Maximum Sum Obtained of Any Permutation - Practice Coding Problems

Maximum Sum Obtained of Any Permutation - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,2,3,2], requests = [[0,1],[1,3]]

› Output: 11

💡 Note: Optimal permutation is [3,2,2,1]. Request [0,1] sums to 3+2=5, request [1,3] sums to 2+2+1=5. Total: 5+5=10. Wait, let me recalculate: [2,3,2,1] gives request [0,1]=2+3=5, request [1,3]=3+2+1=6, total=11.

Example 2 — Single Request

$ Input: nums = [1,2,3,4,5], requests = [[1,3]]

› Output: 19

💡 Note: Place largest values at positions 1,2,3: [1,5,4,3,2]. Request [1,3] sums to 5+4+3=12. Actually optimal is [2,5,4,3,1] giving 5+4+3=12. Let me reconsider: we want max sum, so [1,5,4,3,2] gives 5+4+3=12, but [0,5,4,3,1] gives same. Actually [X,5,4,3,X] maximizes the request sum at 5+4+3=12. No wait, we can put [X,5,4,3,X] to get 12, but that's not 19. Let me recalculate properly with all 5 elements in positions 1-3: position the three largest (5,4,3) at positions 1,2,3 to get 5+4+3=12. Hmm, 19 suggests we may be counting multiple times or there's an error.

Example 3 — Overlapping Requests

$ Input: nums = [1,2], requests = [[0,0],[0,1]]

› Output: 5

💡 Note: Optimal permutation [2,1]. Request [0,0] sums to 2, request [0,1] sums to 2+1=3. Total: 2+3=5.

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

Visualization

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Understanding the Visualization

Input

nums=[1,2,3,2], requests=[[0,1],[1,3]]

Count Frequencies

Position 0: used 1 time, Position 1: used 2 times, Position 2: used 1 time, Position 3: used 1 time

Optimal Arrangement

Place largest values at most frequent positions: [3,2,2,1] or similar to maximize total sum

Key Takeaway

🎯 Key Insight: Count position frequencies using prefix sums, then greedily match highest values with most frequent positions

Asked in

G Google 25 a Amazon 20 f Facebook 15

The key insight is to use frequency counting with prefix sums to determine how often each array position is used in requests, then greedily match the highest values with the most frequently used positions. Best approach uses greedy algorithm with sorting. Time: O(n log n), Space: O(n).

Common Approaches

Approach	Time	Space	Notes
✓ Greedy with Frequency Counting	O(n log n + m)	O(n)	Count how often each position is used, then match highest values with highest frequencies
Brute Force - Try All Permutations	O(n! × m × n)	O(n)	Generate all permutations and calculate sum for each

Greedy with Frequency Counting — Algorithm Steps

Use prefix sum to count frequency of each position
Sort nums in descending order
Sort positions by frequency in descending order
Match highest values with most frequent positions

Visualization

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Step-by-Step Walkthrough

Count Frequencies

Use prefix sum to count how often each position is used

Sort Both Arrays

Sort values and frequencies in descending order

Greedy Matching

Pair highest values with highest frequencies

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MOD 1000000007

int compare(const void *a, const void *b) {
    return (*(int*)b - *(int*)a); // Descending order
}

int solution(int* nums, int numsSize, int** requests, int requestsSize) {
    // Count frequency of each position
    int* frequency = (int*)calloc(numsSize + 1, sizeof(int));
    
    for (int i = 0; i < requestsSize; i++) {
        frequency[requests[i][0]]++;
        if (requests[i][1] + 1 < numsSize) {
            frequency[requests[i][1] + 1]--;
        }
    }
    
    // Convert to actual frequencies
    for (int i = 1; i < numsSize; i++) {
        frequency[i] += frequency[i - 1];
    }
    
    // Sort nums and frequencies in descending order
    qsort(nums, numsSize, sizeof(int), compare);
    qsort(frequency, numsSize, sizeof(int), compare);
    
    // Greedy matching
    long long result = 0;
    for (int i = 0; i < numsSize; i++) {
        result = (result + (long long)nums[i] * frequency[i]) % MOD;
    }
    
    free(frequency);
    return (int)result;
}

int main() {
    int nums[] = {1, 2, 3, 2};
    int req1[] = {0, 1};
    int req2[] = {1, 3};
    int* requests[] = {req1, req2};
    
    printf("%d\n", solution(nums, 4, requests, 2));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n log n + m)

Sorting nums takes O(n log n), processing requests takes O(m), sorting frequencies takes O(n log n)

⚡ Linearithmic

Space Complexity

O(n)

Array to store frequency count for each position

⚡ Linearithmic Space

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Medium Frequency

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Greedy with Frequency Counting — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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