_value(arr, target), res);
    let arr = vec![60864, 25176, 27249, 21296, 20204];
    let target = 56803;
    let res = 11361;
    assert_eq!(Solution::find_best_value(arr, target), res);
}

// Accepted solution for LeetCode #1300: Sum of Mutated Array Closest to Target
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1300: Sum of Mutated Array Closest to Target
// class Solution {
//     public int findBestValue(int[] arr, int target) {
//         Arrays.sort(arr);
//         int n = arr.length;
//         int[] s = new int[n + 1];
//         int mx = 0;
//         for (int i = 0; i < n; ++i) {
//             s[i + 1] = s[i] + arr[i];
//             mx = Math.max(mx, arr[i]);
//         }
//         int ans = 0, diff = 1 << 30;
//         for (int value = 0; value <= mx; ++value) {
//             int i = search(arr, value);
//             int d = Math.abs(s[i] + (n - i) * value - target);
//             if (diff > d) {
//                 diff = d;
//                 ans = value;
//             }
//         }
//         return ans;
//     }
// 
//     private int search(int[] arr, int x) {
//         int left = 0, right = arr.length;
//         while (left < right) {
//             int mid = (left + right) >> 1;
//             if (arr[mid] > x) {
//                 right = mid;
//             } else {
//                 left = mid + 1;
//             }
//         }
//         return left;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.