Maximum Subsequence Score

Maximum Subsequence Score - Problem

Maximum Subsequence Score is a fascinating optimization problem that combines elements of greedy algorithms and heap management. You're given two arrays nums1 and nums2 of equal length n, and you need to select exactly k indices to maximize a special score.

Your score is calculated as: (sum of selected elements from nums1) × (minimum of selected elements from nums2).

For example, if you select indices [0, 2, 3], your score would be:
(nums1[0] + nums1[2] + nums1[3]) × min(nums2[0], nums2[2], nums2[3])

The challenge is finding which k indices give you the maximum possible score. This problem appears frequently in technical interviews at top tech companies and tests your ability to think strategically about optimization under constraints.

Input & Output

example_1.py — Basic Example

$ Input: nums1 = [1,3,3,2], nums2 = [2,1,3,4], k = 3

› Output: 12

💡 Note: The optimal subsequence is indices [0,2,3] giving us (1+3+2) × min(2,3,4) = 6 × 2 = 12.

example_2.py — Larger Values

$ Input: nums1 = [4,2,3,1,1], nums2 = [7,5,10,9,6], k = 1

› Output: 30

💡 Note: With k=1, we choose the single index that maximizes nums1[i] × nums2[i]. Index 2 gives us 3 × 10 = 30.

example_3.py — All Elements

$ Input: nums1 = [2,1,14,12], nums2 = [11,7,13,6], k = 4

› Output: 174

💡 Note: We must select all indices: (2+1+14+12) × min(11,7,13,6) = 29 × 6 = 174.

Constraints

n == nums1.length == nums2.length
1 ≤ n ≤ 10⁵
1 ≤ k ≤ n
0 ≤ nums1[i], nums2[i] ≤ 10⁵

Visualization

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Asked in

G Google 45 a Amazon 38 f Meta 32 ⊞ Microsoft 28 Apple 22

The optimal solution uses a greedy approach with heap. Key insight: sort pairs by nums2 in descending order, then use a min-heap to maintain the k largest nums1 values. This ensures we consider each possible minimum while maximizing the sum efficiently in O(n log n) time.

Common Approaches

Approach	Time	Space	Notes
✓ Greedy with Heap (Optimal)	O(n log n + n log k)	O(n + k)	Sort by nums2 values and use max-heap to greedily select best nums1 values
Brute Force (Try All Combinations)	O(C(n,k) × k)	O(k)	Generate all possible combinations of k indices and calculate score for each

Greedy with Heap (Optimal) — Algorithm Steps

Create pairs of (nums1[i], nums2[i]) and sort by nums2 in descending order
Use a min-heap to track the k largest nums1 values seen so far
For each pair, add nums1 value to heap and maintain heap size ≤ k
When heap has k elements, calculate score using current nums2 as minimum
Track the maximum score across all valid configurations

Visualization

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Step-by-Step Walkthrough

Sort by nums2

Sort pairs in descending order of nums2 values

Process with Heap

For each element, maintain k largest nums1 values in min-heap

Calculate Scores

When heap has k elements, calculate score using current nums2 as minimum

Code -

solution.c — C

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

typedef struct {
    int num1, num2;
} Pair;

typedef struct {
    int* data;
    int size;
    int capacity;
} MinHeap;

MinHeap* createHeap(int capacity) {
    MinHeap* heap = malloc(sizeof(MinHeap));
    heap->data = malloc(capacity * sizeof(int));
    heap->size = 0;
    heap->capacity = capacity;
    return heap;
}

void heapPush(MinHeap* heap, int val) {
    heap->data[heap->size] = val;
    int i = heap->size++;
    
    while (i > 0 && heap->data[i] < heap->data[(i-1)/2]) {
        int temp = heap->data[i];
        heap->data[i] = heap->data[(i-1)/2];
        heap->data[(i-1)/2] = temp;
        i = (i-1)/2;
    }
}

int heapPop(MinHeap* heap) {
    int result = heap->data[0];
    heap->data[0] = heap->data[--heap->size];
    
    int i = 0;
    while (2*i + 1 < heap->size) {
        int minChild = 2*i + 1;
        if (2*i + 2 < heap->size && heap->data[2*i + 2] < heap->data[minChild]) {
            minChild = 2*i + 2;
        }
        
        if (heap->data[i] <= heap->data[minChild]) break;
        
        int temp = heap->data[i];
        heap->data[i] = heap->data[minChild];
        heap->data[minChild] = temp;
        i = minChild;
    }
    
    return result;
}

int comparePairs(const void* a, const void* b) {
    return ((Pair*)b)->num2 - ((Pair*)a)->num2;
}

long long maxScore(int* nums1, int nums1Size, int* nums2, int nums2Size, int k) {
    Pair* pairs = malloc(nums1Size * sizeof(Pair));
    
    for (int i = 0; i < nums1Size; i++) {
        pairs[i].num1 = nums1[i];
        pairs[i].num2 = nums2[i];
    }
    
    qsort(pairs, nums1Size, sizeof(Pair), comparePairs);
    
    long long maxScore = 0;
    long long nums1Sum = 0;
    MinHeap* minHeap = createHeap(k + 1);
    
    for (int i = 0; i < nums1Size; i++) {
        nums1Sum += pairs[i].num1;
        heapPush(minHeap, pairs[i].num1);
        
        if (minHeap->size > k) {
            nums1Sum -= heapPop(minHeap);
        }
        
        if (minHeap->size == k) {
            long long score = nums1Sum * pairs[i].num2;
            if (score > maxScore) {
                maxScore = score;
            }
        }
    }
    
    free(pairs);
    free(minHeap->data);
    free(minHeap);
    return maxScore;
}

Time & Space Complexity

Time Complexity

⏱️

O(n log n + n log k)

O(n log n) for sorting, O(n log k) for heap operations

⚡ Linearithmic

Space Complexity

O(n + k)

O(n) for storing pairs, O(k) for heap

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Greedy with Heap (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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