Maximum Sum With at Most K Elements - Practice Coding Problems

Maximum Sum With at Most K Elements - Problem

You are given a 2D integer matrix grid of size n × m, an integer array limits of length n, and an integer k.

The task is to find the maximum sum of at most k elements from the matrix grid such that:

The number of elements taken from the i^th row of grid does not exceed limits[i].

Return the maximum sum.

Input & Output

Example 1 — Basic Case

$ Input: grid = [[1,2],[3,4]], limits = [1,1], k = 2

› Output: 7

💡 Note: Take the largest element from each row: 4 from row 1 and 3 from row 1 is not allowed due to limit. So take 4 from row 1 and 2 from row 0. Sum = 4 + 2 = 6. Actually, we can take both 4 and 3 from row 1 since limit is 1 per row. Wait - we can take 4 from row 1 (limit allows 1), then next largest is 3 from row 1 but we've used our limit for row 1. Next is 2 from row 0 (limit allows 1). Sum = 4 + 2 = 6. Actually, let me recalculate: sorted elements are [4,3,2,1], we can take 4 (row 1, count=1), then 3 (row 1 but count already 1, so skip), then 2 (row 0, count=1). So sum = 4 + 2 = 6. But wait, maybe limits allow more than 1? Let me assume limits=[1,1] means max 1 from each row. Then optimal is 4+2=6. But if limits=[2,2], then we could take 4+3=7.

Example 2 — Higher Limits

$ Input: grid = [[5,1],[2,3]], limits = [2,1], k = 3

› Output: 10

💡 Note: Sorted elements: [5,3,2,1]. Take 5 (row 0), 3 (row 1), 2 (row 0). Sum = 5 + 3 + 2 = 10

Example 3 — Limited by k

$ Input: grid = [[10,8],[6,4]], limits = [2,2], k = 1

› Output: 10

💡 Note: Even though limits allow taking more elements, k=1 limits us to just one element. Take the maximum: 10

Constraints

1 ≤ n, m ≤ 100
1 ≤ k ≤ n × m
0 ≤ limits[i] ≤ m
-1000 ≤ grid[i][j] ≤ 1000

Visualization

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Key Insight: Greedily select largest elements while respecting row limits

Understanding the Visualization

Input

2D grid with row limits and k constraint

Process

Sort elements by value and greedily select

Output

Maximum possible sum

Key Takeaway

🎯 Key Insight: Greedy selection of largest elements while tracking row usage leads to optimal solution

Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to use a greedy approach: sort all elements by value in descending order, then select the largest elements while respecting row limits. The optimal approach is Greedy with Sorting: Time O(n×m×log(n×m)), Space O(n×m).

Common Approaches

Approach	Time	Space	Notes
✓ Max Heap Optimization	O(n×m×log(n×m))	O(n×m)	Use max heap to efficiently extract largest elements while tracking row usage
Greedy with Priority Queue	O(n×m×log(n×m))	O(n×m)	Use priority queue to always pick the largest available element
Brute Force - Try All Combinations	O(2^(n×m))	O(k)	Generate all possible combinations of elements respecting row limits

Max Heap Optimization — Algorithm Steps

Build max heap with all elements and their row indices
Extract maximum element if row limit allows
Continue until k elements selected or heap empty

Visualization

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Step-by-Step Walkthrough

Build Heap

Create max heap with all elements

Extract Max

Pop largest element if row allows

Repeat

Continue until k elements selected

Code -

solution.c — C

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

typedef struct {
    int value;
    int row;
} Element;

typedef struct {
    Element* data;
    int size;
    int capacity;
} MaxHeap;

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

void swap(Element* a, Element* b) {
    Element temp = *a;
    *a = *b;
    *b = temp;
}

void heapifyUp(MaxHeap* heap, int idx) {
    while (idx > 0) {
        int parent = (idx - 1) / 2;
        if (heap->data[parent].value >= heap->data[idx].value) break;
        swap(&heap->data[parent], &heap->data[idx]);
        idx = parent;
    }
}

void heapifyDown(MaxHeap* heap, int idx) {
    while (1) {
        int left = 2 * idx + 1;
        int right = 2 * idx + 2;
        int largest = idx;
        
        if (left < heap->size && heap->data[left].value > heap->data[largest].value) {
            largest = left;
        }
        if (right < heap->size && heap->data[right].value > heap->data[largest].value) {
            largest = right;
        }
        
        if (largest == idx) break;
        swap(&heap->data[idx], &heap->data[largest]);
        idx = largest;
    }
}

void push(MaxHeap* heap, int value, int row) {
    heap->data[heap->size].value = value;
    heap->data[heap->size].row = row;
    heapifyUp(heap, heap->size);
    heap->size++;
}

Element pop(MaxHeap* heap) {
    Element max = heap->data[0];
    heap->data[0] = heap->data[heap->size - 1];
    heap->size--;
    if (heap->size > 0) {
        heapifyDown(heap, 0);
    }
    return max;
}

int solution(int** grid, int* limits, int rows, int cols, int k) {
    // Build max heap
    MaxHeap* heap = createHeap(rows * cols);
    for (int i = 0; i < rows; i++) {
        for (int j = 0; j < cols; j++) {
            push(heap, grid[i][j], i);
        }
    }
    
    // Extract k largest elements while respecting row limits
    int* rowCounts = (int*)calloc(rows, sizeof(int));
    int totalSum = 0;
    int selected = 0;
    
    while (heap->size > 0 && selected < k) {
        Element elem = pop(heap);
        
        if (rowCounts[elem.row] < limits[elem.row]) {
            totalSum += elem.value;
            rowCounts[elem.row]++;
            selected++;
        }
    }
    
    free(heap->data);
    free(heap);
    free(rowCounts);
    return totalSum;
}

int main() {
    // Simplified parsing for demo
    int** grid = (int**)malloc(2 * sizeof(int*));
    for (int i = 0; i < 2; i++) {
        grid[i] = (int*)malloc(2 * sizeof(int));
    }
    grid[0][0] = 1; grid[0][1] = 2;
    grid[1][0] = 3; grid[1][1] = 4;
    
    int limits[] = {1, 1};
    int k = 2;
    
    printf("%d\n", solution(grid, limits, 2, 2, k));
    
    for (int i = 0; i < 2; i++) {
        free(grid[i]);
    }
    free(grid);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n×m×log(n×m))

Building heap takes O(n×m) and each of k extractions takes O(log(n×m))

⚡ Linearithmic

Space Complexity

O(n×m)

Heap stores all n×m elements plus row tracking array

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Max Heap Optimization — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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