LeetCode #1589 — MEDIUM

Maximum Sum Obtained of Any Permutation

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

The Problem

Problem Statement

We have an array of integers, nums, and an array of requests where requests[i] = [start_i, end_i]. The i^th request asks for the sum of nums[start_i] + nums[start_i + 1] + ... + nums[end_i - 1] + nums[end_i]. Both start_i and end_i are 0-indexed.

Return the maximum total sum of all requests among all permutations of nums.

Since the answer may be too large, return it modulo 10⁹ + 7.

Example 1:

Input: nums = [1,2,3,4,5], requests = [[1,3],[0,1]]
Output: 19
Explanation: One permutation of nums is [2,1,3,4,5] with the following result: 
requests[0] -> nums[1] + nums[2] + nums[3] = 1 + 3 + 4 = 8
requests[1] -> nums[0] + nums[1] = 2 + 1 = 3
Total sum: 8 + 3 = 11.
A permutation with a higher total sum is [3,5,4,2,1] with the following result:
requests[0] -> nums[1] + nums[2] + nums[3] = 5 + 4 + 2 = 11
requests[1] -> nums[0] + nums[1] = 3 + 5  = 8
Total sum: 11 + 8 = 19, which is the best that you can do.

Example 2:

Input: nums = [1,2,3,4,5,6], requests = [[0,1]]
Output: 11
Explanation: A permutation with the max total sum is [6,5,4,3,2,1] with request sums [11].

Example 3:

Input: nums = [1,2,3,4,5,10], requests = [[0,2],[1,3],[1,1]]
Output: 47
Explanation: A permutation with the max total sum is [4,10,5,3,2,1] with request sums [19,18,10].

Constraints:

n == nums.length
1 <= n <= 10⁵
0 <= nums[i] <= 10⁵
1 <= requests.length <= 10⁵
requests[i].length == 2
0 <= start_i <= end_i < n

Step 01

Brute Force Baseline

Problem summary: We have an array of integers, nums, and an array of requests where requests[i] = [starti, endi]. The ith request asks for the sum of nums[starti] + nums[starti + 1] + ... + nums[endi - 1] + nums[endi]. Both starti and endi are 0-indexed. Return the maximum total sum of all requests among all permutations of nums. Since the answer may be too large, return it modulo 109 + 7.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

[1,2,3,4,5]
[[1,3],[0,1]]

Example 2

[1,2,3,4,5,6]
[[0,1]]

Example 3

[1,2,3,4,5,10]
[[0,2],[1,3],[1,1]]

Step 02

Core Insight

What unlocks the optimal approach

Indexes with higher frequencies should be bound with larger values

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 #1589: Maximum Sum Obtained of Any Permutation
class Solution {
    public int maxSumRangeQuery(int[] nums, int[][] requests) {
        int n = nums.length;
        int[] d = new int[n];
        for (var req : requests) {
            int l = req[0], r = req[1];
            d[l]++;
            if (r + 1 < n) {
                d[r + 1]--;
            }
        }
        for (int i = 1; i < n; ++i) {
            d[i] += d[i - 1];
        }
        Arrays.sort(nums);
        Arrays.sort(d);
        final int mod = (int) 1e9 + 7;
        long ans = 0;
        for (int i = 0; i < n; ++i) {
            ans = (ans + 1L * nums[i] * d[i]) % mod;
        }
        return (int) ans;
    }
}

// Accepted solution for LeetCode #1589: Maximum Sum Obtained of Any Permutation
func maxSumRangeQuery(nums []int, requests [][]int) (ans int) {
	n := len(nums)
	d := make([]int, n)
	for _, req := range requests {
		l, r := req[0], req[1]
		d[l]++
		if r+1 < n {
			d[r+1]--
		}
	}
	for i := 1; i < n; i++ {
		d[i] += d[i-1]
	}
	sort.Ints(nums)
	sort.Ints(d)
	const mod = 1e9 + 7
	for i, a := range nums {
		b := d[i]
		ans = (ans + a*b) % mod
	}
	return

# Accepted solution for LeetCode #1589: Maximum Sum Obtained of Any Permutation
class Solution:
    def maxSumRangeQuery(self, nums: List[int], requests: List[List[int]]) -> int:
        n = len(nums)
        d = [0] * n
        for l, r in requests:
            d[l] += 1
            if r + 1 < n:
                d[r + 1] -= 1
        for i in range(1, n):
            d[i] += d[i - 1]
        nums.sort()
        d.sort()
        mod = 10**9 + 7
        return sum(a * b for a, b in zip(nums, d)) % mod

// Accepted solution for LeetCode #1589: Maximum Sum Obtained of Any Permutation
struct Solution;

const MOD: i64 = 1_000_000_007;

impl Solution {
    fn max_sum_range_query(mut nums: Vec<i32>, requests: Vec<Vec<i32>>) -> i32 {
        let n = nums.len();
        let mut count = vec![0; n + 1];
        for r in requests {
            count[r[0] as usize] += 1;
            count[r[1] as usize + 1] -= 1;
        }
        count.pop();
        let mut prev = 0;
        for i in 0..n {
            prev += count[i];
            count[i] = prev;
        }
        count.sort_unstable();
        nums.sort_unstable();
        let mut res = 0;
        for i in 0..n {
            res += nums[i] as i64 * count[i] as i64;
            res %= MOD;
        }
        res as i32
    }
}

#[test]
fn test() {
    let nums = vec![1, 2, 3, 4, 5];
    let requests = vec_vec_i32![[1, 3], [0, 1]];
    let res = 19;
    assert_eq!(Solution::max_sum_range_query(nums, requests), res);
    let nums = vec![1, 2, 3, 4, 5, 6];
    let requests = vec_vec_i32![[0, 1]];
    let res = 11;
    assert_eq!(Solution::max_sum_range_query(nums, requests), res);
    let nums = vec![1, 2, 3, 4, 5, 10];
    let requests = vec_vec_i32![[0, 2], [1, 3], [1, 1]];
    let res = 47;
    assert_eq!(Solution::max_sum_range_query(nums, requests), res);
}

// Accepted solution for LeetCode #1589: Maximum Sum Obtained of Any Permutation
function maxSumRangeQuery(nums: number[], requests: number[][]): number {
    const n = nums.length;
    const d = new Array(n).fill(0);
    for (const [l, r] of requests) {
        d[l]++;
        if (r + 1 < n) {
            d[r + 1]--;
        }
    }
    for (let i = 1; i < n; ++i) {
        d[i] += d[i - 1];
    }
    nums.sort((a, b) => a - b);
    d.sort((a, b) => a - b);
    let ans = 0;
    const mod = 10 ** 9 + 7;
    for (let i = 0; i < n; ++i) {
        ans = (ans + nums[i] * d[i]) % mod;
    }
    return ans;
}

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(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

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.