LeetCode #3281 — MEDIUM

Maximize Score of Numbers in Ranges

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

The Problem

Problem Statement

You are given an array of integers start and an integer d, representing n intervals [start[i], start[i] + d].

You are asked to choose n integers where the i^th integer must belong to the i^th interval. The score of the chosen integers is defined as the minimum absolute difference between any two integers that have been chosen.

Return the maximum possible score of the chosen integers.

Example 1:

Input: start = [6,0,3], d = 2

Output: 4

Explanation:

The maximum possible score can be obtained by choosing integers: 8, 0, and 4. The score of these chosen integers is min(|8 - 0|, |8 - 4|, |0 - 4|) which equals 4.

Example 2:

Input: start = [2,6,13,13], d = 5

Output: 5

Explanation:

The maximum possible score can be obtained by choosing integers: 2, 7, 13, and 18. The score of these chosen integers is min(|2 - 7|, |2 - 13|, |2 - 18|, |7 - 13|, |7 - 18|, |13 - 18|) which equals 5.

Constraints:

2 <= start.length <= 10⁵
0 <= start[i] <= 10⁹
0 <= d <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given an array of integers start and an integer d, representing n intervals [start[i], start[i] + d]. You are asked to choose n integers where the ith integer must belong to the ith interval. The score of the chosen integers is defined as the minimum absolute difference between any two integers that have been chosen. Return the maximum possible score of the chosen integers.

Baseline thinking

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

Pattern signal: Array · Binary Search · Greedy

Example 1

[6,0,3]
2

Example 2

[2,6,13,13]
5

Core Insight

What unlocks the optimal approach

Can we use binary search here?
Suppose that the answer is <code>x</code>. We can find a valid configuration of integers by sorting <code>start</code>, the first integer should be <code>start[0]</code>, then each subsequent integer should be the smallest one in <code>[start[i], start[i] + d]</code> that is greater than <code>last_chosen_value + x</code>.
Binary search over <code>x</code>

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 #3281: Maximize Score of Numbers in Ranges
class Solution {
    private int[] start;
    private int d;

    public int maxPossibleScore(int[] start, int d) {
        Arrays.sort(start);
        this.start = start;
        this.d = d;
        int n = start.length;
        int l = 0, r = start[n - 1] + d - start[0];
        while (l < r) {
            int mid = (l + r + 1) >>> 1;
            if (check(mid)) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        return l;
    }

    private boolean check(int mi) {
        long last = Long.MIN_VALUE;
        for (int st : start) {
            if (last + mi > st + d) {
                return false;
            }
            last = Math.max(st, last + mi);
        }
        return true;
    }
}

// Accepted solution for LeetCode #3281: Maximize Score of Numbers in Ranges
func maxPossibleScore(start []int, d int) int {
	check := func(mi int) bool {
		last := math.MinInt64
		for _, st := range start {
			if last+mi > st+d {
				return false
			}
			last = max(st, last+mi)
		}
		return true
	}
	sort.Ints(start)
	l, r := 0, start[len(start)-1]+d-start[0]
	for l < r {
		mid := (l + r + 1) >> 1
		if check(mid) {
			l = mid
		} else {
			r = mid - 1
		}
	}
	return l
}

# Accepted solution for LeetCode #3281: Maximize Score of Numbers in Ranges
class Solution:
    def maxPossibleScore(self, start: List[int], d: int) -> int:
        def check(mi: int) -> bool:
            last = -inf
            for st in start:
                if last + mi > st + d:
                    return False
                last = max(st, last + mi)
            return True

        start.sort()
        l, r = 0, start[-1] + d - start[0]
        while l < r:
            mid = (l + r + 1) >> 1
            if check(mid):
                l = mid
            else:
                r = mid - 1
        return l

// Accepted solution for LeetCode #3281: Maximize Score of Numbers in Ranges
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3281: Maximize Score of Numbers in Ranges
// class Solution {
//     private int[] start;
//     private int d;
// 
//     public int maxPossibleScore(int[] start, int d) {
//         Arrays.sort(start);
//         this.start = start;
//         this.d = d;
//         int n = start.length;
//         int l = 0, r = start[n - 1] + d - start[0];
//         while (l < r) {
//             int mid = (l + r + 1) >>> 1;
//             if (check(mid)) {
//                 l = mid;
//             } else {
//                 r = mid - 1;
//             }
//         }
//         return l;
//     }
// 
//     private boolean check(int mi) {
//         long last = Long.MIN_VALUE;
//         for (int st : start) {
//             if (last + mi > st + d) {
//                 return false;
//             }
//             last = Math.max(st, last + mi);
//         }
//         return true;
//     }
// }

// Accepted solution for LeetCode #3281: Maximize Score of Numbers in Ranges
function maxPossibleScore(start: number[], d: number): number {
    start.sort((a, b) => a - b);
    let [l, r] = [0, start.at(-1)! + d - start[0]];
    const check = (mi: number): boolean => {
        let last = -Infinity;
        for (const st of start) {
            if (last + mi > st + d) {
                return false;
            }
            last = Math.max(st, last + mi);
        }
        return true;
    };
    while (l < r) {
        const mid = l + ((r - l + 1) >> 1);
        if (check(mid)) {
            l = mid;
        } else {
            r = mid - 1;
        }
    }
    return l;
}

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.